Class 10 Physics - Electricity

The electronic current flows opposite to the conventional current. True/False?

The flow of electrons is termed electron current. Electrons flow from the negative terminal to the positive. Conventional current or simply current, behaves as if positive charge carriers cause current flow. Conventional current flows from the positive terminal to the negative. Perhaps the clearest way to think about this is to pretend as if movement of positive charge carriers constituted current flow.
It is important to realize that the difference between conventional current flow and electron flow in no way effects any real-world behavior or computational results. In general, analyzing an electrical circuit yields results that are independent of the assumed direction of current flow. Conventional current flow is the standard that most all of the world follows.
Hence, the statement is true.

What are insulators? Define on the basis of atomic structure. Give four examples of insulators.

The substances or materials which have very few free electrons, such that they do not easily drift from one end to the other end, when potential difference is applied are called insulators.
The examples of insulators are Mica, glass wool, wood, shellac, sealing wax are all examples of insulators.

State three characteristics of a series circuit?

Characteristics of a Series circuit are:

1) Same current passes through both the resistors.

2) Voltage is divided in the ratio of resistances i.e, $\dfrac{V_1}{V_2}=\dfrac{R_1}{R_2}$

3) Voltage of circuit is equal to sum of all the voltage drops. $V=V_{R_1}+V_{R_2}+......$

A heating element is marked $210 V$ , $630 W$ . Find the resistance of the element when connected to a $210 V$ dc source.

Given: A heating element is marked

$210V, 630W$

To find the resistance of the element when connected to a

$210V$ dc source.

Solution:

The power of the heating element is

$630$ Watt and its voltage is

$210 V$ . Thus, the current drawn can be calculated with the formula

$P = VI$ ,

where

$P$ denotes the power,

$V$ stands for the voltage and

$I$ is the symbol of the current.

So the value of the current drawn

$= \dfrac P V =\dfrac {630}{ 210} = 3A$

And the resistance of the element when connected to a

$210V$ dc source =

$\dfrac VI=\dfrac {210}3=70\Omega$

The wattage of an electric heater which draws 5 A current when connected to a 220 V supply is _____________.

Given:

Current drawn by the heater

$I= 5 \ A$

Voltage supplied to the heater

$V= 220 \ V$

Wattage of the heater, i.e. its power

$P= VI$

$=220 \times 5$

$= 1100 \ W$

...................... does not conduct electricity.

An insulator does not conduct electricity because it has no free electrons or charged carriers.

A 1000 W heater has more resistance than a 100 W bulb.

1000 W has more resistance than 100 W

This statement is false because

$P= \dfrac { V^2 }{ R }$

$P\propto \dfrac { 1 }{ R }$

more resistance has less power. Hence heater has less resistance than bulb,

The current drawn by a bulb rated 40 W, 220 V when it is connected to a 220 V supply is ___________.

Current =?

Power = 40 W

voltage = 220V

Power = VI

40 = 220 I

I =

$\dfrac{4}{22}$

I =

$\dfrac{2}{11}$ A

State the characteristics required in a material to be used as an effective fuse wire.

The material to be used as a fuse wire should have these characteristics:

high resistivity
low melting point
least deterioration due to oxidation during normal use.

Two copper wires are of the same length, but one is thicker than the other.
Which wire will have more resistance?

Thin wire will have more resistance. Since

$R\propto \dfrac {l}{A}$ ; where

$l=$ length of the wire and

$A=$ cross-sectional area of the wire.

As the thickness increases the resistance of the wire decreases, and as the thickness decreases, the resistance of the wire increases.

Match the following:

The SI unit of electric charge is

$Coulomb$ .

The SI unit of electric potential difference is

$Volt$ .

The SI unit of the electric current is

$Ampere$ .

The SI unit of electrical resistance is

$Ohm$ .

Write the differences between static electricity and electric current.

Basis for comparison	Static Electricity	Electric current
Definition	The electricity which is build up on the surface of the substance is known as the static electricity.	The current electricity is because of the flow of electrons.
Causes	It induces because of the movement of the negative charges from one object to another	The current electricity is because of the movement of the electrons.
Material	The static electricity develops both in the conductor and insulator.	The current electricity develops only in conductor
Magnetic Field	Not induces the magnetic field.	It induces magnetic field.
Time period	Exist for short time.	Exist for long time.
Measuring Device	Gold leaf electroscope	Analog and digital meter.
Examples	Lightning strokes, it develop by rubbing the balloons on hair, etc.	The current electricity is used for running the fan, light, T.V etc.

$(i)$ What happens to the resistivity of semi-conductors with the increase of temperature?
$(ii)$ For a fuse, higher the current rating ________ is the fuse wire. (Thicker / Thinner)

$(i)$ The resistivity of semi-conductor decreases with the increase in temperature. It is because the charge carriers in the semiconductors increase flowing with more ease in semiconductors due to rise in temperature.

$(ii)$ For a fuse, higher the current rating thicker is the fuse wire. High current rating means the fuse has a lower resistance. The cross-sectional area (i.e. thickness) of the fuse wire must be higher.

Two resistors of $4\Omega$ and $6\Omega$ are connected in parallel to a cell to draw $0.5\ A$ current from the cell.
Draw a labelled circuit diagram showing the above arrangement

The circuit diagram is shown for two resistors (

$4\Omega$ and

$6\Omega$ ) connected in parallel. The electric current supplied by the cell is

$0.5A$ . The current flows from the positive terminal (longer line in the symbol of the cell) of the cell to the negative terminal (shorter line in its symbol) of the cell.

The SI units of current is .........

The SI unit of current is

$Ampere$ . One ampere is constituted by the flow of one coulomb of charge per second, that is,

$1 \ A = \dfrac{1 \ C}{1 \ s}$

(i) Which particles are responsible for current in conductors?
(ii) To which wire of a cable in a power circuit should the metal case of a geyser be connected?
(iii) To which wire should the fuse be connected ?

(i) Free electrons are responsible for current in conductors.
(ii) In a power circuit, the metal case of a geyser should be connected to Earth wire (Green).
(iii) The fuse should be connected to Live wire (Red).

Write three safety measures in using electricity.

1) Never touch electrical equipment with wet hands.When working with electricity do not stand on metal, wet concrete or wet ground.

2)Always wear rubber glows and slippers while working with electricity board.

3)Disconnect electrical gadgets when not in use and at the time of lightning and thundering.

4) Repair and maintenance of electrical fuses and gadgets should be done time to time.

5) Faulty electrical wires and cords should never be used.

An electric bulb is rated $200V - 100W$ . Find the resistance of the bulb.

Characteristics of the bulb is given as

$V = 200 \ V$ and

$P = 100 \ W$

Resistance of the bulb

$R = \dfrac{V^2}{P}$

$\therefore$

$R = \dfrac{(200)^2}{100} = 400 \ \Omega$

Murugan measured the electric current. What unit should he use?

The

$SI$ unit for measuring an electric current is the ampere (A), which is the flow of electric charge across a surface at the rate of one coulomb per second. Electric current is measured using a device called an ammeter.

Write the S.I. unit of electric power.

Electric power is defined as the energy consumed by the electrical appliance per unit time.

Power

$P = \dfrac{E}{t}=$

Unit

$\dfrac{Joule}{sec}=watt$

Hence, S.I unit of power is watt.

Equivalent resistance when $2$ and $3$ ohm resistances are connected in series is _______.

Given resistance are

$R_1=2\Omega\\R_2=3\Omega$

The equivalent resistance of two resistance connected in series can be expressed as

$R=R_1+R_2$ . Now from the given information

$R=2+3= 5 \Omega$

Give two examples each of conductors and insulators of heat.

Conductors-aluminium, iron Insulators-plastic, wood.

Explore and answer :
Does pure water conduct electricity? If not, what can we do to make it conduct?

Since pure water does not have positive and negative ions, so it does not conduct electricity. When a salt is added to it, salt gets ionised into positive and negative ions and so salt solution conducts electricity as shown in figure.

1136734_617037_ans_71da4f01c7b44c91bcb871be13a6b54e.png

1136734_617037_ans_71da4f01c7b44c91bcb871be13a6b54e.png

Write equivalent resistance of given circuit.

Clearly they are in series, therefore the resistances add up directly to get the equivalent resistance

$\Longrightarrow 1+2+3 =6\Omega$

Though same current flows the wire and the filament of a bulb, yet only the filament glows.Why?

The resistance of electric line wires is very, very less than that of the filament of the bulb and as the wires are the good conductors of electricity which cannot stop electricity so they cannot heat up if the circuit is proper. Therefore, the current through high resistance filament produces more heat which makes it glow.

Calculate the resistance across points AD.

$\underline{\text{Calculating resistance across AD}}:$

To calculate the resistance across $AD$ , we will connect a voltage source across $AD$ . Hence, the equivalent circuit will look like the given figure.

Two $16\ \Omega$ resistances are connected in parallel.

$\therefore$ the equivalent resistance is $\dfrac{1}{R_{eq}}=\dfrac{1}{16}+\dfrac{1}{16}=\dfrac{2}{16}$

$\therefore R_{eq}=\dfrac{16}{2}=8\ \Omega$

1081276_1126214_ans_f7436077d66145e090cd0c56f8cbc03d.png

Define electric current. What is its $SI$ unit?

Electric current is basically the flow or charge of the electric charge in motion in a conductor. It is said to exist when there is a net flow of charge through the region.

the current is given as the flow of charge per second:

$I=\dfrac{charge}{time}$

The SI unit of current is Ampere.

Power of a lamp is $60\ W$ . Find the energy in joules consumed by it in $1\ s$ .?

We know that

$E=Pt$

$E=Pt = 60\times1 = 60 Joules$

Define $1$ ampere.

1 ampere is defined as 1 coloumb of charge per second.

An ampere is a unit of measure of the rate of electron flow or current in an electrical conductor.

Name and define the SI unit of current

Name of the SI unit of current is Ampere.

Ampere can be defined as the one unit of charge flown through a given circuit in one second.

$1 \ Ampere (A) = \frac{1 \ Coulomb (C)}{1 \ Second (s)}$

Name and define unit for electric current.

The SI unit for measuring an electric current is the ampere, which is the flow of electric charge across a surface at the rate of one coulomb per second.

$6\Omega,8\Omega$ and $10\Omega$ resistance are connected in series. Find total resistance of the circuit.

The same current flows through each part of a series circuit. The total resistance of a series circuit is equal to the sum of individual resistances.

Hence,

$R_{eq} = R_1+R_2+R_35$

$\Rightarrow R_{eq} = 6+8+10$

$\Rightarrow R_{eq} = 24\Omega$

Two resistance 3 ohm and 2 ohm are in parallel connection and a potential difference of 12 V is applied across them.Find
a) The equivalent resistance of the parallel combinations.
b) The circuit current and
c) The branch currents.

(a) Req

$= \dfrac{3 \times 2}{3 + 2} = \dfrac{6}{5} = 1.2 \Omega$

(b)

$V = IR \Rightarrow I = \dfrac{V}{R}$

$\Rightarrow I = \dfrac{12}{1.2} = 10 A$

(c)

$3 (10 - i) = 2i$ (

$\because$ Potential drop in parallel combination are same)

$\Rightarrow 30 - 3i = 2i \Rightarrow 30 = 5 i \Rightarrow i = 6A$

Current in

$2 \Omega$ resistor is

$= i = 6A$

Current in

$3 \Omega$ resistor is

$= 10 - i = 4A$

1317933_1141333_ans_d428a16d256d48df86235026ffac8629.png

What are conductors of electricity ?

The materials or substances that allow electricity to flow through them easily are called conductors of electricity.

Some examples of conductors of electricity are:

Copper.

Aluminium.

Silver.

How much current flows through a $10\ \Omega$ resistor when it is connected to a battery of $3\ v$ ?

Formula,

$I=\dfrac{V}{R}$

$=\dfrac{3V}{10\Omega }$

$\therefore I=0.3A$

Define electric current.

Rate Flow of electric charge is called electric current.

It’s represent as I and it’s unit is Ampere.

Name the quantity whose unit is (i) kilowatt , and (ii) kilowatt - hour.

(i) Electric power has the unit of kilowatt.

(ii) Electric energy has the unit of Kilowatt-hour

What is the unit of electric charge ?

The unit of the electric charge is Coulomb

Find the total resistance in the given circuit.

Two resistances are connected in series, the effective resistance is:

$R_{eq}=R_1+R_2$

$R_{eq}=2 + 2 = 4\Omega$

By what name is the physical quantity coulomb / second called ?

$coulomb/sec$ is defined as the rate of flow of charge. It is the current. So, it can also be represented by

$Ampere$ .

What is the unit of electric current ?

The unit of electric current is

$Ampere$ .

What is meant by electrical resistivity of a material?

The resistivity of the material is defined as the resistance offered by a metallic wire having a unit length and a unit area of cross-section. Since the unit length and unit area of cross-section forms a cube, the specific resistance or resistivity can also be defined as the resistance offered by a cube of a material of side

$1\ m$ when current flows perpendicularly through the opposite faces

A wire of length L and resistance R is stretched so that the length is doubled and area of cross section halved. How will (i) resistance change and (ii) resistivity change.

$\Rightarrow$ Length

$(l)$ is doubled & Area of cross section (A) is halved.

$l$ becomes

$2l$ & A becomes

$\frac{A}{2}$

$R = P\frac{L}{A}$

now take

$l = 2l,A = \frac{A}{2}$

$R = \frac{{P2l}}{{A/2}} = 4P\frac{L}{A} \Rightarrow 4R$

Resistance becomes 4 times

1) Resistivity does not depend on its length 02 Area.

Hence, it does not change.

Why does the cord of an electric heater not glow while the heating element does ?

What is the total resistance of 5 resistors each of resistance 5 $\Omega$ connected in parallel?

The voltage - current $(V-I)$ graph of a metallic conductor at two different temperature $T_{1}$ and $T_{2}$ is shown below. At which temperature is the resistance higher?

The mathematical expression for Ohm's law is:

$V=IR$

Thus, graph between V versus I gives a straight line with slope R.

Since, slope at

${T_{2}}$ is more than slope at

${T_{1}}$ .

For metallic conductors, resistance is directly proportional to temperature. Thus, resistance is higher at

${T_{2}}$

Match the entries in column A with appropriate ones from column B.

The electrical potential or voltage (V) between two points is defined to be the work (W) done by the electric force in transporting a unit of positive charge (Q) from one point to another.

The SI unit joule is too small to express very large quantities of energy. Hence we use a bigger unit called kilowatt-hour (kWh) to express energy. 1 kWh is the amount of energy consumed by an electrical gadget in one hour at the rate of 1000 J/s or 1kW.
Resistance of a wire is directly proportional to the length of the wire and inversely proportional to the cross-section of the wire. The constant of proportion is called resistivity.
Superconductor has zero resistance.
The resistance R in ohms is equal to the voltage V in volts (V) divided by the current I in amps (A). 1-ohm resistance means, transmitting a current of one ampere when subjected to a potential difference of one volt.
Transistor is an example of non-ohmic conductor. A transistor, made up of semiconductors, does not follow ohm's law.
Resistance $R=\rho \dfrac {L}{A}$ where $\rho$ resistivity

$\rho=R\dfrac AL$

So its unit comes as ohm-meter.

When cells are connected in series, their EMFs add up to indicate the total EMF of the circuit. ie $E=E_1+E_2$

According to Joule law of heating, $Q=i^2Rt$

Neutral wire is there is no current flow ie $i=0$ .

A current of 8 A flows through a 25 V car headlight bulb. Calculate the electric power (in Watts).

Given

$\displaystyle I= 8A, V= 25V$

We Know,

Power (P)

$\displaystyle = V*I$

$\displaystyle = 25*8= 200 watt$

Two resistors of resistances $10 \Omega$ and $20 \Omega$ are connected in parallel. A battery supplies $6 A$ of current to the combination. Calculate the current in each resistor.

The current in the

$10 \Omega$ resistance is

$\displaystyle i_1 = \frac {R_2}{R_1 + R_2}i = \frac {(20) \times (6)}{20 + 10} = 4A$

The current in the

$20 \Omega$ resistor is

$\displaystyle i_2 = \frac {R_1}{R_1 + R_2}i = \frac {10 \times 6}{10 + 20} = 2A$

Or

emf across

$10 \Omega$ resistor

$V_1 = i_1 \times 10$
emf across

$20 \Omega$ resistor

$V_2 = i_2 \times 20$
Since

$i_1 + i_2 = 6 A$ and

$V_1 = V_2$ , we have

$i_1 = 4 A$

$i_2 = 2 A$

Explain the following.
(a) Why is the tungsten used almost exclusively for filament of electric lamps?
(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross section?
(e) Why are copper and aluminium wires usually employed for electricity transmission?

(a)

Electric lamps heat up with time as they are used. Since, tungsten has very high melting point, it does not melt even at very high temperatures. Hence, the lamp in not destroyed.

(b)

Electrical heating devices are required to produce a lot of heat and maintain high temperatures. Since, heat generated is proportional to resistivity of device, materials of higher resistivities are preferred in the construction of heating devices. Alloys have more resistance than metals and hence are preferred for the same.

(c)

When resistors are connected in series, potential difference across each resistor is smaller than the applied voltage. Hence, the power delivered to each device is reduced and hence they don't work at full efficiency.

(d)

$R=\rho l/A$

Hence, resistance is inversely proportional to the cross section of the resistance of the wire.

(e)

Cables used for electrical transmission should use up as small power as possible so that most of the power is delivered to the electrical equipment. Power consumed in the wire is proportional to resistivity. Copper and aluminium have very low resistivity and hence well suited for the purpose of being used as wires.

A certain household has consumed $250\ units$ of energy during a month. How much energy is this in Joules?

Energy consumed by certain households in a month

$E = 250$ units

$=250$

$kWh$

$\therefore E = 250 \times 3.6 \times 10^6$

$J$

$(\because 1kWh = 3.6 \times 10^6 J)$

$\implies E = 9\times 10^8$ Joules

How many $176\ \Omega$ resistors (in parallel) are required to carry 5A on a 220V line?

$\textbf{Step 1 : Equivalent resistance}$

Let

$n$ be the number of resistors to be connected in parallel

Then,

$R_1=R_2=......=R_n=R=176\ \Omega\quad$

Now,

$\cfrac{1}{R_{eq}}=\cfrac{1}{R_1}+\cfrac{1}{R_2}+.....\cfrac{1}{R_n}\quad$

$\Rightarrow \cfrac{1}{R_{eq}} = \cfrac{n}{R_1} = \cfrac{n}{176}$

$\Rightarrow R_{eq}=176/n\quad$

$....(1)$

$\textbf{Step 2 : Applying Ohm's law}$

$V=IR_{eq}$

$\Rightarrow 220=5R_{eq}$

$\Rightarrow 220=5 \times 176/n$

$($ Using Equation

$(1))$

$\Rightarrow n=4$

Hence, four resistors are required.

A toaster produces more heat than an electric lamp when they are connected in parallel. Which of the two has the greater resistance?

Given: a toaster and an electric lamp, such that toaster produces more heat than an electric lamp when they are connected in parallel.

To find which of the two has greater resistance.

Solution:

The light bulb has greater resistance.

Heat produced,

$H = \dfrac {V^2t}R$

$\implies H \propto \dfrac 1R$

Toster produces more heat. Hence it has less resistance.

Bulb produces less heat and therefore it has greater resistance.

Which of the following bulbs offers more resistance?
(A) 60 W - 220 V(B) 100 W - 250 V

Given: two bulbs of rating as (i)

$60W-220V$ (ii)

$100W-250V$

To find which of these two bulbs offers more resistance

Solution:

We know by Ohm's law equation:

$V = I\times R$ and by the power law equation:

$P = I \times V$ , where

$V$ is the voltage,

$I$ is the current,

$R$ is the resistance and

$P$ is the power.

Dividing both equations we get,

$\dfrac VP=\dfrac RV\\\implies R=\dfrac {V^2}P$

Now for first bulb,

$R_1=\dfrac {(220)^2}{60}=806.7\Omega$

And for second bulb,

$R_2=\dfrac {(250)^2}{100}=625\Omega$

By comparing

$R_1$ and

$R_2$ first bulb, i.e., bulb with rating

$60W-220V$ offers more resistance.

A resistance of $3\ ohms$ is connected in series with another resistance of $5\ ohms$ . A potential difference of $6$ volts is applied across the combination. Calculate the current through the circuit and potential difference across the $3\ ohm$ resistance.

Given: A resistance of

$3\Omega$ is connected in series with another resistance of

$5\Omega$ . A potential difference of

$6V$ is applied across the combination.

To find the current through the circuit and potential difference across the

$3\Omega$ resistance.

Solution:

as the resistance are connected in series

$R=R_1 + R_2 = 3 +5 = 8\Omega$

now according to Ohm's law,

$V = IR$

$\implies I= \dfrac VR\\\implies I=\dfrac {6}{8}\\\implies I=0.75A$

So,

$0.75A$ of current flow through circuit.

Now , Potential Difference across the

$3\Omega$ resistance is

$V_1=IR_1\\\implies V_1=0.75\times 3\\\implies V_1=2.25V$

Now the corresponding current will be,

$I_1 =\dfrac {V_1}{R_1}\\\implies I_1=\dfrac {2.25}{3}\\\implies I_1=0.75A=I$

and we know that the same current flow through each of the resistance in series.

Give examples for good conductors solid and liquid.

Solid conductor's are iron copper gold silver graphite platinum etc.
liquid conductor's are sodium Argento cynide, copper sulphate solution which should be acidified with sulphuric acid and water is the most common conductor

The potential difference between the ends of a conductor is 40V. If the conductor draws a current of 4A from the source, what current will the conductor draw if the potential difference is increased to 100V ?

Given:

Voltage = 40V

Current = 4A

From Ohm's law,

$V=IR$

$\therefore R=\dfrac{V}{I}=\dfrac{40}{4}=10\ \Omega$

When

$V=100\ volt$ and

$R=10\ \Omega$ we need to find the current.

Again using Ohm's law we can write,

$I=\dfrac{V}{R}=\dfrac{100}{10}=10\ A$

The figure shows a circuit
When the circuit is switched on, the ammeter reads $0.5 A$ .
(i) Calculate the value of the unknown resistor $R$ .
(ii) Calculate the charge passing through the $3 \Omega$ resistor in $120 s$ .
(iii) Calculate the power dissipated in the $3 \Omega$ resistor.

Given:

Ammeter reading = Current (I) =

$0.5\ A$

Voltage (V) =

$6.0\ volt$

(i)

$\text{Calculating the unknown resistance (R):}$

In the circuit, two resistors

$R$ and

$3\ \Omega$ are in series. Thus the equivalent resistance is

$R_{eq}=R+3$

We know,

Electric current

$\left(I\right) = \dfrac{Total \quad potential \quad difference \left(V\right)}{Total \quad resistance \left(R_{eq}\right)}$

$\Rightarrow 0.5 = \dfrac{6}{3 + R}$

$\Rightarrow R + 3 = \dfrac{60}{5}$

$\Rightarrow R = 12 - 3 = 9\ \Omega$

The unknown resistor,

$R = 9 \Omega$

(ii)

$\text{Calculating the charge (q) passing through the 3 ohm resistor:}$

We know from the definition of current,

$i=\dfrac{q}{t}$

$\therefore q=i\times t = 0.5\times 120 = 60\ C$
Charge passing through the resistor,

$q = 60C$

(iii)

$\text{Calculating the power dissipation in the 3 ohm resistor:}$

We know from the definition of power,

$P = V \times I$

$= I^2 R$

$\therefore P= {\left(0.5\right)}^{2} \times 3 = 0.25 \times 3$

The power dissipated,

$P = 0.75W$

Deduce the expression for the the equivalent resistance of three resistors connected in series.

Series combination of resistors : If a number of resistors are joined end to end so that the same current flows through each of them in succession, then the resistors are said to be connected in series.

As shown in the figure, consider three resistors

$R_1, R_2, R_3$ connected in series. Suppose a current

$I$ flows through the circuit when a cell of

$V$ volt is connected across the combination.

By Ohm’s law, the potential differences across the three resistors will be,

$V_1 = IR_1, V_2 = IR_2, V_3 = IR_3$

$R_s$ be the equivalent resistance of the series combination, then on applying a potential difference

$V$ across it, the same current

$I$ must flow through it.

Therefore,

$V=IR_s$

But

$V=V_1+V_2+V_3\\\therefore, IR_s=IR_1+IR_2+IR_3\\\implies IR_s=I(R_1+R_2+R_3)\\\implies R_s=R_1+R_2+R_3$

Laws of resistances in series:

(i) Current through each resistance is same

(ii) Total voltage across the combination = sum of the voltage drops

(iii) Voltage drop across any resistor is proportional to its resistance

(iv) Equivalent resistance = sum of the individual resistances

(v) Equivalent resistance is larger than the largest individual resistance.

How do you verify that resistance of a conductor is proportional to the length of the conductor for constant cross section area and temperature?

Aim : To verify that resistance of a conductor is proportional to the length of the conductor for constant cross-section area and temperature.

Procedure :

Collect iron spokes of different lengths ; with same cross sectional areas.
Make a circuit leaving gas between P and Q with a battery, ammeter and plug key.
Connect one of the iron spoke say 10 cm length between P and Q.
Measure the value of current using ammeter and note the value.
Repeat this for other lengths of the iron spokes and note the corresponding values of currents.
You will observe that the current decreases with increase in the length of the spoke.
Thus the resistance of each spoke increases with increase in the length for a constant potential difference.
From this, we conclude that the resistance of a conductor is directly proportional to its length (l) for constant cross-sectional area. Therefore, $R \propto L$
We notice that the resistance was more when the cross - section area was less. Therefore , $R \propto\dfrac LA$

Thus we verify

$L$ and

$A$

(i) Name the device used to protect the electric circuits from overloading and short circuits.
(ii) On what effect of electricity does the above device work?

(i) The device used to protect the electric circuits from overloading and short circuits is electric fuse.
(ii) Electric fuse works on the principle of heating effect of electrical current.

Deduce the expression for the the equivalent resistance of three resistors connected in Parallel.

In parallel we know that voltage remains same

$I={ I }_{ 1 }+{ I }_{ 2 }+{ I }_{ 3 }\\ \dfrac { V }{ { R }_{ eq } } =\dfrac { V }{ { R }_{ 1 } } +\dfrac { V }{ { R }_{ 2 } } +\dfrac { V }{ { R }_{ 3 } } \\ \dfrac { 1 }{ { R }_{ eq } } =\dfrac { 1 }{ { R }_{ 1 } } +\dfrac { 1 }{ { R }_{ 2 } } +\dfrac { 1 }{ { R }_{ 3 } }$

An unknown circuit draws a current of 2A from a 12V battery equivalent resistance is.......

Current = 2A

Voltage = 12V

We know V = IR

$\dfrac { V }{ R }$

= 6

$\Omega$

Three resistors of values $2 \Omega, 4\Omega, 6 \Omega$ are connected is series. The equivalent resistance of combination of resistors is............

${ R }_{ eq }={ R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }$ (In series)

$=2\Omega +4\Omega +6\Omega \\ =12\Omega$

Difference between resistances in series and parallel.

Sr.No.	Resistances in Series	Resistances in Parallel
1	Current remains constant in series connection.	Potential difference remains constant in parallel connection.
2	$R_{eq} = R_1 + R_2 + R_3$	$\dfrac{1}{R_{eq}} =\dfrac{1}{R_{1}} +\dfrac{1}{R_{2}} +\dfrac{1}{R_{3}}$
3	If one component burns out, no current will flow through the circuit.	If one connection burns out, still same current will flow through the circuit.

Find the expression for the resistors connected in series. Write any two characteristics of series combination of resistors.

Resistors in Series : If the number of resistance are connected one after another in such a way that the same current flows through each resistance, then the arrangement is called resistance in series.

$I$ is the current,

$V$ is the PD across

$C$ and

$D$ , the three resistors

${R}_{1}$ ,

${R}_{2}$ and

${R}_{3}$ are connected in series between points

$C$ and

$D$ .

${R}_{s}$ is the effective resistance in the circuit and

${V}_{1}$ ,

${V}_{2}$ and

${V}_{3}$ are the potentials across the three resistors.

$V = {V}_{1} + {V}_{2} + {V}_{3}$ ........(1)
By Ohm's Law,

$V = I{R}_{s}$

${V}_{1} = I{R}_{1}$ ,

${V}_{2} = I {R}_{2}$ ,

${V}_{3} = I{R}_{3}$
Substituting these values in equation no. 1

$I{R}_{s} = I{R}_{1} + I{R}_{2} +I{R}_{3}$

$I\left({R}_{s}\right) = I \left({R}_{1}+{R}_{2}+{R}_{3}\right)$

$\therefore$

${R}_{s}={R}_{1}+{R}_{2}+{R}_{3}$
For '

$n$ ' number of resistors connected in series

${R}_{s}={R}_{1}+{R}_{2}+{R}_{3}+$ ..........

${R}_{n}$ .

Characteristics of series combination of resistors :
(1) In series combination current through each resistor is constant.
(2) In series combination potential difference across each resistance is different depending upon the value of resistance.

With a neat labeled diagram, derive the equation for three resistance connected in parallel.

The above figure shows the circuit diagram for a parallel combination of three resistors with resistance

$R_1, R_2, R_3$ connected between the points C and D. The Voltmeter measures the potential difference V between C and D and the ammeter measure the current in a given branch.

Let

$I$ be the total current passing through the circuit.

$I_1, I_2$ and

$I_3$ be the current passing through the resistor of resistance

$R_1, R_2$ and

$R_3$ respectively.

$I=I_1+I_2+I_3$

$R_p$ be resultant resistance of parallel combination, we have by Ohm's Law.

$I=\dfrac{V}{R_p}$ and

$I_1=\dfrac{V}{R_1}, I_2 = \dfrac{V}{R_2}$ and

$I_3=\dfrac{V}{R_3}$ .

from equation (i)

$I=I_1+I_2+I_3=\dfrac{V}{R_1}+\dfrac{V}{R_2}+\dfrac{V}{R_3}$

$\therefore I=V(\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3})$

$\therefore \dfrac{V}{R_p}=V(\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3})$

$\therefore \dfrac{1}{R_p}=\dfrac{1}{R_1} + \dfrac{1}{R_2}+\dfrac{1}{R_3}$

Find the expression for the resistors connected in series and state its two characteristics. (Draw figure).

In various electrical appliances, resistors are often used in combinations. Usually, they are connected either in series or in parallel.

$V=V_1 +V_2 + V_3$ eq. 1
According to Ohm's law,

$V_r=IR_s$
(where

$Rs$ is the effective resistance in series combination)

$V_1=IR_1$

$V_2=IR_2$

$V_3=IR_3$
Substituting these values in equations 1,

$IRs = IR_1 +IR_2 +IR_3$

$IRs =I(R_1+R_2 +R_3)$
Striking off '

$I$ ' from the both sides of the equation,

$R_s=R_1 +R_2 +R_3$
Thus, the effective resistance of a series combination of resistance is the sum of individual resistances of the resistors in the combination.
Characteristics of the series combination of resistors are:
(a) The value of the current

$'I'$ flowing through the resistors connected in series is the same through each resistor.
(b). The potential difference

$'V'$ across the resistors connected in series is equal to the sum of individual potential differences across each resistor.

A student connected two bulbs in two different combinations as shown in figure A and B. Observe the figure and answer the following questions:
a. Name the type of combination of the bulbs in figure A.
b. Name the type of combination of the bulbs in figure B.
c. State the combination in which the resistance will be decreased.

a. In figure A bulbs are connected as a series combination as they are connected one by one.
b. In figure B bulbs are connected as a parallel combination as they are connected across the same voltage.
c. In the parallel combination, the net resistance will be less.

$\text{Proof:}$

Let two identical resistances (

$R$ and

$R$ ) are connected in series. Then the equivalent resistance will be

$R_s=R+R=2R$

Now, let the resistances are connected in parallel. So, the equivalent resistance is

$R_p=\dfrac{R\times R}{R+R}=\dfrac{R}{2}$

So, it is clear that

$R_p<R_s$ .

When do you say that the resistors are connected in series? Find the expression for the resistors connected in this way. (Draw circuit diagram).

$I$ is the current,

$V$ is the PD across

$C$ and

$D$ , the three resistors

${R}_{1}$ ,

${R}_{2}$ and

${R}_{3}$ are connected in series between points

$C$ and

$D$ .

${R}_{s}$ is the effective resistance in the circuit and

${V}_{1}$ ,

${V}_{2}$ and

${V}_{3}$ are the potentials across the three resistors.

$V = {V}_{1} + {V}_{2} + {V}_{3}$ ........(1)
By Ohm's Law,

$V = I{R}_{s}$

${V}_{1} = I{R}_{1}$ ,

${V}_{2} = I {R}_{2}$ ,

${V}_{3} = I{R}_{3}$
Substituting these values in equation no. 1

$I{R}_{s} = I{R}_{1} + I{R}_{2} +I{R}_{3}$

$I\left({R}_{s}\right) = I \left({R}_{1}+{R}_{2}+{R}_{3}\right)$

$\therefore$

${R}_{s}={R}_{1}+{R}_{2}+{R}_{3}$
For '

$n$ ' number of resistors connected in series

${R}_{s}={R}_{1}+{R}_{2}+{R}_{3}+$ ..........

${R}_{n}$ .
Characteristics of series combination of resistors :
(1) In series combination current through each resistor is constant.
(2) In series combination potential difference across each resistance is different depending upon the value of resistance.

Three resistances having the values 5 $\Omega$ , 10 $\Omega$ , 30 $\Omega$ are connected in parallel with each other. Calculate the total circuit resistance.

$R_1 = 5 \Omega, \,\,\,\,\,\, R_2 = 10 \Omega\,\,\,\,\,\, R_3 = 30 \Omega,$
In parallel connection, the resultant resistance is found by:

$\displaystyle \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$
=

$\displaystyle \frac{1}{R_p} = \frac{1}{5} + \frac{1}{10} + \frac{1}{30} = \frac{10}{30}$
Therefore, total resistance,

$\displaystyle R_p = \frac{30}{10} = 3 \Omega.$

A domestic electric circuit has a fuse of $5A$ . What is the maximum number of $100 W ( 220 V )$ bulbs that can be safely used in the circuit?

Power (P)=

$voltage(v) \times current (i)=100W$

$i=$

$\cfrac{100}{220}=0.45 A$
Number of bulbs that can be safely operated through a

$5A$ fuse

$=$

$\cfrac{5}{0.45}$

$=11$

Two lamps one rated $60\ W$ , $220\ V$ and the other $40\ W$ , $220\ V$ are connected in parallel to a $220\ V$ electric supply mains. What is the total current drawn from the electric mains if the voltage of the electric supply is $220\ V$ ?

Relationship between power

$(P)$ , voltage

$(V)$ , and Resistance

$(R)$

$P=\dfrac{V^2}{R}$

$\Rightarrow R=\dfrac{V^2}{P}$

Resistance of lamp of

$60W, 220V$ is

$R_1$ ,

$R_1=\cfrac{220^2}{60}=806.67\ \Omega$

Resistance of

$40W, 220V$ lamp be

$R_2$ ,

$R_2=\cfrac{220^2}{40}=1210\ \Omega$

These resistances are connected in parallel.

Equivalent resistance,

$R = \dfrac{ R_1R_2 }{ R_1+R_2 }$

$R=$

$\cfrac{(806.67\times 1210)}{(806.67 + 1210)}=484 \Omega$

Current drawn from

$220 V$ ,

$I=$

$\cfrac{220}{484}$

$=0.45 A$

Three resistors are connected in series with $10 V$ supply as shown in the figure. Find the voltage drop across each resistor, and effective resistance of series combination.

Given,

${ R }_{ 1 }=5\Omega$ ,

${ R }_{ 2 }=3\Omega$ ,

${ R }_{ 3 }=2\Omega$ ,

$V=10$ Volt
Effective resistance of series combination.

${ R }_{ s }={ R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }=10\Omega$
Now,
Current in Circuit

$I = I = \dfrac { V }{ { R }_{ s } } =\dfrac { 10 }{ 10 } =1A$
Voltage drop across

${ R }_{ 1 }$ ,

${ V }_{ 1 }=I{ R }_{ 1 }=1\times 5=5V$
Voltage drop across

${ R }_{ 2 }$ ,

${ V }_{ 2 }=I{ R }_{ 2 }=1\times 3=3V$
Voltage drop across

${ R }_{ 3 }$ ,

${ V }_{ 3 }=I{ R }_{ 3 }=1\times 2=2V$

My mother was operating the washing machine on Sunday morning. Suddenly, she saw a spark coming out of the electric board and the electric current in the house failed. An electrician was called to look into the matter. What should he have noticed?

Fuse protects circuits and appliances by stopping the flow of any unduly high electric current. The fuse is placed in series with the device. It consists of a piece of wire made of a metal or an alloy of appropriate melting point, for example aluminium, copper, iron, lead etc. If a current larger than the specified value flows through the circuit, the temperature of the fuse wire increases. This melts the fuse wire and breaks the circuit.

In the above mentioned case the electrician must have noticed that the fuse has burnt due to overheating as large amount of current has flown through it.

Show that the effective resistance of series combination in a circuit is equal to the sum of their resistances.

$I$ is the current,

$V$ is the PD across

$C$ and

$D$ , the three resistors

${R}_{1}$ ,

${R}_{2}$ and

${R}_{3}$ are connected in series between points

$C$ and

$D$ .

${R}_{s}$ is the effective resistance in the circuit and

${V}_{1}$ ,

${V}_{2}$ and

${V}_{3}$ are the potentials across the three resistors.

$V = {V}_{1} + {V}_{2} + {V}_{3}$ ........(1)
By Ohm's Law,

$V = I{R}_{s}$

${V}_{1} = I{R}_{1}$ ,

${V}_{2} = I {R}_{2}$ ,

${V}_{3} = I{R}_{3}$
Substituting these values in equation no. 1

$I{R}_{s} = I{R}_{1} + I{R}_{2} +I{R}_{3}$

$I\left({R}_{s}\right) = I \left({R}_{1}+{R}_{2}+{R}_{3}\right)$

$\therefore$

${R}_{s}={R}_{1}+{R}_{2}+{R}_{3}$
For '

$n$ ' number of resistors connected in series

${R}_{s}={R}_{1}+{R}_{2}+{R}_{3}+$ ..........

${R}_{n}$ .

A wire of resistance $4 R$ is bent in the form of a circle. What is the effective resistance between the ends of the diameter?

When a wire of resistance

$4R$ is bent in the form of circle, then the resistance of each segment across the diameter becomes

$2R$ as shown in the figure.

Equivalent resistance between A and B:

$\dfrac{1}{R_{AB}} = \dfrac{1}{R_1} + \dfrac{1}{R_B}$ (in parallel combination)

$\therefore$

$\dfrac{1}{R_{AB}} = \dfrac{1}{2R} + \dfrac{1}{2R}$

$\dfrac{1}{R_{AB}} = \dfrac{2}{2R}$

$\implies$

$R_{AB} =R$

Explain series connections of Resistors and drive the formula of equivalent resistance.

The resistors are connected across two points in the circuit in such a way that the current flowing through each resistor is the same and only one path is available for it to flow, then the resistors are said be connected in series.
In the figure above, three resistors

$R_1, R_2$ and

$R_3$ are connected in series across the points A and B. Hence, the current (I) flowing through each of the resistors is same but the total voltage of the battery gets divided.
If the voltage drops across

$V_1, V_2$ and

$V_3$ are

$R_1, R_2$ and

$R_3$ respectively then,

$V = V_1 + V_2 + V_3$ .(i)

Now, if the resistor R instead of these three resistances

$R_1, R_2$ and

$R_3$ is connected in such a way that the current flowing through the circuit remains the same, then R is called the equivalent resistance of the circuit.For an equivalent resistor, V = IR.(ii)
According to Ohms law,

$R_1$ voltage drops across,

$V_1 = I R_1$ ..(iii)

$R_2$ voltage drops across,

$V_2 = I R_2$ ..(iv)
R3 voltage drops across,

$V_3 = I R_3$ ..(v)
Adding (iii), (iv) and (v):

$V_1 + V_2 + V_3 = I R_1 + I R_2 + I R_3$ (vi)
Or,

$V = IR = I (R_1 + R_2 + R_3)$
Or,

$R = R_1 + R_2 + R_3$
Thus, in a series connection, the equivalent resistor R is equal to the sum of all the resistors connected in the series.

With the help of a circuit diagram, obtain the expression for equivalent resistance of two resistors connected in parallel.

Lets resistance

$R_1, R_2,$ are connected in parallel resistance circuit. Then the equivalent resistance of the circuit is given by

$\dfrac{1}{R_{eq}}= \dfrac{1}{R_1} + \dfrac{1}{R_2}$

So in the given circuit the equivalent resistance is

$\dfrac{1}{R}= \dfrac{1}{2}+\dfrac{1}{6}$

$R_{eq}= \dfrac{3}{2} \Omega$

Nichrome and copper wire of same length and same radius are connected in series. Current $I$ is passed through them. Which wire gets heated up more? Justify your answer.

Resistance of nichrome is greater than copper wire of same length and same radius, because the former has greater resistivity.
In series, Power loss P=

$I^2 R$ . Because of higher resistance power loss and heating up of nichrome wire is more than that of copper.

Name the device used to protect the electric circuits from overloading and short circuits.

The device used to protect the electric circuits from overloading and short circuits is electric fuse. Wires made from some special materials melt quickly and break when large electric currents are passed through them. These materials are used for making electric fuses which prevent fires and damage to electric appliances.

Vani's hairdryer has a resistance of $50 \Omega$ when it is first turned on.
i) How much current does the hairdryer draw from the $230\ V$ line in Vani's house?
ii) What happens to the resistance of the hairdryer when it runs for a long time?

Given: Vani's hairdryer has a resistance of

$50\ Ω$

According to the question:

$R=50\ \Omega$ ,

$V=230\ V$

(i) According to Ohm's law,

$V=IR$

$\Rightarrow I = \dfrac{V}{R}$

$\Rightarrow I=\dfrac {230}{50}$

$\Rightarrow I=4.6\Omega$

ii) When it runs for long time, the resistance of the conductor increases.

The commercial unit of electrical energy is _____.

Power is the rate of energy consumed or delivered. Kilowatt-hour

$kWh$ is the commercial unit of electrical energy.

One Kwh is equal to ___________J.

$1 \ Kwh$ =power consumption of

$1000\ watt$ for

$1 \ hour$ ,

Since,

$1 \ hour = 3600 \ sec$

Hence,

$1 \ Kwh=1000\times 3600J=3.6\times 10^6 \ J$

Give the definition of electric current and define its unit.

A stream of electrons moving through a conductor constitutes an electric current. Conventionally, the direction of current is taken opposite to the direction of the flow of electrons. The SI unit of electric current is ampere.

What is equivalent resistance? Derive the expression for effective resistance of two resistors connected in parallel.

Equivalent resistance is a different way of indicating 'total' resistance, which we calculate differently for series and parallel circuits.

Lets resistance

$R_1, R_2,$ are connected in parallel resistance circuit. Then the equivalent resistance of the circuit is given by

$\dfrac{1}{R_{eq}}= \dfrac{1}{R_1} + \dfrac{1}{R_2}$

Three resistors $R_1, R_2, R_3$ are joined in series. Draw labelled diagram and obtain expression of its equivalent resistance.

In Series combination of resistances, the second end to each resistance is connected to the first end of the next resistance and so on.
In a series combination of resistances, the current flowing through each resistance will be the same and will be equal to the current supplied by the battery. As resistance is different and the current flowing through each resistance is the same, hence potential differences across different resistances will be different.
Let three resistance

$R_1, R_2, R_3$ are connected in series with a cell of potential difference

$V$ . Let

$I$ current supplied by the battery in this combination.
Let,

$V_1=$ Potential difference across the ends of resistance

$R_1$

$V_2=$ Potential difference across the ends of resistance

$R_2$ .

$V_3=$ Potential difference across the ends of resistance

$R_3$ .

$V=V_1+V_2+V_3$ ........(1)
If equivalent resistance of the combination be

$R$ then by Ohm's law,

$V=IR, V_1=IR_1, V_2=IR_2, V_3=IR_3$
Put in Eq.(1)

$I=IR_1+IR_2+IR_3$

$IR=I(R_1+R_2+R_3)$

$R=R_1+R_2+R_3$

Therefore, in a series combination of resistance equivalent resistance is equal to the sum of all individual resistances.

The safety device based on the heating effect of electric current is called a ___________(fuse, heater).

The safety device based on the heating effect of current is called a

$\text{fuse}$ .

Commercial unit of electric energy is kilowatt hour (kWh). Convert 1 kWh into Joule.

$1 kWh = (1 \ kilowatt) \ (1 \ hour)$

$= (1 \times 10^3 \ watt) \ (1 \ hour)$

We know

$1 \ watt= 1 \ Js^{-1}$

So,

$1 kWh= (1000 Js^{-1}) (1 \times60 \times 60 s)$

$= 36 \times 10^5 J$

$= 3.6 \times 10^6 J$

Thus,

$1 kWh$ in Joules is equal to

$3.6 \times 10^6 \ J.$

Resistors are sometimes joined together and they have several applications in electronics. Obtain the expression for the effective resistance in the combination of two resistors ${R}_{1}$ and ${R}_{2}$ in parallel.

When two resistor are connected in parallel, they will have common terminals A and B as shown in figure. Thus, there is same potential V across each resistor.

In parallel ,

$I=I_1+I_2$

$\dfrac{V}{R_{eff}}=\dfrac{V}{R_1}+\dfrac{V}{R_2}$ (by Ohm's law)

$\dfrac{1}{R_{eff}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}$

$R_{eff}=\dfrac{R_1R_2}{R_1+R2}$

Name any two effects of electric current.

Explanation: Electric current has the following effect :
(i) Electric current can give rise to heating and lighting.
Most common examples are bulb, electric heater etc.
(ii) Electric current can convert a straight conductor into a temporary magnet.
Most common examples are electromagnets used to pick up metal pieces from garbage.

What do you understand by the thermal effect of electric current?

Thermal effect of electric current - The transformation of electric energy to heat energy is called thermal effect of electric current.
The substances of a conductor is made up of atoms. In a metallic conductor, there are a large number of free electron moving with different velocity. When electric current flows through the conductor, the free electrons collide with their atoms sue to which number of electrons increase and the major portion of Kinetic energy gets converted to heat energy and the temperature of the conductor increases. heating effect of current is used in home appliances like heater, iron, water heater, oven, etc.

Calculate the resultant resistance when three resistances of $3\Omega$ are connected in parallel?

Suppose

$R$ is the total resistance of parallel combination then it is given by

$\dfrac {1}{R} = \dfrac {1}{R_{1}} + \dfrac {1}{R_{2}} + \dfrac {1}{R_{3}}$
Given -

$R_{1} = R_{2} = R_{3} = 3\Omega$
or

$\dfrac {1}{R} = \dfrac {1}{3} + \dfrac {1}{3} + \dfrac {1}{3} = 1$
Thus

$R = 1\ Ohm$ .

Resistors are sometimes joined together and they have several applications in electronics. Draw a series combination of three resistors ${R}_{1}$ , ${R}_{2}$ and ${R}_{3}$ .

In series combination, the resistors are connected back to back as shown in the figure. Here,

$V=V_1+V_2+V_3$ or

$IR_{eq}=IR_1+IR_2+IR_3$ or

$R_{eq}=R_1+R_2+R_3$ , the equivalent resistance of the whole circuit.

On which factors does the heat produced depend when the current is flown through a conducting wire and explain it.

The heat produced in conducting wire is given by joules law as,

$H=I^2 Rt$

i ) The amount of heat produced in current conducting wire, is proportional to the square of the amount of current that is flowing through the circuit, when the electrical resistance of the wire and the time of current flow is constant.

$H\propto I^2$ (when R, t are constant)

II) Heat generated due to the flow of current is proportional to the time of current flow, when the electrical resistance and amount of current flow is constant.

$H\propto t$ (when I,R are constant)

III) The amount of heat produced is proportional to the electrical resistance of the wire when the current in the circuit and the time of current flow is constant.

$H\propto R$ (when I, t are constant)

You have three resistors of values $2\Omega, 3\Omega$ and $5\Omega$ . How will you join them so that the total resistance is more than $7\Omega$ ? Draw a diagram for the arrangement.

In order to get more than

$7\Omega$ resistance from three resistors

$2\Omega, \ 3\Omega$ and

$5\Omega$ , we need to connect them in series as shown in the diagram.

The net resistance for this series connection is

$2+3+5=10\ \Omega$ , which is greater than

$7\ \Omega$ .

You have three resistors of values $2\Omega, 3\Omega$ and $5\Omega$ . How will you join them so that the total resistance is more than $7\Omega$ ? Calculate the equivalent resistance.

In order to get more than

$7\Omega$ resistance from three resistors

$2\Omega, \ 3\Omega$ and

$5\Omega$ , we need to connect them in series as shown in the diagram.
Equivalent resistance in series combination is given by

$R_{eq} = R_1 + R_2 + R_3$

$\implies \ R_{eq} = 2+3+5 = 10\Omega$ (which is greater than

$7\Omega$ )

An electric lamp of resistance $100\Omega$ , a toaster of resistance $50\Omega$ and a water filter of resistance $500\Omega$ are connected in parallel to a $220V$ source. What is the resistance of an electric iron connected to the same source and takes as much current as all three appliances and what is the current through it ?

$I$ in electric lamp

$=\dfrac{220}{100}$

$I$ in toaster

$=\dfrac{220}{50}$

$I$ in filter

$=\dfrac{220}{500}$

$\Rightarrow$ Total current

$=\dfrac{220}{100}+\dfrac{220}{50}+\dfrac{220}{500}$

$=\dfrac{220}{50}\left( \dfrac{1}{2}+1+\dfrac{1}{10}\right)$

$=\dfrac{22}{5}\times \dfrac{8}{5}$

$=\dfrac{176}{25}$

$=7A$

$\Rightarrow R$ if iron

$=\dfrac{V}{I}=\dfrac{220}{\dfrac{176}{25}}$

$=31.25\Omega$

Hence, the answer is

$31.25\Omega .$

A resistance coil is made by joining in parallel two resistances each of 10 $\Omega$ . An emf of 1V is applied between the two ends of coil for 5 minutes. Calculate the heat produced in calories.

$R_1=R_2=10\Omega$

$\therefore R_{net}=\dfrac{R_1R_2}{R_1+R_2}=\dfrac{100}{20}=5\Omega$

$V=1v$

$\therefore P=\dfrac{V^2}{R}=\dfrac1{5}\Omega =0.2\Omega$

$\therefore$ Heat produced

$=0.2=0.05$ calories per second.

$\therefore$ Heat produced for

$5min=0.05\times 5\times 60cals$

$=15cal$

If the value of resistance is $10.845 \Omega$ and the value current is $3.23$ A, the potential difference is $35.02935\,\,V$ . Find its value in significant number.

The product of two no.'s can be precise as the least precise no. So the significant digits after the decimal are two.

$\therefore$ In

$V=IR$

$I=3.23A$ and

$R=10.845\Omega \Rightarrow { V }=35.03V$

As we can report up to only two significant figures after the decimal so the value of

$V$ is rounded off.

A copper wire has diameter $0.5 \ mm$ and resistivity of $1.6\times { 10 }^{ -8 }\Omega m$ . What will be the length of this wire to make its resistance $10\ \Omega$ ?

Given:

$d=0.5\quad mm=0.5\times { 10 }^{ -3 }\quad m\\ r=\dfrac { d }{ 2 } =\dfrac { 0.5\times { 10 }^{ -3 } }{ 2 } =0.25\times { 10 }^{ -3 }m\\ A=\pi { r }^{ 2 }=3.14\times (0.25\times { 10 }^{ -3 }{ ) }^{ 2 }\\ \rho =1.6\times { 10 }^{ -8 }\Omega m\\ R=10\Omega \\ We\quad know\quad R=\dfrac { \rho l }{ A } \\ or\quad l=\dfrac { RA }{ \rho } \\ \quad \quad =\dfrac { 10\times 3.14\times (0.25{ ) }^{ 2 }\times { 10 }^{ -6 } }{ 1.6\times { 10 }^{ -8 } } \\ \quad \quad =1.227\times { 10 }^{ 2 }\\ \quad \quad =122.7\quad m\\$

Explain and derive equivalent Resistance formula for 3 Resistors connected in Pure Series. Generalize the formula?

When three resistors are connected in series, then the same current passes through each resistor but the voltage drop is different for each resistor.

We know that applied voltage is

$\left( V \right)$

$\therefore V= {V}_{ 1 }+{ V }_{ 2 }+{ V }{ 3 }$

we also know that,

$V= I\cdot R \quad$ (from ohm's law)

$\therefore I\cdot { R }_{ eq }=I\cdot { R }_{ 1 }+I\cdot { R }_{ 2 }+I\cdot { R }_{ 3 }$

$\Rightarrow I\cdot { R }_{ eq}= I\cdot \left( { R }_{ 1 }{ +R }_{ 2 }+{ R }_{ 3 } \right)$

$\Rightarrow$

$R_{eq}= { R }_{ 1 }{ +R }_{ 2 }+{ R }_{ 3 }$

It can be extended for

$n$ resistors

$R_{eq}=$

$R_1 +R_2 ....................................R_n$

So, the equivalent resistance or the total resistance of the circuit can be defined as a single value of resistance that can replace any number of resistors connected in series without altering the vlaues of the current or the voltage in the circuit.

953423_1028177_ans_8fdc5416435649daafc9e6b3229a5b5c.png

When an electric current is passed through an electric bulb, it becomes warm when touched. Give reasons.

When an electric current is passed through an electric bulb, energy is dissipated in form of heat due to its resistance, that's why it becomes warm when touched.

Amount of heat dissipated when

$I$ current follow through resistance

$R$ in time

$t$ .

$\Rightarrow H = I^2Rt$

Heat energy is being produced in a resistance in a circuit at the rate of $100\ W$ . The current of $2\ A$ is flowing in the circuit. what must be the value of the resistance (in $\Omega$ )?

Rate of heat energy produced is power.

$\Rightarrow$ P =

$I^2\times R$

$\Rightarrow$ R =

$\dfrac{P}{I^2}$

$\Rightarrow$ R =

$\dfrac{100}{4}= 25 \Omega$

Two resistances when connected in parallel give resultant value of $2\Omega$ . When connected in series the value becomes $9\Omega$ . Calculate the value of each resistance.

Let the value of two resistors are

$R_1$ and

$R_2$ respectively.

$\underline{\text{When connected in series,}}$

The equivalent resistance:

$R_1+R_2=9$

$\therefore R_2=9-R_1$

$\underline{\text{When connected in parellel,}}$

The equivalent resistance:

$\dfrac{1}{R_1} + \dfrac{1}{R_2}=\dfrac{1}{2}$

$\Rightarrow \dfrac{1}{R_1} + \dfrac{1}{9-R_1}=\dfrac{1}{2}$ (Since,

$R_2 = 9-R_1$ )

$\Rightarrow \dfrac{9-R_1+R_1}{R_1(9-R_1)}=\dfrac{1}{2}$

$\Rightarrow\dfrac{9}{9R_1-R_1^2}=\dfrac{1}{2}$

$\Rightarrow R_1^2-9R_1+18=0$

$\Rightarrow (R_1-3)(R_1-6)=0$

$\Rightarrow R_1=3,6$
So,

When

$R_1=3\Omega , R_2=9-3=6\Omega$

When

$R_1=6\Omega , R_2=9-6=3\Omega$

Find the reading of ammeter. Is this the current thought $6\Omega$ ?

1311752_1034519_ans_d0274c5c1fd9404384679b82e0dd65b3.jpg

Calculate combined resistance of resistances $3 \Omega$ and $6 \Omega$ connected in parallel.

The resistances are connected is parallel then the resultant resistance is

$R_{eq} = \dfrac{{{R_1} \times {R_2}}}{{{R_1} + {R_2}}}$

$R_{eq} = \dfrac{{3 \times 6}}{{3 + 6}}$

The resultant resistance is $2\Omega$ .

Calculate the power of soldering iron with a hot resistance $4$ $K\Omega$ and operating voltage of $200$ $V$ .

Given:

$R=4$

$K\Omega$

$V=200$

$V$

$Power=\cfrac{V^2}{R}=\cfrac{(200)^2}{(4\times 10^3)}$ ,

$P$

$=\cfrac{4\times 10^4}{4\times 10^3}$

$P$

$=10$

$W$

In the above question, Find individual power consumed.
Given $R_1=1 \ k \Omega$ and $R_2=0.5\ k\Omega$

The equivalent resistance in the given circuit is

$R_{eq}=R_1+R_2$ (

$\because$ series combination)

$\Rightarrow R_{eq}=1+0.5=1.5 k\Omega=1500 \Omega$

Now, the current through the

$R_1$ and

$R_2$ will be same as that in the equivalent resistance

$R_{eq}.$

The voltage across it is

$V=220 \ V$

Applying Ohm's law for

$R_{eq}$ ,

$V=IR_{eq}$

$\Rightarrow 220= I \times 1500$

$\Rightarrow I = \dfrac{220}{1500}=0.146 \ A$

Now we apply Ohm's law individually on

$R_1$ and

$R_2$ to find

$V_1$ and

$V_2$ respectively,

$V_1=IR_1 =0.14666 \times 1000=146.66 \ V$

$V_2=IR_2 =0.14666 \times 500=73.33 \ V$

Now we can calculate the individual power by either of the formula

$P=\dfrac{V^2}{R}$ or

$P=I^2R$

Let's use the first relation.

Power consumed by resistance $R_1$ :

${P}_{1}=\dfrac{(V_1)^2}{R_1}= \dfrac{(146.66)^2}{1000}=21.509 \ W$

Power consumed by resistance $R_2$ :

${P}_{2}=\dfrac{(V_2)^2}{R_2}= \dfrac{(73.33)^2}{500}=10.755 \ W$

Judge the equivalent resistance when the following are connected in parallel
(i) $1 \Omega$ and $10^6 \Omega$
(ii) $1 \Omega, 10^3 \Omega$ and $10^6 \Omega.$

$(a)$ The net resistance in parallel is given by

$\dfrac{1}{R}=\dfrac{1}{{R}_{1}}+\dfrac{1}{{R}_{2}}$

Here

${R}_{1}=1\Omega$ and

${R}_{2}={10}^{6}\Omega$

So,

$\dfrac{1}{R}=\dfrac{{R}_{1}+{R}_{2}}{{R}_{1}{R}_{2}}$

$\Rightarrow R=\dfrac{{R}_{1}{R}_{2}}{{R}_{1}+{R}_{2}}$

$=\dfrac{1\times{10}^{6} }{{10}^{6}+1}$

$=\dfrac{1000000‬}{1000001}$

$=0.999\approx 1\Omega$

$(b)$ The net resistance in parallel is given by

$\dfrac{1}{R}=\dfrac{1}{{R}_{1}}+\dfrac{1}{{R}_{2}}+\dfrac{1}{{R}_{3}}$

Here

${R}_{1}=1\Omega,\,\,\,{R}_{2}={10}^{3}\Omega$ and

${R}_{3}={10}^{6}\Omega$

So,

$\dfrac{1}{R}=\dfrac{1}{1}+\dfrac{1}{{10}^{3}}+\dfrac{1}{{10}^{6}}$

$=\dfrac{{10}^{6}+{10}^{3}+1}{1000000}$

$=\dfrac{1000000‬+1000+1}{1000000}$

$=\dfrac{1001001‬}{1000000}$

$\Rightarrow R=\dfrac{1000000}{1001001‬}=0.999\Omega\approx 1\Omega$

$B_1, B_2$ and $B_3$ are three identical bulbs connected as shown in Figure 12.When all the three bulbs glow, a current of 3A is recorded by the ammeter A.
What happens to the reading of $A_1, A_2, A_3$ and A when the bulb $B_2$ gets fused?

Hint : Try to find the resistance of individual bulb and then use Ohm's Law to get current in each arm.

Step 1 : Find resistance of individual bulb.

Given: Three bulbs connected in parallel to a source of

$4.5\ V$ and net current in the circuit is

$3A$

Assume every bulb has a resistance

$R$

Net resistance of all three together in parallel (case

$1$ )

$\dfrac {1}{R_{net1}}=$

$\dfrac{1}{R_1} + \dfrac {1}{R_2} + \dfrac{1}{R_3}$

$R_1 = R_2 = R_3 = R$

$\implies$

$R_{net1} = \dfrac{R}{3}$

Current in whole circuit =

$3A$ (Given).

Using Ohm's Law

$V = IR_{net}$

$4.5 = 3 \times \dfrac{R}{3}$

$\therefore$

$R = 4.5 \Omega$

Step 2 : Find the net resistance and current in the circuit when bulb $B_{2}$ is fused.

Now, as

$B_{2}$ is fused

$B_{1}$ and

$B_{3}$ are in parallel (case

$2$ )

$\therefore$

$\dfrac {1}{R_{net2}}=$

$\dfrac{1}{R} + \dfrac {1}{R}$

${R_{net2}}=$

$\dfrac{R}{2} =$

$2.25 \Omega$

$\therefore$ Current in the circuit

$I = \dfrac{V}{R_{net2}} = \dfrac {4.5}{2.25} =$

$2A$

Reading of ammeter

$A = 2A$

Step 3 : Find current in individual branches

$B_{2}$ is fused no current flows through

$B_{2}$

$\therefore$

$A_2 =0A$

The net current of

$2A$ divides equally in all two branches as the resistance of both the bulbs is same.

$A_1 = 1A$ and

$A_2 = 1A$

Final Step : When Bulb

$B_2$ fuses,

The reading of

$A_1$ becomes

$1A,$

The reading of

$A_2$ becomes

$0A,$

The reading of

$A_3$ becomes

$1A,$

The reading of

$A$ becomes

$2A,$

Resistors having resistances of $15\Omega$ , $20\Omega$ , and $10\Omega$ are connected in parallel. What is the effective resistance in the circuit.

$\cfrac{1}{R_{eff}}=\cfrac{1}{R_1}+\cfrac{1}{R_2}+\cfrac{1}{R_3}+\cfrac{1}{R_4}+......+\cfrac{1}{R_n}$

$\cfrac{1}{R_{eff}}=\cfrac{1}{15}+\cfrac{1}{20}+\cfrac{1}{10}\times \left(\cfrac{1}{2}\right)=\cfrac{13}{60}$

$R_{eff}=\cfrac{60}{13}==4.615\Omega$

$B_1, B_2$ and $B_3$ are three identical bulbs connected as shown in Figure 12.When all the three bulbs glow, a current of 3A is recorded by the ammeter A.
How much power is dissipated in the circuit when all the three bulbs glow together?

Hint: Try to establish relation between Power consumed, voltage of source and net current flowing in the circuit.

Step 1 : Find power consumed by the combination

Given that,

A current of

$3A$ flows in the circuit and voltage of source is

$4.5\ V$

We know,

$P = V\times I_{net}$

$\therefore P = 4.5\ V \times 3\ A= 13.5\ W$

A wire of resistance 5 ohm is bend in the form of closed circle. What is the resistance between 2 points at the ends of any diameter of the circle?

The diametrically opposite points divides the resistance into half of the total resistance.

Resistance of each semi circle

$R_1 = R_2 = \dfrac{5}{2} = 2.5 \ \Omega$

These resistance are connected in parallel.

So, equivalent resistance

$R_{eq} = \dfrac{R_1R_2}{R_1+R_2} = \dfrac{2.5\times 2.5}{2.5+2.5} = 1.25\Omega$

A combination consists of three resistors in series. Four similar sets are connected in parallel. If the resistance of each resistor is $2\ \Omega$ . Find the equivalent resistance of the combination.

${\Large \underline {\textbf{Concept used:}}}$

${\large{\bullet \ \textbf{Series Resistance}}}$ :

$R_{series}=R_1+R_2+ . . . +R_n$

${\large{ \bullet \ \textbf{Parallel Resistance}}}$ :

$\dfrac{1}{R_{parallel}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+...+\dfrac{1}{R_n}$

${\large \underline {\textbf{Step 1: Identify the knowns and unknowns}}}$

$\bullet$ Resistance of one resistor

$R=2 \ \Omega$

$\bullet$ Three such resistors are in series to make one set.

$\bullet$ Four such sets are in parallel.

$\bullet$ The equivalent resistance of the combination

$R_{eq}=?$

${\large \underline {\textbf{Step 2: Determine the total resistance of one set}}}$

Three resistors of

$R=2 \ \Omega$ each are in series to make one set.

$R_{set}=R+R+R=3R$

$\Rightarrow R_{set}=3 \times 2 \ \Omega = 6 \ \Omega$

${\large \underline {\textbf{Step 3: Determine the total resistance of 4 such sets in parallel}}}$

Four resistors of

$R_{set}=6 \ \Omega$ each are in parallel to make one combination.

$\dfrac{1}{R_{eq}}=\dfrac{1}{R_{set}}+\dfrac{1}{R_{set}}+\dfrac{1}{R{set}}+\dfrac{1}{R_{set}}=\dfrac{4}{R_{set}}$

$\Rightarrow \dfrac{1}{R_{eq}}=\dfrac{4}{6 \ \Omega}$

$\Rightarrow R_{eq}=\dfrac{6}{4} \ \Omega$

$\Rightarrow R_{eq}=1.5 \ \Omega$

${\large \underline{\textbf{Answer:}}}$

The equivalent resistance of the combination is

$R_{eq}=1.5 \ \Omega$

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A conductor connected to a battery is heated, explain how.

Potential difference applied across two terminals of the conductor sets the electrons in motion to flow the current through it. To maintain the current, the source has to keep expending its energy. A part of the source energy in maintaining the current may be consumed into useful work (like in rotating the blades of an electric fan). Rest of the source energy may be expended in heat to raise the temperature of gadget.

The conductor gets heated up due to its resistance.
The moving electrons collide with atoms of the conductor.
There is an exchange of kinetic energy, heat is produced as a form of kinetic energy.
Hence the conductors heats up.

The energy supplied to the circuit by the source in time

$t$ is

$Power$

$(P) \times$

$Time$

$t$ , that is,

$VIt$ . This energy expended by the source gets dissipated in the conductor as heat. Thus for a steady current

$I$ , the amount of heat

$H$ produced in time

$t$ is

$H = VIt$

By Ohm's law,

$H=I^2Rt$

which is the Joule's law of heating.

Find the resistor each $3 \Omega$ are connected as shown, calculate the resistance
a) between the paint p and q
2) between the paint x and y

1948425_1102496_ans_b1148233f8ed4102a8de3ac9c08fe5f6.JPG

How many bulbs of power $60\ W$ can be connected in parallel across a $240\ V$ source if current rating of the circuit is $5A$ ? What is the resistance of each bulb?

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Two lamps of rating $100 W$ , $220V$ and $60 W$ , $220V$ are joined in parallel with $220 V$ line. How much current will flow through the circuit?

Given

$P_1=100W$

$V_1=220V$

$P_2=60W$

$V_2=220V$

We know that

$P=\dfrac{V^2}{R}$

$R=\dfrac{V^2}{P}$

$R_1=\dfrac{V_1^2}{P_1}=\dfrac{220^2}{100}$ =

$484\Omega$

$R_2=\dfrac{V_2^2}{P_2}=\dfrac{220^2}{60}$ =

$\dfrac{2420}{3}\Omega$

In parallel

$\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}$ =

$\dfrac{1}{484}+\dfrac{3}{2420}=\dfrac{8}{2420}$

$R=302.5\Omega$

$V=IR$

Supply=

$V=220V$

$220=I\times 302.5$

$I=0.727A$

What is the ratio of heat generation in two resistors having resistances $R$ and $2R$ in same time $t$ .

Since the time interval to be considered is same, we can compare power of both the resistors.

The ratio of powers will be same as the ratio of heat ( $\because$ heat = Power $\times$ time $\Rightarrow H \propto P$ )

If we assume the potential difference V across both of them is equal, then

Power $P = \dfrac{V^2}{r} \Rightarrow P \propto \dfrac{1}{r}$

So $\dfrac{P_1}{P_2}= \dfrac{r_2}{r_1}=\dfrac{2R}{R} \Rightarrow P_1:P_2=2:1$

If we assume the current through both of them is equal, then

Power $P = {I^2}r \Rightarrow P \propto r$

So $\dfrac{P_1}{P_2}= \dfrac{r_1}{r_2}=\dfrac{R}{2R} \Rightarrow P_1:P_2=1:2$

Find the total resistance in the circuit.?

Equivalent resistance of

$8\ \Omega$ and

$12\ \Omega$ in parallel.

$\dfrac{1}{R_{parallel}} = \dfrac 18 + \dfrac {1}{12}$

$\dfrac{1}{R_{parallel}} = \dfrac {3+2}{24}$

${R_{parallel}} = 4.8\ \Omega$

Equivalent resistance of

$7.2\ \Omega$ and

$4.8\ \Omega$ in series.

$R_{eq}=7.2+4.8$

$R_{eq}=12\ \Omega$

The total resistance in the circuit

$=12\ \Omega$

2108776_1123442_ans_4a9ad466475f4c4bb763d115332e1837.png

An electric heater of power 3 kW is used for one minute.Find the energy supply by the heater.

Given that,

Power $P=3\,kw$

Time $t=1\,\min =60\,s$

We know that,

$P=\dfrac{E}{t}$

$E=P\times t$

$E=3\times {{10}^{3}}\times 60$

$E=180\times {{10}^{3}}$

$E=18\times {{10}^{4}}\,J$

Hence, is $18\times {{10}^{4}}\,J$

A lamp consumes $1000\ J$ of electrical energy in $10\ s$ . What is its power?

1317370_1123745_ans_3833c270dfb84ae39445efb4b9d51509.jpg

What is the resistance of an ideal voltmeter?

The resistance of an ideal voltmeter is infinity so that there is no current flow from it and the resistance of an ideal ammeter is zero. Ammeter is connected in series and voltmeter is connected in parallel with the electric appliance

A piece of wire of resistance R is cut into five equal parts. When these parts are connected in parallel. Then the equivalent resistance of this combination is R' .The ratio $\frac{R}{R'}$ is

Resistance of piece of wire is proportional to its length. A piece of wire has a resistnce

$R$ . The wire is cut into five equal parts

Resistance of each part is

$\cfrac{R}{5}$

$\cfrac{1}{R'}=\cfrac{5}{R}+\cfrac{5}{R}+\cfrac{5}{R}+\cfrac{5}{R}+\cfrac{5}{R}$

$=\cfrac{5+5+5+5+5}{R}$

$\Rightarrow$

$\cfrac{1}{R'}=\cfrac{25}{R}$

$\cfrac{R}{R'}=25$

An electric bulb consumes $7.2\ kJ$ of electrical energy in $2\ min$ . What is the power of the electric bulb?

Given,

Energy consumes, $$E=7.2\ kJ= 7.2\times 1000\ J$$

$E=7200\ J$

Time,

$T=2\ min=2\times 60\ s$

$t=120\ s$

Power of the bulb,

$P = \dfrac{E}{t}$

So,

$P = \dfrac{7200}{120} = 60\ W$

State the relation between the commercial unit of energy and SI unit of energy.

The Commercial unit of energy is

$kilowatt \ hour (kWh)$ .

The SI unit of energy is

$Joule (J)$ .

$1 \ kWh = 1000 \ \dfrac{J}{s} \times 3600 \ s$

$1 \ kWh = 3.6\times 10^6 \ J$

Find the followings -
a) total resistance of the circuit,
b) the value of resistor R
c) the current flowing through R.

1479897_1260388_ans_4e23945799fd4f769d49c16de9ca96ce.jpg

A constantan wire of length $50\ cm$ and $0.4\ mm$ diameter is used in making a resistor. If the resistivity of constantan is $5\times 10^{-7} \ \Omega-m$ , calculate the value of the resistance.

Given,

length

$l=50 \ cm = 0.50 \ m$

resistivity

$\rho = 5 \times 10^{-7} \ \Omega -m$

diameter

$d=0.4 \ mm$

$\therefore$ radius

$r=0.2 \ mm = 0.2 \times 10^{-3} \ m$

Now,

Area

$A =\pi {r^2}$

$A =\pi { \left( { 0.2\times { { 10 }^{ -3 } } } \right) ^{ 2 } } \ m^2$

$R=\dfrac { { \rho l } }{ A } =\dfrac { { 5\times { { 10 }^{ -7 } }\times 0.5 } }{ { \pi \times 0.2\times 0.2\times { { 10 }^{ -6 } } } }$

$R =1.99 \ \Omega$

Twelve wires, each of resistance 'r' are joined to from a skeleton cube. Find the effective resistance between two diagonally opposite ends of the cube.

Hence,

${R_{AB}} = \frac{{5r}}{6}$

1380131_1246036_ans_ee9bca8ee8154dd48b74a0e00c82a7e6.jpg

A common device which can supply small electric current continuously for a sufficient time

A battery is a device which can supply small electric current continuously for a sufficient time

An electric current is a flow of electric charge. In an electric circuit, this charge is often carried by moving electrons in a wire. A battery and cells provides the potential difference between two terminals for the flow of current.

Classify the following into good conductors, resistors and insulators :
Rubber, Mercury, Nichrome, Polythene, Aluminium, Wood, Manganin, Bakelite, Iron, Paper, Thermocol, Metal coin.

Good conductor- Aluminium, Iron, Manganin,Nichrome, metal coin

Resistors - Mercury

Insulator- polythene, wood, bakelite, paper, thermocol, rubber

Give S.I. unit and commercial unit of energy

Commercial Unit of Energy. The SI unit joule is too small to express very large quantities of energy. Hence we use a bigger unit called kilowatt hour (kWh) to express energy.

$1 kWh$ is the amount of energy consumed by an electrical gadget in one hour at the rate of

$1000\ J/s$ or

$1kW.$

Commercial Unit Of Energy

$= kWh$

SI Unit Of Energy = Joule (J)

$1 kWh = 1 kW \times 1 h$

$=> 1000 W\ \times 3600 s$

1 watt - second = 1 joule

$1 kWh = 3600000 J$

A voltage of 30 V is applied across a colour coded carbon resister with first, second and third rings of blue, black and yellow colours. What is the current flowing through the resistor?

Value of first digit (blue ring)=6
Value of second digit (black ring)=0
Multiplier (yellow ring)=

$10^4$

$\therefore$ Resistance ,

$R=60 \times 10^4 \Omega,voltage,V=30V$

Current

$I=\dfrac{V}{R}=\dfrac{30}{60 \times 10^4}=0.5 \times 10^{-4}A=0.05 m$

An electric heater which is connected to a 220 V supply line has two resistance coils. A and B of $24 \Omega$ resistance each. these coils can be used separately (one at a time), in series or in parallel. calculate the current drawn when:
Only one coil A is used
coils A and B are used in series

➡When Series

R = R1 + R2

R = 24 + 24

R = 48 ohm

V = R × I ( Ohm's law )

I = V / R

I = 220 / 48 = 4.583333333 Amp

➡When parallel

1 / R = 1 / 24 + 1 / 24

1 / R = 2 / 24

1 / R = 1 / 12

R = 12 ohm

I = V / R

I = 220 / 12

I = 18.33333333 Amp

A uniform wire with a resistance of $27 \Omega$ is divided into three equal pieces and then they are joined in parallel . find the equivalent resistance of the parallel combination.

R=27 ohm
after dividing them into equal parts,
resistance of each part =9 ohm
when connected in parallel,
1/9+1/9+1/9=1/R
3(1/9)=1/R
1/3=1/R
R=3 ohm

If two resistors with resistance 5 $\Omega$ and 10 $\Omega$ respectively , are connected to a battery of emf 6V, then determine the strength of the total resistance and current in the circuit when connected in
(a) parallel combination and (b) series combination

Given,

Two resistors with resistances,

$R_1=5\ \Omega$

$R_2=10\ \Omega$

Voltage,

$V=6\ V$

In parallel combination,

Equivalent resistance,

$R_p$

$\dfrac{1}{R_p}=\dfrac{1}{R_1}+\dfrac{1}{R_2}$

$R_p=\dfrac{R_1R_2}{R_1+R_2}$

$R_p=\dfrac{5\times 10}{5+10}=3.33\,\Omega$

Current,

$I_p=\dfrac {V}{R_p}$

$I_p=\dfrac {6}{3.33}$

$I_p=1.80\,A$

In series combination,

$R_s=R_1+R_2$

$R_s=5+10$

$R_s=15\,\Omega$

Current,

$I_s=\dfrac {V}{R_s}$

$I_s=\dfrac{6}{15}$

$I_s=0.4\,A$

An electric bulb is rated $220\ V$ and $100\ W$ . When it is operated on $110\ V$ , what will be the power consumed?

Given,

Power rating of the electric bulb,

$P=100\ W$

Voltage supplied,

$V=220\ V$

Let the resistance of the bulb be

$R$

We know,

Power consumed,

$P = \dfrac{{V}^{2}}{R}$

$\implies R=\dfrac{V^2}{P}$

$R=\dfrac{220^2}{100}$

$R=484\ V$

When it is operated on voltage,

$V'= 110\ V$

Power consumed,

$P = \dfrac{{V'}^{2}}{R}$

$P = \dfrac{{110}^{2}}{484}$

$P=25\ W$

What is an electric fuse? State its purpose in the household electrical circuit?

FUSE: It is a safety device which is used to limit the current in an electric circuit.
PURPOSE OF FUSE IN HOUSE HOLD CIRCUIT: It safe guards the circuit and the appliances connected in the circuit from being damaged if the current in the circuit exceeds the specified value due to voltage fluctuation or short circuiting.

(i)What is an electric fuse?
(ii)An electric fuse of rating 3 ampere is connected in a circuit in which an electric iron of power 1 kw is connect which operates at 220 V. What would happen, Explain?

$Part (a):$ Fuse is a piece of wire of a material with a very low melting point. When a high current flows through the circuit due to overloading or short circuit, the wires gets heated and melts. As a result, the circuit is broken and current stops flowing

$Part (b):$

Given

$Power, P = 1kW = 1000 W$

and $Voltage, V = 220 V$

we know,

Power $, P =VI$

$I =\frac{P}{V} =\frac{1000}{220} = 4.54 A$

Hence, the current drawn by the electric circuit is $4.54A$ , which exceeds the safe limit $(3 A)$ of the circuit. Fuse element of the electric fuse will melt and break the circuit

How does the heat $H$ produced by a current passing through a fixed resistance wire depend on the magnitude of current $I$ ?

The heat produced by the flow of current through a resistor is given as:

$H=I^2\ R\ t$

Here,

$H$ is the heat produced,

$I$ is the current and

$t$ is the time for which the current flows.

With the help of labelled circuit diagram derive a formula to find combined resistance (R) when two or more resistance $(R_1,R_2,R_3.....)$ are connected in parallel taking symbols potential difference $(V)$ and current $(I)$ .

Consider two resistors connected in parallel, as shown in figure. It is clear, from the figure, that the potential drop 'V' across the two resistors is the same. In general, however, the currents I1 and I2 which flow through resistors R1 and R2, respectively, are different. According to Ohm's law, the equivalent resistance Req between B and A is the ratio of the potential drop V across these points and the current I which flows between them. This current must equal the sum of the currents I1 and I2 flowing through the two resistors, otherwise charge would build up at one or both of the junctions in the circuit.

Thus,

I=I1+I2
It follows that
1/Req= I/V= (I1+I2)/V= I1/V +I2/V
giving,
1/Req=1/R1 + 1/R2

Here, we have made use of the fact that the potential drop V is common to all three resistors. Clearly, the rule is
'The reciprocal of the equivalent resistance of two resistances connected in parallel is the sum of the reciprocals of the individual resistances. '

1201739_1508490_ans_30357e46ad274b079f0fffe0d2b5b084.png

1201739_1508490_ans_30357e46ad274b079f0fffe0d2b5b084.png

An electric lamp of resistance $20 \, \Omega$ and a conductor of resistance $4\, \Omega$ are connected to a $6\, V$ battery as shown in the circuit. Calculate:
(a) The total resistance of the circuit
(b) The current through the circuit
(c)The potential difference across the (i) Electric lamp and (ii) Conductor, and
(d) Power of the lamp.

a). the total resistance of the circuit =

$20\Omega +4 \Omega = 24\Omega$

b). The current through the circuit= current through the bulb= current through the conductor.

The current through the circuit=

$\dfrac{voltage\ applied}{Total\ Resistance}.$

the current through the circuit

$=\dfrac{6}{24}amp$

The current through the circuit =

$0.25amp$

c). The potential difference across the lamp and the conductor.

(i) The potential difference across the electrical lamp

$=0.25\times 20= 5 volt$

(ii) The potential difference conductor

$=0.25 \times 4= 1 \ volt$

d). Power can be calculated as

$\dfrac{voltage \ across\ the \ lamp}{current\ through\ the\ lamp}$

power=

$\dfrac{5}{0.25}\rightarrow 20 W$ .

Two Metallic resistors are connected first in series and then in parallel across a d c supply. Plot of I-V graph is shown for the two cases. Which one represents a parallel combination of the resistors and why?

Line A represents the parallel combination.
Reason: At a given potential difference V,current in the combination A
Is more than in the combination B.
i.e.,

$l_A > l_B$
Since

$R_A=\dfrac{V}{l_A}$ and

$R_B=\dfrac{V}{l_B}$

$\Rightarrow R_A < R_B$
1663416_1534395_ans_1874d9c9186b49578945d56754d2471e.png

Define resistivity.

Resistance of a conductor is given by

$R=\rho \dfrac{L}{A}$

$L$ is length of conductor

$A$ is area of cross section.

And

$\rho$ is resistivity of material of conductor and it is the measure of opposition offered to current flow by that conductor material.

(i)Draw a circuit diagram to show three resistors connected in parallel with a cell and a key. (ii)Write formula for the equivalent resistance $R$ of these combination if the individual resistance of the resistance are $R_1, R_2$ and $R_3$ .

$Part(i)$ $\textbf{Draw the required circuit diagram.}$

$Part(ii)$

$\textbf{Calculate formula for the three resistors connected in parallel}$

Given that,

First resistance =R₁

Second resistance =R₂

Third resistance = R₃

According to figure, Three resistors connected in parallel with a cell and key.

The equivalent resistor R will be

$\frac{1}{{{{\mathop{\rm R}\nolimits} _{}}}} = \frac{1}{{{{\mathop{\rm R}\nolimits} _1}}} + \frac{1}{{{{\mathop{\rm R}\nolimits} _2}}} + \frac{1}{{{{\mathop{\rm R}\nolimits} _3}}}$

$\frac{1}{{{{\mathop{\rm R}\nolimits} _{}}}} = \frac{{{R_2}{R_3} + {R_1}{R_3} + {R_1}{R_2}}}{{{R_1}{R_2}{R_3}}}$

${{\mathop{\rm R}\nolimits} _{}} = \frac{{{R_1}{R_2}{R_3}}}{{{R_1}{R_2} + {R_2}{R_3} + {R_1}{R_3}}}$

Hence, The formula for the equivalent resistor R of this combination is :

${{\mathop{\rm R}\nolimits} _{}} = \frac{{{R_1}{R_2}{R_3}}}{{{R_1}{R_2} + {R_2}{R_3} + {R_1}{R_3}}}$

Three bulbs, each having a resistance of $180 \Omega$ , are connected in parallel to an ideal battery of emf $60 V$ . Find the current delivered by the battery when
(a) all the bulbs are switched on,
(b) two of the bulbs are switched on and
(c) only one bulb is switched on.

a) All are swtched on

$\frac{1}{R_{eff}}$ =

$\frac{1}{180}$ +

$\frac{1}{180}$

$R_{eff} = \dfrac{180}{3} = 60\Omega$

$V=IR_{eff}$

$60=i \times 60$

$i = 60/60 = 1A$

b) When two bulbs are switched on only two resistors are considered so similar to case (a) calculation

$R_{eff} = \dfrac{180}{2} = 90\Omega$

$V=iR_{eff}$

$60=i\times 90$

$i = 60/90 = 0.67A$

c) When only one bulb is switched on then only one resistor considered

$R_{eff} = 180\Omega \Rightarrow i = \frac{V}{R_{eff}}=60/180 = 0.33A$

1547978_1721748_ans_6a5aabf4572c46b793b65390090f8b90.png

Consider the circuit shown in figure. Find the current through the $10 \Omega$ resistor when the switch $S$ is (a) open (b) closed.

a) When

$S$ is open, the resistors will be connected in series, as the current will flow through the

$20\Omega$ resistor.

$R_{eq} = (10 + 20)\Omega = 30\Omega$

The current flowing through the circuit,

$i=\left(\dfrac{3}{30}\right)A=0.1A$

b) When

$S$ is closed,current will flow through the switch that offers the least resistive path. So, the

$20\Omega$ resistance will be ineffective in this case. The effective resistance in this case,

$R_{eq} = 10\Omega$

$i = \left(\dfrac{3}{10}\right)A = 0.3A$

1549948_1721757_ans_831ab934aa2c4779b9df785b9e0d32c5.png

A resistor $R_1$ consumes electrical power $P_1$ when connected to e.m.f. e. when resistor $R_2$ is connected to the same e.m.f. it consumes electrical power $P_2$ in terms of $P_1$ and $P_2$ what is the total electrical power consumed when they are both connected to this e.m.f. source.
a. in parallel
b. in series ?

$P_1= \epsilon^2/R_1, so R_1 = \epsilon^2 /P_1$

$P_2 = \epsilon^2 /R_1 so R_2 = \epsilon^2 /P_2$
a. When the resistors are connected in parallel to e.m.f. , the voltage across each resistor and the power dissipated by each resistor are the same as if only one resistor was connected.

$P_{tot} =P_1 +P_2$
b. When the resistors are connected in series, the equivalent resistance is

$R_{eq} = R_1 +R_2$

$P_{tot} = \dfrac { \epsilon^2}{R_1+R_2} = \dfrac { \epsilon^2}{ \epsilon^2 /P_1 + \epsilon^2 / P_2 }= \dfrac {P_1P_2}{P_1 +P_2 }$

Find the equivalent resistance of the network shown in figure between the points $a$ and $b$ .

Given,

Three resistors of resistance

$r$ are connected in parallel.

Equivalent resistance of three resistors connected in parallel

$\dfrac {1}{R_{ab}}=\dfrac {1}{R_{1}}+\dfrac {1}{R_{2}}+\dfrac {1}{R_{3}}$

$\dfrac {1}{R_{ab}}=\dfrac {1}{r}+\dfrac {1}{r}+\dfrac {1}{r}$

$\dfrac {1}{R_{ab}}=\dfrac {3}{r}$

$R_{ab}=\dfrac r3$

2113608_1721753_ans_b22b53c4561846d88f9115ef8928e662.png

(a) What is meant by the "resistance of a conductor" ? Write the relation between resistance, potential difference and current.
(b) When a $12 \,V$ battery is connected across an unknown resistor , there is a current of $2,5 \,mA$ in the circuit. Calculate the value of the resistance of the resistor.

(a) The property of a conductor due to which it opposes the flow of current through it is called resistance of the conductor.
Work done = Potential difference

$\times$ charge moved.

(b)

$V = 12 \,volt$

$I = 2.5 \times 10^{-3} \,A$

We know that

$V = IR$

$R = V/I$

$R = 12 / (2.5 \times 10-3)$

$R = 4.8 \times 10^3 \,ohm = 4800 \,ohm.$

(a) Give two examples of substances which are good conductors of electricity. Why do you think they are good conductors of electricity ?
Calculate the resistance of a copper wire $1.0 \,km$ long and $0.50 \,mm$ diameter if the resistivity of copper is $1.7 \times 10^{-8} \,\Omega m.$

(a) Silver and copper are good conductors of electricity because they have free electrons available for conduction.

(b)

$l = 1 \,km = 1000 \,m$

$r = \dfrac {d}{2} = \dfrac {0.5}{2}mm = 0.25 \,mm = 0.25 \times 10^{-3} m$

$\rho = 1.7 \times 10^{-8} \,\Omega m$

$R = \rho \dfrac {l}{A} = \rho \dfrac {l}{\pi r^2}$

$R = 1.7 \times 10^{-8} \times \dfrac {1000}{3.14 \times (0.25 \times 10^{-3})^2} = 86.6 \,\Omega$

Distinguish between good conductors , resistors and insulators . Name two good conductors , two resistors and two insulators.

Those substances which have very low electrical resistance are called as good conductors.
E.g., Copper and aluminium.

Those substances which have comparatively high resistance than conductors are known as resistors .

E.g., nichrome and manganin.

Those substances which have infinitely high electrical resistance are called insulators.
E.g., rubber and wood.

A resistance of $20 \,ohms$ has a current of $2$ amperes flowing in it. What potential difference is there between its ends ?

$R = 20 \,ohms$

$I = 2 \,amp$

We know that

$V = IR$
Thus ,

$V = 2 \times 20$

$V = 40 \,V$

Which of the following equation shows the correct relationship between electrical units ?
$1 \,A = 1 \,C/s$ or $1 \,C = 1 \,A/s$

Current is denied as the rate of flow of charge through the conductor. So, the correct relation is:

$I=\dfrac{q}{t}$ , where

$q$ is the charge and

$t$ is the time.

The unit of charge is coulomb (C) and the unit of current is ampere (A),

$\therefore$

$1 \,A = 1 \,C/s$

(a) In which direction does conventional current flow around a circuit ?
(b) In which direction do electron flow ?

(a) Conventional current flows from positive terminal of a battery to the negative terminal, through the outer circuit.

(b) Electrons flow from negative terminal to positive terminal of the battery (opposite to the direction of conventional current).

(a) Give one example to show how the resistance depends on the nature of material of the conductor.
(b) Calculate the resistance of an aluminium cable of length $10 \,km$ and diameter $2.0 \,mm$ if the resistivity of aluminium is $2.7 \times 10^{-8} \,\Omega m.$

(a) If we take two similar wires of same length and same diameter , one of copper metal and other of nichrome alloy , we will find that the resistance of nichrome wire is about

$60$ times more than that of the copper wire. This shows that the resistance depends on the nature of material of the conductor.

(b)

$l = 10 \,km = 10000 \,m$

$d = 2 \,mm$

$r = 1 \,mm = 10^{-3} \,m$

$\rho = 2.7 \times 10^{-8} \,\Omega m$

$R = \rho \dfrac {l}{A}$

$= 2.7 \times 10^{-8} \times \dfrac {10000}{3.14 \times (10^{-3})^2}$

$= 0.859 \times 10^2 \,\Omega$

$\approx 86 \,\Omega$

In the circuit given below :
(a) What is the combined resistance ?
(b) What is the p.d. across the combined resistance ?
(c) What is the p.d. across the $3 \,\Omega$ resistor ?
(d) What is the current in the $3 \,\Omega$ resistor ?
(e) What is the current in the $6 \,\Omega$ resistor ?

Total current flowing through circuit ,

$I = 6 \,A$

$R_1 = 3 \,ohm , R_2 = 6 \,ohm$

(a) Combined resistance

$R$ is

$1/R = 1/3 + 1/6$

$1/R = 3/6$

$R = 2 \,ohms$

(b) p.d. across the combined resistance

$V= IR = 6 \times 2 = 12 \,V$

$= 12 \,V.$

(d) Current flowing through the 3 ohm resistor

$I_{3\ \Omega}= V/R_1 = 12/3 = 4 \,A$

(e) Current flowing through the 6 ohm resistor

$I_{6\ \Omega}= V/R_2 = 12/6 = 2 \,A$

The electrical resistivities of four materials A, B, C, and D are given below :
Which material is : (a) good conductor (b) resistor (c) insulator, and (d) semiconductor?

$A=110\times 10^{-8}\ \Omega m$
$B=1.0\times 10^{10}\ \Omega m$
$C=10.0\times 10^{-8}\ \Omega m$
$D=2.3\times 10^{3}\ \Omega m$

(a) Good conductors have low resistivity, Hence its resistivity will be

$C= (10 \times 10^{-8} \,ohm-m)$

(b) Resistor: The material which is used for resistance which have resistivity higher than conducting material Hence its resistivity will be

$= A (110 \times 10^{-8} \,ohm-m)$

(c) Insulators have high value of resistivity because they have less number of free electrons, Hence its resistivity will be

$B= (1 \times 10^{10} \,ohm-m)$

(d) Semiconductor: Resistivity of semiconductor always found between higher than conductor but more than insulator, Hence its resistivity will be =

$D (2.3 \times 10^3 \,ohm-m)$

Which of the following resistor arrangement , A or B , has the lower combined resistance ?

Resistance of arrangement A is

$10 \,ohm.$
Combined resistance of arrangement B is calculated as follows:

$1 / R = 1 / 10 + 1 / 1000 = (100 + 1) / 1000$

$R = 1000 / 101 = 9.9 \,ohm$

Therefore , arrangement B has lower combined resistance.

How should the two resistances of $2$ ohms each be connected so as to produce an equivalent resistance of $1$ ohm ?

Since the resultant resistance is less than the individual resistances , so the resistances should be connected in parallel.

If five resistances , each of value $0.2 \,ohm ,$ are connected in series , what will be the resultant resistance ?

As per the law of combination of resistances in series,

$R = R_1 + R_2 + R_3 + R_4 + R_5$

$R = 0.2 + 0.2 + 0.2 + 0.2 + 0.2$

$= 1 \,ohm$

State the law of combination of resistances in parallel.

According to the law of combination of resistance in parallel , the reciprocal of the combined resistance of a number of resistances connected in parallel is equal to the sum of the reciprocals of all the individual resistances.

If $3$ resistances of $3$ ohm each are connected in parallel , what will be their total resistance ?

The three resistances are equal.

So,

$R_1 = R_2 = R_3 = 3 \Omega$

$\dfrac {1}{R} = \dfrac {1}{R_1} + \dfrac {1}{R_2} + \dfrac {1}{R_3}$

$= \dfrac {1}{3} + \dfrac {1}{3} + \dfrac {1}{3} = \dfrac {3}{3} = 1$

$\therefore R = 1 \,\Omega$

The figure given below shows an electric circuit in which current flows from a $6 \,V$ battery through two resistors.
(a) Are the resistors connected in series with each other or in parallel ?
(b) For each resistor , state the p.d. across it.
(c) The current flowing through from the battery is shared between the two resistors. Which resistor will have bigger share of the current ?
(d) Calculate the effective resistance of the two resistors.
(e) Calculate the current that flows from the battery.

$V = 6 \,V$

$R_1 = 2 \,ohms$

$R_2 = 3 \,ohms$

(a) Resistors are connected in parallel.

(b) p.d. across each resistor is same and is equal to

$6 \,V.$

(d) Effective resistance = R

$1/R = 1/2 + 1/3$

$1/R = 5/6$

$R = 1.2 \,ohms$

(e) Current flowing through battery ,

$I = \dfrac VR = \dfrac{6}{1.2} = 5 \,amps$

In the circuit shown below , the voltmeter reads $10 \,V.$
(a) What is the combined resistance ?
(b) What current flows ?
(c) What is the p.d. across $2 \,\Omega$ resistor ?
(d) What is the p.d. across $3 \,\Omega$ resistor ?

Given,

$V=10 \ V$

$R_1 = 2 \,\Omega , R_2 = 3 \,\Omega$

(a) Combined resistance ,

$R_{eq} = R_1+R_2 = 2 + 3 = 5 \,\Omega$

(b) Using Ohm's law

$V=iR$

$I = \dfrac {V}{R_{eq}} = \dfrac {10}{5} = 2 \,A$

$2 \,\Omega$ resistor

$= I \times R_1 = 2 \times 2 = 4 \,V$

(d) p.d. across

$3 \,\Omega$ resistor

$= I \times R_2 = 2 \times 3 = 6 \,V$

A potential difference of 6 V is applied to two resistors of 3 $\Omega$ and 6 $\Omega$ connected in parallel. Calculate : (a) the combined resistance (b) the current flowing in the main circuit (c) the current flowing in the 3 $\Omega$ resistor

$V = 6 \,V$

$R_1 = 3 \,ohm , R_2 = 6 \,ohm$ (in parallel)

(a) Combined resistance,

$1/R = 1/R_1 + 1/R_2$

$1/R = 1/3 + 1/6 = 3/6 = 1/2$

$R = 2 \,ohm$

(b) Current flowing in the main circuit ,

$I = V/R = 6/2 = 3 \,A$

$3$ ohm resistor

$= V / R_1 = 6/3 = 2 \,A$

(a) Explain with the help of a labelled circuit diagram, how you will find the resistance of a combination of three resistors of resistances $R_1 , R_2$ and $R_3$ joined in parallel.
(b) In the diagram shown below , the cell and the ammeter both have negligible resistance . The resistors are identical.
With the switch K open , the ammeter reads $0.6 \,A$ . What will be the ammeter reading when the switch is closed ?

Suppose total current flowing in the circuit is I. then the current passing through the resistance

$R_1$ will be

$I_1.$ current passing through resistance

$R_2$ will be

$I_2$ and current passing through resistance

$R_3$ will be

$I_3.$
Total current

$= I = I_1 + I_2 + I_3$
Let resultant resistance of this parallel combination is R. By applying the ohm's law to the whole circuit , we get that

$I_1= V/R_1$

$I_2 = V/R_2$

$I_3 = V/R_3$
Putting these eqs. in the above once , we get

$V/R = V/R_1 + V/R_2 + V/R_3$

$1/R = 1/R_1 + 1/R_2 + 1/R_3$
If two resistance are connected in parallel, then the resultant resistance will be

$1/R = 1/R_1 + 1/R_2 + 1/R_3$

(b) If switch is open then only upper two resistances (connected in parallel) are in the circuit.
Effective resistance is

$1/R_{eq} = 1/R + 1/R = 2/R$

$R_{eq} = R/2$
So the current

$= I = V/(R/2) = 0.6 \,A$ (given)

$V/R = 0.3 \,A$
When the switch closes , the third resistance also comes in the circuit. The effective resistance of the circuit becomes

$R/3$
Hence , current

$= I = V/(R/3) = 3(V/R) = 3 \times 0.3 = 0.9 \,A$

1605405_1761182_ans_11dda15a0070407cb156da9f2dd689f7.png

If the current passing through a conductor is doubled , what will be the change in heat produced ?

Heat produced is directly proportional to the square of current.
If current

$I$ is doubled , heat

$H$ will be four times.

A resistor of 8 ohms is connected in parallel with another resistor X. The resultant resistance of the combination is $4.8\,ohms$ . What is the value of the resistor X ?

Given :
A resistor of 8 ohm is connected in parallel with a resistor of X.
And resultant is

$4.8.$
Then X = ?

We know that for parallel case

$1/R = 1/R_1 + 1/X$

$1/4.8 = 1/8 + 1/x$

$1/4.8 - 1/8 = 1/x$
After solving we get that

$X = 12 \,ohms$

(a) With the help of a diagram , derive the formula for the resultant resistance of three resistors connected in series.
(b) For the circuit shown in the diagram given below :
Calculate :
(i) the value of current through each resistor.
(ii) the total current in the circuit.
(iii) the total effective resistance of the circuit.

(a)

Fig shows three resistances

$R_1 , R_2$ and

$R_3$ connected in series with a battery of V volts.
Let the p.d. across

$R_1 , R_2$ and

$R_3$ is

$V_1 , V_2$ and

$V_3$ respectively.

$s.t. \,V = V_1 + V_3 + V_2$ --------------(1)
Let the equivalent resistance be R and the current flowing through the whole circuit is I.
By ohm's law,

$\dfrac {V}{I} = R$

$V = I \times R$ ------------------(2)
Applying ohm's law to both

$R_1 , R_2$ and

$R_3.$

$V = I \times R_1$ ------------------(3)

$V = I \times R_2$ ------------------(4)

$V = I \times R_3$ ------------------(5)

From eqs. (1) , (2) , (3) , (4) and (5) , we get

$I \times R = I \times R_1 + I \times R_2 + I \times R_3$

$I \times R = I \times (R_1 + R_2 + R_3)$

$R = R_1 + R_2 + R_3$

(b)

Let

$5 \,\Omega = R_1 , 10\,\Omega = R_2 , 30\,\Omega = R_3$

(i) Current through

$R_1 = I_1 = \dfrac {V}{R_1} = \dfrac {6}{5} = 1.2 \,A$

Current through

$R_2 = I_2 = \dfrac {V}{R_2} = \dfrac {6}{10} = 0.6 \,A$

Current through

$R_3 = I_3 = \dfrac {V}{R_3} = \dfrac {6}{30} = 0.2 \,A$

(ii) Total current in the circuit

$= 1.2 + 0.6 + 0.2 = 2 \,A$

(iii) Effective resistance

$R$ is given as

$\dfrac {1}{R} = \dfrac {1}{R_1} + \dfrac {1}{R_2} + \dfrac {1}{R_3}$

$= \dfrac {1}{5} + \dfrac {1}{10} + \dfrac {1}{30}$

$\dfrac {6 + 3 + 1}{30} = \dfrac {10}{30}$

$R = \dfrac {30}{10} = 3 \,\Omega$

1605107_1760907_ans_b453e14709da4d2daa76d84a30d10c1e.png

(a) With the help of a circuit diagram , deduce the equivalent resistance of two resistances connected in series.
(b) Two resistances are connected in series as shown in the diagram:
(i) What is the current through the 5 ohm resistance ?
(ii) What is the current through R ?
(iii) What is the value of R ?
(iv) What is the value of V ?

(a) Fig. shows two resistances

$R_1$ and

$R_2$ connected in series with a battery of V volts.
Let the p.d. across

$R_1$ is

$V_1$ and the p.d. across

$R_2$ is

$V_2.$

$s.t. \,V = V_1 + V_2$ -----------------------(1)
Let the equivalent resistance be

$R$ and current flowing through the whole circuit is

$I.$
By Ohm's law,

$\dfrac {V}{I} = R$

$V = I \times R$ -------------------------------(2)
Applying ohm's law to both

$R_1$ and

$R_2,$

$V_1 = I \times R_1$ -----------------------(3)

$V_2 = I \times R_2$ ----------------------(4)
From eqs. (1) , (2) , (3) and (4) , we get

$I \times R = I \times R_1 + I \times R_2$

$I \times R = I \times (R_1 + R_2)$

$R = R_1 + R_2$

(b)
(i) Current through

$5 \,\Omega$ resistor =

$\dfrac {10}{5} = 2 \,A$

(ii) Since

$5 \,\Omega$ resistor and R are connected in series , so same current through them. So, current through

$R = 2\,A$

(iii)

$V = IR$

$R = \dfrac {V}{I} = \dfrac {6}{2} = 3 \,\Omega$

(iv)

$V = 10 + 6 = 16 \,V$

1605059_1760900_ans_9a8b27e3c0dc4b7896076ede9e48ad26.png

State whether an electric heater will consume more electrical energy or less electrical energy per second when the length of its heating element is reduced. Give reasons for your answer.

By reducing the length of element the resistance will decrease.

Power is inversely proportional to resistance. So, this will result in more consumption of energy.

(a) With the help of a circuit diagram , obtain the relation for the equivalent resistance of two resistances connected in parallel.
In the circuit diagram shown below , find :
(i) Total resistance
(ii) Current shown by the ammeter A

Suppose total current flowing the circuit is I. Then the current passing through resistance

$R_1$ will be

$I_1$ and the current passing through resistance

$R_2$ will be

$I_2$
Total current

$= I = I_1 + I_2$
Let resultant resistance of this parallel combination is R. By applying ohm's law to the whole circuit, we get that

$I = V/R$
Since the potential difference across both the resistance is the same. So applying the ohm's law to each resistance we get that

$I_1 = V/R_1$

$I_2 = V/R_2$
Putting these eq in the above one, we get that

$V/R = V/R_1 + V/R_2$

$1/R = 1/R_1 + 1/R_2$
If two resistance are connected in parallel than the resultant resistance will be

$1/R = 1/R_1 + 1/R_2$

(b)
(i) Total resistance = R

$1/R = 1/R_1 + 1/R_2$

$R_2 = 3 + 2 = 5 \,ohms$

$R_1 = 5 \,ohms$

$1/R = 2/5$

$R = 2.5 \,ohms$

(ii) Current through the circuit

$I = V/R = 4/(2.5)$

$= 1.6 \,amps$

1605196_1761162_ans_c150acd4e78c45c38e1d12913a2b8c3f.png

The resistors $R_1 , R_2 , R_3$ and $R_4$ in the figure given below are all equal in value.
What would you expect the voltmeters A , B and C to read assuming that the connecting wires in the circuit have negligible resistance ?

The equivalent resistance of the series connection is:

$R_{eq} = R + R + R + R = 4R \,ohm$

Total current in the circuit ,

$I = V/R = 12/4R = 3/R$

Reading of voltmeter A = Voltage across

$R_1 = I \times R_1 = 3/R \times R = 3 \,V$

Reading of voltmeter B = Voltage across

$R_2 = I \times R_2 = 3/R \times R = 3 \,V$

Reading of voltmeter C = Voltage across the series combination of

$R_3$ and

$R_4 = I \times (R_3 + R_4) = 3/R \times 2R = 6 \,V$

Four resistances of 16 ohms each are connected in parallel. Four such combinations are connected in series.
What is the total resistance ?

Resultant resistance of a parallel combination of four 16 ohm resistance is

$1/R = 1/16 + 1/16 + 1/16 + 1/16 = 4/16$

$R = 4 \,ohm$

Four such combinations are connected in series ,

so total resistance

$= 4 + 4 + 4 + 4 = 16 \,ohm$

Name two effects produced by electric current.

Two effects of produced by electric current are :
(a) Heating effect of current
(b) Magnetic effect of current.

$(a)$ What current is taken by a $3\ kW$ electric geyser working on $240\ V$ mains?
$(b)$ What size fuse should be used in the geyser circuit?

$(a)$ Given,

Power rating of the electrical geyser,

$P=3\ kW=3000\ W$

Voltage supply,

$V=240\ V$

Current,

$I$

We know,

$P=V\times I$

$\implies I=\dfrac PV$

$I=\dfrac{3000}{240}$

$I=12.5\ A$

$(b)$ Current,

$I =12.5\ A \approx 13\ A$

$13\ A$ fuse should be used in the geyser circuit.

What is the usual capacity of an electric fuse used in the lighting circuit.

The lighting circuit uses a fuse of capacity

$5A$ . Fuses are deliberately the weak link in a circuit and will 'blow' if an appliance or cable draws too much power, to prevent a fire occurring.

Three appliances are connected in parallel to the same source which provides a voltage of $220\ V$ . A fuse connected to the source will blow if the current from the source exceeds $10\ A$ . If the three appliances are rated at $60\ W, 500\ W$ and $1200\ W$ at $220\ V$ , will the fuse blow?

$P_1=60\ W$

$P_2=1200W$

$P_3=500W$
Fuse rating

$=10A$

$V=220V$

We have,

$P=VI$

Total power

$=60+1200+500=1760W$

Therefore,

$1760=220\times I$

$I=8A$
The required current is

$8A$ and fuse rating is

$10A$ . SO, all the appliances will work normally and the fuse will not blow.

Which of the following circuits will still be dangerous even if the fuse blows off and electric iron stops working during a short circuit?

Circuit

$A$ is not dangerous after the fuse blows because the fuse is fitted in the live wire, but circuit

$B$ is dangerous even if the fuse blows because the fuse is in the neutral wire.

A vacuum cleaner draws a current of $2\ A$ from the mains supply. What will happen if a $13\ A$ fuse is fitted in its circuit?

$13\ A$ fuse is fitted in the circuit, it will not protect the vacuum cleaner against the very high current flowing through it. This may damage the appliance.

A vacuum cleaner draws a current of $2\ A$ from the mains supply.
What is the appropriate value of the fuse to be fitted in its circuit?

A fuse is an electrical safety device which provides protection to our circuit from large value of currents.

Hence, for a vacuum cleaner that draws

$2 \ A$ froms the mains supply. The appropriate value of the fuse to be fitted in the circuit is

$2.5A$ or

$3A$ .

What is the usual capacity of an electric fuse used
In the power, circuit, of a small house?

The fuse used in the power circuit, of a small house is

$15\ A$ .

State three factors on which the heat produced by an electric current depends. How does it depend on these factors ?

Heat produced by an electric current depends on the following factors :
(i) Heat produced is directly proportional to square of current.
(ii) Heat produced is directly proportional to resistance.
(iii) Heat produced is directly proportional to the time for which current flows.

Is electric current a scalar or vector quantity? State the standard unit of electric current.

Current is a vector because it have a magnitude and a direction. But the thing is a vector always obey the law of addition of vectors.sincecurrent doesn't obey it and it follows algebraic addition,currents a scalar. Also, current is written as the dot product of current density vector and the area vector. Cgs unit is biot.

An electric fan/ motor becomes warm when continuously used for a long time. Why?

This is because an electric fan/motor converts electrical energy into mechanical energy. So there are moving parts in both fan and motor, which means there will be friction between moving parts and it generates heat.

Present the relationship between the voltage applied across a conductor and the current flowing through it graphically.

The potential difference across a conductor is proportional to the current through the resistor.

$V \propto I$

$V = IR$ , where

$R$ is the proportionality constant, named resistance.
This is similar to the equation of a straight line passing through the origin.
So the graph between voltage and current is a straight line with a slope of R.
1628856_1773168_ans_0f0c9e65f10640c79d5fadd0aa6b7e98.bmp

1628856_1773168_ans_0f0c9e65f10640c79d5fadd0aa6b7e98.bmp

Identify the type of combination in which $R_1$ and $R_2$ are connected in the given circuit diagram. Find the equivalent resistance in the given circuit.

Given resistance

$R_{ 1 }$ and

$R_{ 2 }$ are connected in parallel combination as voltage across both resistance are same.

Equivalent resistance is given by

$\dfrac { 1 }{ R_{ eq } } =\dfrac { 1 }{ R_{ 1 } } +\dfrac { 1 }{ R_{ 2 } } \\ \Rightarrow \dfrac { 1 }{ R_{ eq } } =\dfrac { 1 }{ 3 } +\dfrac { 1 }{ 6 } \\ \Rightarrow \dfrac { 1 }{ R_{ eq } } =\dfrac { 1+2 }{ 6 } \\ \Rightarrow \dfrac { 1 }{ R_{ eq } } =\dfrac { 3 }{ 6 } \\ \Rightarrow R_{ eq }=2\Omega$

The potential difference between the terminals of an electric heater is $60\ V$ when it draws a current of $4\ A$ from the source. Find the resistance of the heater.

Given that
Potential Difference

$(V)=60\ V$
Current Drawn

$(I)=4\ A$
By ohm's Law

$V=IR$

$\Rightarrow \ 60=4\times R$

$\Rightarrow \ R=\dfrac {60}{4}=15$
Hence the resistance of the electric heater is

$15\ \Omega$

What is the commercial unit of electrical energy ? Convert it into joule . What is the other name of the commercial unit ? What is meant by a parallel combination of resistors ? Deduce an expression for the equivalent resistance when two resistors are connected in parallel.

Commercial unit of electrical energy is the unit in which all our household electrical consumptions are measured .

$1\, kWh = 1000 \, watt \times 3600$ seconds

$= 3.6 \times 10^6$

The commercial unit of current is kilowatt hour (kWh).

In parallel combination of resistors , the voltage across all the resistors are constant but the total current is divided in all the branches.

As the current is given by

$i = \dfrac{V}{R}$
So , in the circuit

$i = i_1 + i_2 + i_3$

$\dfrac{V}{R} = \dfrac{V}{R_1} + \dfrac{V}{R_2} + \dfrac{V}{R_3}$

$\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_1} + \dfrac{1}{R_3}$

Hence , in parallel connection the reciprocal of equivalent resistance is equal to the sum of reciprocal of individual resistances .

1628841_1773164_ans_8960f9c15ebc4a70bed16bd37c571f36.png

1628841_1773164_ans_8960f9c15ebc4a70bed16bd37c571f36.png

Relate $Watt$ to $Joule$ .

$1\, Watt$ is the power of an instrument that does the work at the rate of

$1\, Joule$ per second.

$1\, W$

$=$

$1\, J/s$

When a high resistance voltmeter is connected across a resistance R and its reading is $2\ V$ . An electric cell is sending the current of $0.4\ A$ in the electric circuit in which a rheostat is connected to vary the current.
Find the resistance of the resistor.

Since the voltmeter reading shows

$2\ V$ . Thus

$V=2$ .
Also, Current send by electric cell

$=0.4\ A$
Thus

$I=0.4\ A$
By Ohm's law

$V=IR$

$\Rightarrow 2=0.4\times R$

$\Rightarrow R=\dfrac {2}{0.4}=\dfrac {20}{4}=5\ \Omega$
The resistance of the resistor is

$5\ \Omega$ .
1637485_1774165_ans_05c03b1dfd604561bc0604a5ba344714.png

Heat is generated continuously in an electric heater but the temperature of its element becomes constant after some time. Why?

When the temperature of the heater becomes greater than the temperature of the surrounding, some of the heat is lost to the surroundings in the form of thermal radiations. After some time, rate at which heat is being produced becomes equal to the rate of which heat is lost. Hence, the temperature of the element becomes constant.

Why is resistance less when resistors are joined in parallel?

We know that

$\dfrac 1R= \dfrac {1}{r_1}+\dfrac{1}{r_2}$

In parallel combination of resistors, the effective area of cross-section of the conductor increases, to the resistance decreases.

How is the fuse connected in an electric circuit?

Figure shows three $20.0\Omega$ resistors. Find the equivalent resistance between points (a) A and B, (b) A and C, and (c) B and C. (Hint:
Imagine that a battery is connected between a given pair of points.)

2098581_1778841_ans_02528bfd11414d8bb898f69945988662.jpg

Two heated wires of the same dimensions are first connected in series and then in parallel to a source of supply.What will be the ratio of heat produced in the two cases?

The wires are of the same material and of the same dimensions. Therefore, both wires will have the same resistance (R).

When the wires are connected in series the equivalent resistance is

$R_1=R+R=2R$

When the wires are connected in parallel the equivalent resistance is

$R_2=\dfrac{R\times R}{R+R} = \dfrac{R^2}{2R} = \dfrac{R}{2}$

We know that the heat produced by a wire of resistance

$R'$ when it is connected across a voltage

$V$ is

$H=\dfrac{V^2}{R'}t \propto \dfrac{1}{R'}$

$\dfrac{H_{series}}{H_{parallel}}=\dfrac{R_{2}}{R_{1}}=\dfrac{R/2}{2R}=\dfrac{1}{4}$

In India, domestic power supply is at $220 \ V$ , $50 \ Hz$ ; while in USA it is $110 \ V$ , $50 \ Hz$ . Give one advantage and one disadvantage of $220 \ V$ supply over $110 \ V$ supply.

Advantage of $220 \ V$ over $110 \ V:$

Power loss due to heating effect of current in the transmission lines is low when power is transmitted at a high voltage.

Power loss

$=I^2R$ . . . (i)

We know

$P=VI \Rightarrow I=\dfrac{P}{V}$ . . . (ii)

Using (ii) in (i),

Power loss

$={\left( \dfrac{P}{V} \right)}^2R=\dfrac{P^2}{V^2}R$

Thus, for same power to be transmitted over a same conductor,

$P$ and

$R$ are constant.

So power loss

$\propto \dfrac{1}{V^2}$

$\implies$ Higher the voltage, lower the power loss due to resistance.

Disadvantage of $220 \ V$ over $110 \ V:$

High voltage is more dangerous and the cost increases due to more insulation used.

Which physical quantity remains constant when resistances are connected in parallel?

when resistances are connected in parallel Potential difference (voltage) remains constant.

A uniform wire of resistance $50\ \Omega$ is cut into $5$ equal parts. These parts are now connected in parallel. What is the value of the equivalent resistance of the combination?

Given,

A uniform wire of resistance

$50\ \Omega$ is cut into

$5$ equal parts.

Resistance of each part,

$R=\dfrac {50}{5}$

$R=10\ \Omega$

Equivalent resistance of

$5$ resistors connected in parallel,

$R_{eq}=\dfrac R5$

$R_{eq}=\dfrac {10}{5}$

$R_{eq}=2\ \Omega$

Mention the commercial unit of energy. Express it in terms of joules. Calculate the energy in joule consumed by a device of $60\,W$ in $1\,hour$ .

Commercial unit of energy is

$kiloWatt\,per\,hour$ denoted by

$kWh$ .

$1\,kWh = 1\times 1000\times \dfrac{1J}{1s}\times 60\times 60\,s = 3600000 \,J$

Energy consumed by bulb in an hour:

$60\,W = 60 \,J/s$

$1 \,hour = 60\times 60= 3600\,s$
Energy consumed in 1 hour =

$60\times 60\times 60 = 216000\,J$

In series combination which remains constant, current or voltage?

In series combination the current flowing through the circuit is constant.

Draw an appropriate schematic diagram showing common domestic circuits and discuss the importance of fuse. Why is it that a burnt out fuse should be replaced by another fuse of identical rating?

A fuse in a circuit prevents damage to the appliances and the circuit due to overloading. Otherwise, the appliances or the circuit may be damaged. When current in the circuit exceeds the value of fuse rating, the fuse wire burns due to overloading. This causes a gap in the circuit and the current stops flowing in the circuit. This is done due to the reason so that the circuit or the appliances to be connected in the circuit continue functioning without any damage in future.
1660836_1787880_ans_8ddf72bde2074c4f88bf0f5ec5466d2c.png

1660836_1787880_ans_8ddf72bde2074c4f88bf0f5ec5466d2c.png

Differentiate between watt and watt hour.

Watt (W) is the unit of power. while watt hour (Wh)is the unit of work, since

$power \times time =work$ .

What does the slope of an $I-V$ graph of a linear resistor represent?

$\Large \underline{\textbf{Concept used:}}$

$\large {\textbf{Ohm's Law}}$

According to Ohm's Law, the current passing through a conducting element is directly proportional to the voltage applied across its ends. The resistors which follow this relation are called linear or

$ohmic \ resistors$ .

$V\propto I$

$V=RI$ where

$R$ is called the resistance of the resistor.

$\Large \underline{\textbf{Answer/Explanation}}$

$I$ is plotted along Y-axis for changing voltage,

$V$ plotted along X-axis, then by

$\text{Ohm's Law}$ we can expect the curve to be linear.

Slope of the curve

$\large {=\frac{\Delta I}{\Delta V}=\frac{1}{R}=\frac{1}{Resistance}}$

$\textbf{Note:}$ Reciprocal of resistance is also called

$\textbf{conductance}$ .

Name the factors on which the heat produced in a wire depends when current is passed in it,and state how does it depend on the factors stated by you.

When current is passed in a wire,the heat produced in it depends on the three factors; i) on the amount of current passing through the wire. (ii) on the resistance of wire and (iii) on the time for which current is passed in the wire.
i) Dependence of heat produced on the current in wire: The amount of heat H produced in the wire is directly proportional to the square of current I passing through the wire, i.e.

$H \propto I^2$
ii) Dependence of heat produced on the resistance of wire: The amount of heat H produced in the wire is directly proportional to the resistance R of the wire , i.e.,

$H \propto R$
iii) Dependence of heat produced on the time: The amount of heat H produced in the wire is directly proportional to the time t for which current is passed in the wire , i.e.

$H \propto t$

Paheli does not have a night lamp in her room. She covered the bulb of her room with a towel in the night to get dim light. Has she taken the right step? Give one reason to justify your answer.

No, the step taken by the Paheli was not right. The electric bulb wastes electrical energy in the form of heat. As she covered the bulb of her room with a towel in the night to get dim light, is a total wastage of energy.

A fixed wire AB carries current I. An electron is moving parallel to the wire, in which direction does the electron tend to move?

By right-hand thumb rule, the magnetic field in the vicinity of the wire is downward perpendicular to the plane of the paper. The conventional direction of electric current is opposite to the direction of motion of the electron. By Fleming's left-hand rule, the force on the electron is away from the wire.
1692096_1787899_ans_bb7eb62e94cc4ed88f25ac6d0d2b09fb.png

1692096_1787899_ans_bb7eb62e94cc4ed88f25ac6d0d2b09fb.png

Describe an experiment to show that electricity flows only if the circuit is complete and it does not flow if the circuit is incomplete.

Take two torch bulbs A and B. Connect them to a cell through a switch as shown in fig. The bulbs are said to be connected in series. Close the switch (i,e, the circuit it completed). You will see that both the bulbs glow.

Now take out the connection of the bulb B as shown in fig. Now close the switch. You will observe that the bulb A does not glow, because the circuit is now incomplete.

Now replace the bulb B by a fused bulb and close the switch. Again you will see that the bulb A does not glow. This is because the circuit being in series, is still incomplete.

1680872_1800836_ans_3480f20d46114a4aa42ae687c445b8a8.png

Draw a $I-V$ graph for a linear resistor.

$I-V$ graph for a linear resistor

The relation between voltage

$(V)$ and current

$(I)$ is-

$V = IR$

$R$ ; resistance of the wire

1705913_1805201_ans_defb9c3dd198467394da06d107b5d2d7.png

What does the slope of $V-I$ graph for a conductor represent?

$V=IR$

slope of V-I curve,

$\dfrac{dV}{dI}=R$

Slope of

$V-I$ graph represents the Resistance.

1705522_1805167_ans_f911c346fbe0432ea5e005a1b36f629d.png

Paheli took a wire of length 10 cm. Boojho took a wire of 5 cm of the same material and thickness. Both of them connected the wires as shown in the circuit given in Figure 14.The current flowing in both the circuits is the same.
i) Will the heat produced in both the cases be equal? Explain.
ii) Will, the heat produced be the same if the wires taken by them are of equal lengths but of different thickness? Explain.

i) Total sum of heat produced in the respective wire depends upon the length (end to end distance) of the wire. So, the total sum of heat produced in both cases will not be equal.
ii) No. Thickness or width of the respective wire decides the total sum of heat produced by the wire.
No, the amount of heat produced in the wire depends upon the thickness of the wire.

Calculate the total equivalent resistance for two resistors connected in series.

If current

$I$ is drawn from the battery, the current through each resistor will also be

$I$ .

On applying Ohm's law to the two resistors separately, we further have,

$V_{1} = IR_{1}$

$V_{2} = IR_{2}$

$V=V_1+V_2$

$IR=IR_{1}+IR_{2}$

$R=R_{1}+R_{2}$

Total resistance on series

$R$

$R=R_{1}+R_{2}$

'A fuse is rated 8A'. Can it be used with an electrical appliance of rating 5kW,200 V?

$P$ =

$5000Watts$

$V=200Volts$

Power is given by

$P=VI$

The safe limit of current which can flow through the electrical appliance is

$I=P/V=5000/200=25A$ ; which is greater than 8A.So, such fuse cannot be used.

State how are the two resistors joined with a battery when potential difference is same across each resistor

The required connection will be a parallel combination because in above combination same potential will drop across both resistor as seen in diagram both resistor have same terminal ends that's why they have same potential difference.
1707316_1805823_ans_0ad327444eef42f797dcd7ba43af30cb.png

1707316_1805823_ans_0ad327444eef42f797dcd7ba43af30cb.png

State how are the two resistors joined with a battery when equivalent resistance is less than either of the two resistances

When two resistance are connected in parallel, then equivalent resistance will be lower than both the resistances.

$\underline{\text{Show:}}$

Let us consider two identical resistors with resistance

$R$ . These resistors are connected in parallel.

The equivalent resistance is

$\dfrac{1}{R_{eq}}=\dfrac{1}{R}+\dfrac{1}{R}=\dfrac{2}{R}$

$\Rightarrow R_{eq}=\dfrac{R}{2}$

$\therefore R_{eq}<R$

Name the device used to protect the electric circuits from over loading and short circuit. On what effect of current does it work?

An electric fuse is a safety device, which is used to limit the current in an electric circuit. The use of the fuse safeguards the circuit and appliances connected in that circuit from being damaged. An alloy of lead and tin is used as a material of fuse because it has low melting point and high resistivity.

The fuse works on the heating effect of current. Whenever the current exceeds a rated value in the circuit, it causes the heating of the fuse wire. As a result, it melts and breaks the circuit.

Two resistors having resistance $4 \Omega$ and $6\Omega$ are connected in parallel. Find their equivalent resistance.

Let

$R'$ be the equivalent resistance of the

$4 \Omega$ and

$6 \Omega$ resistors connected in parallel.

Then,

$R' = \dfrac{1}{4} + \dfrac{1}{6} = \dfrac{3+2}{12} = \dfrac{5}{12} \Omega^{-1}$

or,

$R' = \dfrac{12}{5} = 2.4 \Omega$

Draw a graph to show current-voltage relationship of non-ohmic resistor.

Graph to show current-voltage relationship of non-ohmic resistor.

Example: diode valve

1706063_1805295_ans_6da74ed0a48040e7b1b1e17b193aa209.png

State how are the two resistors joined with a battery when equivalent resistance is more than either of the two resistances.

When two resistance are connected in series then equivalent resistance will be higher both resistance.

Write the expression for the equivalent resistance $R$ of three resistors $R_1$ , $R_2$ and $R_3$ joined in series.

Equivalent Resistance in series is given by:

$R_{eq}=R_{1}+R_{2}+R_{3}$

State how are the two resistors joined with a battery when same current flows in each resistor

According to definition of series combination. Current flow in each resistor will be same.

Hence the two resistors should be series joined to flow same current in each resistors.

1707293_1805820_ans_cdb0a32d4ebf4b56b1ea6758851558c7.png

A resistor of $6\Omega$ is connected in series with another resistor of $4 \Omega$ . A potential difference of $20 \ V$ is applied across the combination. Calculate the current in the circuit.

$R_1=6\ ohm$

$R_{2}=4\ ohm$

$R=R_1+R_2=6+4=10\ ohm$

$V=20V$

$I=V/R=20/10=2A$

$\therefore$ The current in the circuit is

$2A$

State two differences between a conductor and an insulator of electricity.

The two differences between a conductor and an insulator of electricity are as follows:
1. Insulators do not conduct electricity but conductors allow the current to flow through them.
2. Insulators have no free electrons and offer a very high resistance to current flow but conductors have a large number of free electrons and offer a very small resistance to current flow.

n electrons flow through a cross section of a conductor in time t. If charge on an electron is e, write an expression for the current in the conductor.

The expression for current (I) in a conductor is as follows:

$I = \dfrac{ne}{t}$
Where: n is the electron flow, t is the time taken for electron flow, e is the charge on an electron e

A particular resistance wire has a resistance of $3.0$ ohm per meter. Find the total resistance of three lengths of this wire each $1.5 m$ long joined in parallel.

Resistance of

$1m$ of wire =

$3\Omega$

Resistance of

$1.5m$ of wire =

$3 \times 1.5 = 4.5\Omega$

$\dfrac{1}{R} = \dfrac{1}{4.5}+\dfrac{1}{4.5}+\dfrac{1}{4.5}$

$\Rightarrow \dfrac{1}{R} =\dfrac{3}{4.5}$

$\Rightarrow R=1.5 \Omega$

Four resistors each of $2 \Omega$ are connected in parallel. What is the effective resistance?

$R_1 = 2 \Omega$

$R_2 = 2 \Omega$

$R_3 = 2 \Omega$

$R_4 = 2 \Omega$

For parallel connection, the effective resistance would be -

$\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2} +\dfrac{1}{R_3} + \dfrac{1}{R_4}$

$=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2} ={2}$

$\therefore R=0.5 \Omega$

The following figure shows the symbols for components used in the accompanying electrical circuit . Place them at proper places and complete the circuit.
Which law can you prove with the help of above circuit ?

Ohm's law can be proved with the help of above circuit.
1748637_1823175_ans_8d9206df23d84af38fc272484923b76b.png

The following table shows current in Amperes and potential difference in Volts. What will be the nature of the graph between the current and potential difference? (Do not draw a graph)
V (Volts) I (Amp)
4 9
5 11.25
6 13.5

V ( Volts)	I (Amperes)	R (Ohms) $=\dfrac{V}{I}$
4	9	0.44
5	11.25	0.44
6	13.5	0.44

We can see the ratio

$\dfrac{V}{I}$ in all the cases is a constant value,

$0.44$ .

So this means

$V \propto I$

$\Rightarrow V=IR$

where

$R=0.44 \Omega$ is the resistance. So the graph between the current and potential difference will be a straight line passing through the origin with a constant slope of magnitude

$0.44.$ Here

$V$ should be on

$y$ -axis and

$I$ should be on the

$x$ -axis.

What is thermal effect of electric current ? Write down the names of any four appliances based on this effect.

Whenever an electric current flows through a conductor, some of it gets converted into heat energy. This is the heating effect of electric current. The greater the resistance offered by the wire to the flow of current, the greater would be the heat generated. That is why copper wires, which offer very little resistance, do not get heated up. Nichrome and tungsten offer high resistance to the flow of electric current and get heated up. Besides the material, the amount of heat generated in a wire also depends of the length and thickness of the wire. The heating effect of electric current is used in a large number of domestic and industrial appliances for example, room heater, electric kettle, geyser, toaster, electric iron, hair dryer, oven, etc.

Give reason -
In practice the unit kWh is used for the measurement of electrical energy rather than joule.

In practice, the unit kWh is used for the measurement of electrical energy, rather then joule. This is because joule is a very small unit and the energy consumption in day to day life is very large i. e. it comes in figures of

$10^6$ to

$10^8$ . Thus, to reduce the complexity of handling such large figures, a bigger unit was required. This bigger unit used for the measurement of electrical energy is kWh and is related to joule as

$1$
kWh

$= 3.6 \times 10^6$ J Hence, the energy reading commercially became simpler by using this bigger unit instead of joule

The following table shows current in Ampere and potential difference in Volts
V (Volts) I (Amp)
4 9
5 11.25
6 13.5
.
Which law will the graph prove? Explain the law.

The graph proves Ohm's law. Ohm's law states that, under constant temperature and physical conditions, the current flowing through a conductor is always proportional to the potential difference across it i.e.

$V \propto I$
Or, V = RI
Where R is a constant of proportionality called resistance of the resistor. It tends to resist the flow of charge through a conducting wire. Its Si unit is ohm .

Solve the following example.
If a 1200 W electric iron is used daily for 30 minute, how much total is consumed in the month of April?

Total electric consumed in the moth of April is

$E = P \times T\times$ ( Total days in April month)

$E$ ; used energy

$P$ ' power

$T$ ; total time

$E = 1200\times 30\times 60\times 30$

$\Rightarrow E=64.8\times 10^{6}J$

we know,

$1\ unit = 3.6\times 10^{6}J$ ,

So,

$E = 18\ units$

Solve the following problem.
The resistance of a 1m long nichrome wire is 6 $\Omega$ . If we reduce the length of the wire to 70 cm. what will its resistance be ?

Resistance of a wire is directly proportional to its length.
Let

$R_{1}\; and\; R_{2}$ be the resistance of the nichrome wire at 100 cm (1 m) and 70 cm, respectively. Therefore,

$R_{1} \propto 100$ ........... (i)
and

$R_{2} \propto 70$ ............(ii)
Dividing (ii) by (i), we get

$\dfrac{R_{2}}{R_{1}} = \dfrac{70}{100}$

$\implies\; R_{2} = \dfrac{70}{100} \times R_{1}$

Now,

$R_{1} = 6$

$\implies\; R_{2} = \dfrac{70}{100} \times 6 = 4.2\Omega$

If $10$ milliampere current flows through a wire produces a potential difference of $2.5 \,V$ between its ends, then find the resistance of wire.

Here,

$I = 10$ milliampere =

$10\times 10^{-3}A$ =

$.01 A$ ,

$V = 2.5 \, V$

Using Ohm's law:

$V=IR$

$R = \dfrac{V}{I}$

$= \dfrac{2.5 \,V}{0.01 \,A} = 250 \Omega$

With the help of suitable diagram derive equivalent resistance of resistors in series combination.

Let us make a circuit as shown in the figure. There are three resistors

$R_1, R_2$ and

$R_3$ in series combination. Current

$I$ is flowing through the circuit. The potential difference between A and B is

$V_1,$ between B and C is

$V_2$ and between C and D is

$V_3.$
According to ohms law

$V_1 = IR_1$

$V_2 = IR_2$

$V_3 = IR_3$
If V is the potential difference in battery, then

$V = V_1 + V_2 + V_3$
Or

$V = IR_1 + IR_2 + IR_3$
Or

$V = I(R_1 + R_2 + R_3)$
If R is the equivalent though the circuit then;

$V = IR$
From above equations, it is clear that

$R = R_1 + R_2 + R_3$

What is the difference between series and parallel combinations of resistors ?

1878336_1831690_ans_a15ba095a7ce4012824c04466ca1378e.jpg

When the given resistances are connected in series, which physical quantity does not change.

Current.

In series combination, there is no alternative path for current flow, So all resistances will have same value of current flowing through them.

1754119_1831804_ans_1aae0a2e3f1d4c0e97d1101491993f11.png

The combined resistance in the given circuit is $5 \Omega.$ What is the value of R ?

In parallel combination of resistances, equivalent resistance is given by,

$\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$

hence

$\dfrac{1}{5} = \dfrac{1}{10} + \dfrac{1}{R_2}$

$\therefore \dfrac{1}{R_2} = \dfrac{1}{5} - \dfrac{1}{10}$

$\therefore R_2 = 10 \ \Omega$

Find the maximum and minimum resistance by combination of three resistors of $1\Omega, 2\Omega$ and $3\Omega.$

Maximum resistance can be obtained by series combination and minimum resistance can be obtained by parallel combination.

For series combination; equivalent resistance

$R = R_1 + R_2 + R_3$

$= 1 + 2 + 3 = 6\Omega$

For parallel combination; equivalent resistance:

$\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}$

$= \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3}$

$= \dfrac{6 + 3 + 2}{6} = \dfrac{11}{6}$

Or,

$R = \dfrac{6}{11} \Omega$

For the circuit shown in the diagram given above:
Calculate total effective resistance of the circuit.

Given resistances,

$R_1=5\Omega\\R_2=10\Omega\\R_3=30\Omega$

$EMF=6\ V$

$\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}$

$= \dfrac{1}{5} + \dfrac{1}{10} + \dfrac{1}{30} = \dfrac{6 + 3 + 1}{30} = \dfrac{10}{30} = \dfrac{1}{3}$

$\therefore R_p = 3 \,Q.$

A circuit diagram is given below.
Calculate the total effective resistance.

$R_1 = 2\Omega$

$R_2 = 5\Omega$

$R_3 = 10\Omega$

Potential difference across each resistor

$V = 10 \,V$

Current through each resistor

$I_1 = \dfrac{V}{R_1} + \dfrac{10}{2} = 5 \,A$

$I_2 = \dfrac{V}{R_2} + \dfrac{10}{5} = 2 \,A$

$I_3 = \dfrac{V}{R_3} + \dfrac{10}{10} = 1 \,A$

Total current

$I = I_1 + I_2 + I_3$

$I= 5 + 2 + 1$

$I= 8 \,A$

$R_{eq}=\dfrac{V}{I}=\dfrac{10}{8}$

Give the difference in unit of electric energy and commercial unit of electric energy.

Since electric energy is a type of energy so it is also measured in

$Joules$ Hence SI unit of Electric Energy is

$Joules$ .

But there is another unit of electric energy used commercially. This commercial unit is

$kWh$ .

Which is unit of energy consumed when an appliance of

$1000W$ is used for

$1hr$ .

Give two disadvantages of the heating effect of current.

The current when flows through any conductor produces heat. This heat produced cannot be utilized at all the places, rather it damages the circuits causing high temperature and melting of other materials can cause damage. To keep the temperature low and cool the circuit, fans are set up e.g. In computers, fans are inbuilt to keep the components cool.

A battery of $10 \,V$ is connected in a circuit with $3 \Omega, 4\Omega, 6\Omega$ resistors connected in series. How much current will flow through $6\Omega$ resistor ?

Total resistance in series

$R_s = R_1 + R_2 + R_3$

$= 3 + 4 + 6$

$R_s = 13 \Omega$

$V = 10 \,V$

$I = ?$

$\therefore I = \dfrac{V}{R} = \dfrac{10}{13}$

$I = 0.76 \,Ampere.$
The current flowing through each resistor is same i.e.

$0.76 \,A.$

For the circuit shown in the diagram given above:
Calculate: the value of current through each resistor.

Data

$V = 6 \,V$

$R_1 = 5\Omega$

$R_2 = 10\Omega$

$R_3 = 30\Omega$

$I_1 = \dfrac{V}{R_1} = \dfrac{6}{5} = 1.2 \,A.$

$I_2 = \dfrac{V}{R_2} = \dfrac{6}{10} = 0.6 \,A.$

$I_3 = \dfrac{V}{R_3} = \dfrac{6}{30} = 0.2 \,A.$

Study the above circuit and answer the questions that follows:
(a) State the type of combination of the two resistors in the circuit.
(b) How much current is flowing through
(i) $10 \Omega$ and
(ii) $15 \Omega$ resistor ?
(c) What is the ammeter reading ?

(a) Two resistors are in parallel combination.
(b) Current through
(i)

$10 \Omega \rightarrow I = \dfrac{V}{R} = \dfrac{3}{10} = 0.3 \,A$
(ii)

$15 \Omega \rightarrow I = \dfrac{V}{R} = \dfrac{3}{15} = 0.2 \,A$
(c) Ammeter reading = total current flowing through the circuit

$= 0.5 \,A.,$

Give the differences between series and parallel connections of resistors.

1754184_1831863_ans_75edbea7011a435d9728152b76055d11.PNG

Calculate the area of cross section of a wire of length $1.0 \,m,$ its resistance is $23\Omega$ and the resistivity of the material of the wire is $1.84 \times 10^{-6} m$

$R = 23\Omega$

$l = 1.0 \,m$

$\rho = 1.84 \times 10^{-6} \Omega m$

$A = ?$

$A = \dfrac{\rho l}{R}$

$\therefore A = \dfrac{1.84 \times 10^{-6} \times 1}{23}$

$A= 0.08 \times 10^{-6} m^2 = 8.0 \times 10^{-8} m^2$

What is heating effect of electric current ? Find the expression for calculating 'Heat'.

When current passes through the wire conductor, the temperature of the conductor increases and heat is emitted. This is called the heating effect of electric current. Consider a current I flowing through a resistor of a resistance R. Let the potential difference across it be V. Let 't' be the time during which a charge Q flows across the resistor.
The work done in moving charge Q from one point to another through a potential difference V is VQ. Therefore the source must supply energy which is equal to VQ in time t.
Hence the power,

$P = \dfrac{VQ}{t}$ .......(1)

$\because I = \dfrac{Q}{t}$ ...........(2)

$\because P = VI$ . (3)
energy supplied in time

$'t' = P \times t$ .....(4)

$\because P \times t = VIt$
From (3) and (4) equation
The energy supplied gets utilised in the form of heat.

$H = VIt$
Ohm's law

$V = IR$
On substituting V in

$H = VIt$
We get

$H = IR \times It$

$H = I^2Rt.$

What is the commercial unit of energy? Define it.

The commercial unit of energy is kWh [kilowatt-hour]. 1 kWh is the energy used in one hour at the rate of 1000 J/s.

Give advantages and disadvantages of series and parellel connection of resistors.

Advantages and disadvantages of series connection of resistors:

$1.$ To control the current in a circuit, series connection is useful. On connecting the resistors in series the total circuit resistance increases and thus the current decreases.

$2.$ Damage of electrical appliance and short circuit can be prevented by connecting the fuse in series with the mains as well as the electrical appliance.

$3.$ Series circuits have a single path that connects the electric source to the output devices. These circuits have limited uses because if a fault occurs in one appliance or the circuit breaks, the current does not flow in the circuit and all the other components in the circuit stop working.

$4$ If the electrical appliances are connected in series, the applied voltage is divided and the appliance cannot give the light efficiently because of less voltage.
Advantages and disadvantages of parallel connection of resistors:

$1.$ Parallel circuits provide the same voltage to every source and appliance in the circuit, thus all appliances work efficiently

$2$ In a parallel circuit if one appliance is fused, the current continues to flow through the others. Hence other appliances continue to work.

$3$ The equivalent resistance in the parallel connection of resistors decreases thus, more current than one source can produce is obtained. Unless all appliances in parallel have the same voltage, current would flow from one source into the other. This will result in loss of power.

$4$ Even if we have several power sources, the voltage across a parallel circuit will remain the same. This is because the potential difference, which is the source of voltage, is split across the circuit as a whole; so increasing the power will not alter the voltage.

Bigger unit of energy is _______.

Bigger unit of energy is kilowatt hour (kW h).

$1\ kWh= 3.6\times 10^6\ J$

What would be the values of $\dfrac V I$ ratios when the potential difference is $0.5 \,V, 1.5 \,V$ and $2.0\,V$ respectively ? What conclusion do you draw from these value..

1881127_1832072_ans_67cae05259f24037acac4b5fe4ae95e4.jpg

In a residence, $4$ tube lights: each of them of $40$ W are operated daily for $5$ hours and $3$ fans: each of $120$ W are operated daily for $4$ hours. Calculate the electricity bill at Rs. $5$ per unit for September month.

Power consumed by 1 tube light

$= 40\,W$
Power consumed by 4 tube lights

$= 4 \times 40 = 160\,W = 0.16\,kW$
Energy consumed by

$4$ tube lights each day

$= 0.16 \ kW \times 5 \ h= 0.8\,kWh$

Power consumed by 1 fan

$= 120\,W$

Power consumed by 3 fans

$= 3 \times 120 = 360\,W = 0.36\,kW$

Energy consumed by

$3$ fans each day

$= 0.36 \ kW \times 4 \ h= 1.44\,kWh$

Total electrical energy consumed per day

$= 0.8 + 1.44 = 2.24\,kWh$

$\therefore$ Total number of units consumed per day

$= 2.24\,units$
Cost of one unit

$=Rs.5/unit$
Number of days in the month of September

$= 30$ days

$\therefore$ Electricity bill for the month of September

$= 2.24 \ units \times 5 \ Rs./unit \times 30 =Rs . 336$

An electrical appliance of 100 W is used for 3 h per day. Calculate the units of energy consumed per day and for a month.

Power of electrical appliance = 100 W = 0.1 kW

Time used = 3 h

Energy = power x time taken

=0.1 kW x 3 h

= 0.3 kWh

= 0.3 units

$(\because 1 kWh = 1 unit)$

The units of energy consumed per day = 0.3 units and for a month = 0.3 x 30 = 9 units.

Define $1\ kWh$ .

$1\ kWh$ is the electrical energy consumed by an electrical appliance of power

$1\ kW$ when it is used for

$1\ hour$ .
It is used as the commercial unit of electricity.

$1\ kWh =3.6\times 10^6\ J$

Show how you would connect three resistors, each of resistance $6\Omega$ , so that the combination has a resistance of
(i) $9\Omega$ ,
(ii) $4\Omega$ .

If resistor are connected n series

$6\Omega+6\Omega+6\Omega=18\Omega$ This is not correct

When they are connected in parallel

$\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=3$ This is also wrong

(i)When they are connected in parallel

Two

$6\Omega$ resistors are connected in parallel.

$Resistance=\frac{1}{\frac{1}{6}+\frac{1}{6}}$

$=\frac{6\times 6}{6+6}=3\Omega$

$3rd$ resistor of

$6\Omega$ and

$3\Omega$ are connected in series, it becomes

$6\Omega+3\Omega=9\Omega$

(ii)When they are connected in series

$Resistance=6\Omega+6\Omega$

$=12\Omega$

$3rd$ resistor

$6\Omega$ and is connected to

$12\Omega$ in parallel

Total resistance =

$4\Omega$

What is meant by heating effect of electric current?

When an electric current passes through a conductor (like a high resistance wire) the conductor becomes hot after some time and produces heat. This is called heating effect of Electric Current. Example 1. A bulb becomes hot after its use for some time.This is because of heating effect of electric current.

Give examples where we find heating of electric current.

When a resistor is connected to a battery, the source energy continually gets dissipated entirely in the form of heat. This is known as the heating effect of electric current. This effect is utilised in devices such as electric heater, electric iron etc.

What factors does the heat produced in a wire depend on?

From Joule's heating law:

$H=i^2Rt$ , where

$H$ is the heat produced,

$R$ is the resistance of the wire.

We also know

$R=\rho \dfrac{l}{A}$ , where

$\rho$ is the resistivity of the material,

$l$ is the length and

$A$ is the cross-sectional area of the wire.

Thus, The amount of heat produced in a wire depends on the material, length, and thickness of the wire.

Fill in the blanks:
When current is switched on in a room heater, it .......................

When “current is switched” on in a “room heater” it produces

$\textbf{heat}$ due to “the heating effect” of the electric current. Explanation: The “heating effect of electric current” is its ability to produce heat when passing through a conductor.

Is the direction of current in the circuit between $A$ and $B$ from $A$ to $B$ or from $B$ to $A$ ?

Judge the equivalent resistance when the following are connected in parallel
(a) $1\Omega$ and $10^{6}\Omega$
(b) $1\Omega$ and $10^{3}\Omega$ and $10^{6}\Omega$

(a) If two resistance are connected in parallel

Given resistance

$R_1=1\Omega\\R_2=10^6\Omega$

Effective resistance when these are connected in parallel

$\dfrac 1R_{eff}=\dfrac 1R_1+\dfrac 1R_2$

On putting values,

$\dfrac{1}{R_{eff}}=\dfrac{1}{1}+\dfrac{1}{10^{6}}=1+10^{-6}\approx 1$

$\therefore R_eff{}=\dfrac 11=1\Omega$

Equivalent resistance

$=1\Omega$

(b)If three resistance

$1\Omega,10^{3}\Omega$ and

$10^{6}\Omega$ are connected in parallel

$\dfrac1{R_{eff}}=\dfrac{1}{1}+\dfrac{1}{10^{3}}+\dfrac{1}{10^{6}}$

$\dfrac1{R_{eff}}=\dfrac{10^{6}+10^{3}+1}{10^{6}}$

$\therefore R_{eff} =\dfrac{1000000}{1001001}=0.999 \Omega$

Equivalent resistance

$=0.999\Omega$

Heat is generated in accordance with the equation $H = I^2Rt.$ If the current $I$ is reduced to half, how much will be the reduction in heat?

According to Joule's law of heating,

Heat produced,

$H=I^2Rt$

where

$I=$ current,

$R=$ resistance, and

$t=$ time

If the current

$I$ is reduced to half

$I'=\dfrac I2$

Then,

Heat produced,

$H'=I'^2Rt$

$H'=\left(\dfrac I2\right)^2Rt$

$H'=\dfrac 14\times I^2Rt$

$H'=\dfrac H4$

The heat produced will reduced by

$one-fourth$ .

Heat is generated in accordance with the equation $H = I^2Rt.$ In that case what are the methods to reduce the heat generated ?

Since the heat generated

$H$ is directly proportional:

to the square of current
resistance
time duration,

So to reduce the heat generated, the following methods can be followed :

Reduce current.
Reduce resistance.
Reduce the time taken.

Convert $1$ Kilowatt-hour of electric energy into joules.

$1\,k \,Wh = 1000\,W \times 1\,hr$

$= 1000\,W \times 60 \times 60\,s = 3600000 \,Ws$

$= 3600000 \,J$

What is the unit of current ?

Electric current is basically the rate of flow of the electric charge in motion in a conductor. It is said to exist when there is a net flow of charge through a region in the conductor. The current is given as the flow of charge per second:

$I=\dfrac{Charge\quad flown}{Time \quad taken}$

The SI unit of current is Ampere.

$A$

Heat is generated in accordance with the equation $H = I^2Rt.$ In that case
How can we reduce the current without change in power ? Find out on the basis of the equation $P = V \times I.$

Heat is generated in accordance with the equation

$H = I^2Rt.$ In that case we reduce the current without change in power by increasing voltage on the basis of the equation

$P = V \times I.$

In the given graphs which is the graph depicting Ohm's Law ? justify your answer.

1879736_1849862_ans_6cbcddcf686f4442a82448f2784c707d.jpg

In a household electric circuit, where is the MCB (Miniature Circuit Breaker) connected?

Miniature Circuit Breaker (MCB) is created to safeguard electrical circuits from excessive current and overheating. So the MCB is connected to the main fuse board to disconnect the household circuit from any power overload.

According to Ohm's Law, V/I must be a constant (resistance R). If so, what is the reason for the change in the ammeter readings?

The reason for the change in the ammeter reading is the variation of resistance of the conductors due to increasing its temperature.

When the circuit was switched on, was the temperature of the bulb low or high?

The temperature of the bulb is high because heat energy is released.

Heat is generated in accordance with the equation $H = I^2Rt.$ In that case if current (I) is reduced to $\dfrac{1}{10}$ times how much will be the reduction in heat?

Heat is generated in accordance with the equation

$H = I^2Rt$

$I$ ; current

$R$ ; resistance

$t$ ; time

In that case, if current

$(I)$ is reduced,

$I' = \dfrac{I}{10}$

So heat will be-

$H' = (\dfrac{I}{10})^2Rt$

$\Rightarrow H' = \dfrac{H}{100}$

In this case which form of energy was converted into heat energy?

In this case,

$Electrical\ energy$ was converted into heat energy.

A current is passing through the resistance. According Joule's heating law, the heat energy produced per second in the resistor is

$H=i^2R$ , where

$i$ is the current through the resistance

$R$ .

Complete the following table based on Ohm's Law

1780405_1849751_ans_4b02923b24784db7b3341eb80f72b69b.jpg

Is there any difference in the amount of heat energy thus obtained and the work done by the current?

The amount of heat energy obtained and the work done by the current are same.

According to Joule's law of heating,

Heat produced,

$H=I^2Rt$

where

$I=$ current,

$R=$ resistane,

$t=$ time

We know,

Power,

$P=VI$

According to Ohm's law,

$V=IR$

$\therefore P=I^2R$

Work done,

$W=Pt$

$W=I^2Rt$

$\implies Heat\ produced=Work\ done$

According to Joule's Law, more heat will be produced when an electric current is increased. What happens to the fuse wire due to this?

The fuse protects circuits and appliances by stopping the flow of any unduly high electric current. The fuse is placed in series with the device. It consists of a piece of wire made of a metal or an alloy of low melting point. If a current larger than the specified value flows through the circuit, the temperature of the fuse wire increases. This melts the fuse wire and breaks the circuit.

A current of 0.4 A flows through an electric bulb working at 230 V. What is the power of the bulb?

According to the question,

$V= 230 \ V$

$I= 4 \ A$

$P= VI$

$P= 230 \times 4 = 92 \ W$

V=230 V

0.5 A current flows through an electric heating device connected to 230 V supply.
How much is the power of the heating device connected to the circuit if we ignore the resistance of the circuit wire?

Power of the heating device

$P= VI$

$=230 \times 0.5 \ W$

$= 115 \ W$

$10$ resistors of $2 \ \Omega$ each are connected in parallel. Calculate the effective resistance.

As we know that in parallel connection

$\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} +\dfrac{1}{R_3} +\dfrac{1}{R_4} +...............+\dfrac{1}{R_n}$

$R_1 = R_2 = R_3 =........=R_n= 2 \Omega$ and

$n= 10$

Then,

$\dfrac{1}{R_{eq}} = \dfrac{1}{2} + \dfrac{1}{2} +\dfrac{1}{2} +\dfrac{1}{2} +...............+\dfrac{1}{2}$

$R_{eq} = \dfrac{2 \Omega}{10}= 0.2 \ \Omega$

Complete the following table on the basis of Joule's Law.
Resistance of conductor $R \ (\Omega)$ Intensity of Current $I \ (A)$ Time for which current flows $t \ (s)$ Heat generated $I^2R t \ (J)$ Change in Heat $(H)$
$2 \ R$ $I$ $t$ $2 \ I^2 Rt$ Twice $(2 H)$
$R$ $2 \ I$ $t$ ................. ...................
$\dfrac{R}{2}$ $I$ $t$ ................. ...................
$R$ $\dfrac{I}{2}$ $t$ ................. ...................
$R$ $I$ $2 t$ ................. ...................
$R$ $I$ $\dfrac{t}{2}$ ................ ...................

Resistance of conductor $R \ (\Omega)$	Intensity of Current $I \ (A)$	Time for which current flows $t \ (s)$	Heat generated $I^2R t \ (J)$	Change in Heat $(H)$
$2 \ R$	$I$	$t$	$(2 \ I)^2 Rt$	Twice $(2 H)$
$R$	$2 \ I$	$t$	$I^2 \dfrac{R}{2} t$	Four times $(4H)$
$\dfrac{R}{2}$	$I$	$t$	$\left(\frac{t}{2} \right)^2 RT$	Quarter $(\frac{H}{4})$
$R$	$\dfrac{I}{2}$	$t$	$I^2 R 2t$	Twice $(2H)$
$R$	$I$	$2 t$	$I^2 R 2 t$	Twice $(2 H)$
$R$	$I$	$\dfrac{t}{2}$	$I^2 R \frac{t}{2}$	Half $(\frac{H}{2})$

When ni-chrome wire become red hot while passing electricity through the circuit, how does this energy change occur?

This based on the concept that energy can neither be created nor destroyed. It can only be converted from one form to another (Law of Conservation of Energy) As the resistance of Ni-chrome is higher it produced more heat.

Name the part in which electrical energy changes into heat energy.

Heating coil changes the electrical energy intoheat energy.

0.5 A current flows through an electric heating device connected to 230 V supply.
How much is the resistance of the circuit?

As we know that

$R= \dfrac{V}{I}= \dfrac{230}{0.5}= 460 \ \Omega$

A potential difference of 6V is applied across a conductor having $12\Omega$ . How much current will pass through it? How many times will the current increase if length if length of resistor is halved and potential difference is double ?

Given,

Initially,

Potential difference of

$V_i=6$ V

Resistence of conductor

$R_i=12\ \Omega$

Ohm's law,

$i=\dfrac{V}{R}=\dfrac{6}{12}=0.5A$

Finally,

Lenght of resistor is half of initial value, ie

$l_f=l_i$

$R\propto l$

If length is halved

$R_f=12\times 1/2=6$

Potential difference is double, Potential difference

$V_f=2\times 6=12V$
Ohm's law,

$i=\dfrac{V}{R}=\dfrac{12}{6}=2A$

Electrical device Operating voltage (V) Power of the device (P) Current through the circuit
$I = P/V$ The amperage of the fuse to be used in the circuit (A)
Water heater
230V
4370 W 20 A
Air conditioner (AC)
230V

3335 W 15 A
Television (LED-TV)
230V

57.5 W 0.3 A
Computer (Laptop)
230V

28.75 W 0.2 A
Complete the table based on the amperage of the fuse wire.

Electrical device	Operating voltage (V)	Power of the device (P)	Current through the circuit $I = P/V$	The amperage of the fuse to be used in the circuit (A)
Water heater	230V	4370 W	19 A	20 A
Air conditioner (AC)	230V	3335 W	14.5 A	15 A
Television (LED-TV)	230V	57.5 W	0.25 A	0.3 A
Computer (Laptop)	230V	28.75 W	0.125 A	0.2 A

We know that

$I = P/V$

Putting all the values in the equation we get the value of current for each appliances. The amperage of the fuse is chosen such that it is slightly higher than the normal current (

$I=P/V$ ) passing through the appliance under normal conditions.

If the resistance of two soldering irons working in 250 V are 500 $\Omega$ and 750 $\Omega$ .
a. Calculate which one of these will carry more current.
b. Find out which soldering iron has more power.
c. Calculate the heat produced in 5 minutes in the soldering iron having resistance 750 $\Omega$ .

$I= \dfrac{V}{R}$

Current through 500

$\Omega$

$I= \dfrac{250\ V}{500\ \Omega}= \dfrac12 \ A$

Current through 750

$\Omega$

$I=\dfrac{250 \ V}{750 \ \Omega}= \dfrac13 \ A$

b. Soldering iron has less resistance

c.

$H= I^2Rt$ Or

$H= \dfrac{V^2 t}{R}$

$H= \dfrac{250^2 \times 300}{750} = 25000 \ J$

When a 12 V battery is connected to resistor, 2.5 mA current flows through the circuit. If so what is the resistance of the resistor?

Given: Battery voltage =

$12V$

Current =

$2.5 mA$

$R= \dfrac{V}{I}= \dfrac{12}{2.5\ mA}= \dfrac{12 \times 1000}{2.5}= 4800 \ \Omega$

Operating voltage
Resistance of the device (R) Currents flowing in the device
$I = V/R$ Heat generated per second, Heat
$H = V \times I \times t$ Power given by the device
$P = V \times I$ or $P = H/t$ Reason for the change in power
230V 57.5 $\Omega$ 4A 920J 920W Difference in resistance
230V 11.5 $\Omega$ 2A 460J 460W Difference in resistance
230V 230 $\Omega$ 1A 230J 230W Difference in resistance
115V 57.5 $\Omega$ 2A 230J 230W Difference in resistance
460V 57.5 $\Omega$ 8A 3680J 3680W Difference in resistance
The table shows details of an electric heating device designed to work in 230 V. Complete the table by calculating the change in the heat and power on changing the voltage and resistance of the device. Analyze the table and answer the following question.
What change happens to power on increasing the resistance without changing the voltage?

On increasing the resistance without changing the voltage, power decreases.

$P=\dfrac{V^2}{R}$

Since V is constant,

$P \propto \dfrac{1}{R}$

An electric heater conducts 4 A current when 60 V is applied across its terminals. What will be the current if the potential difference is 120 V?

According to the question,

$R= \dfrac{V}{I}= \dfrac{60}{4}= 15 \ \Omega$ if the voltage is changed to

$120$

$VI= \dfrac{V}{R}= \dfrac{120}{15}= 8\ A$

Given below questions are related with heating of an electric iron box. Answer them.
a. Which is the part that produces heat in an electric iron?
b. Which nature of this part is made use in the above situation?
c. What is the relation between intensity of electric current and heat energy generated?
d. What are the factors that affect the heat generated in such heating appliances?
e. What is the relation connecting these factors with the heat generated?
f. What is this law known as?
g. Name a device that works on this law used for ensuring safety in electric circuits?

$a.$ Heating coil

$b.$ High resistivity, High melting point

$c.$ As electric current increases, Heat energy increases. The heat energy generated is directly proportional to the square of current for a given resistance.

$d.$ Electric current, Resistance, time

$e.$

$H= I^2 Rt$

$f.$ Joule's law of heating.

$g.$ Safety fuse.

Electric appliance Energy change
(electrical energy $\to$ ) Effect of electric current
Bulb .......a....... Lighting effect
Electric Fan Mechanical Energy ........b.........
Electric Iron box ........c....... Heating effect
Strong battery
(While charging) Chemical Energy .........d.........
Complete the table based on the effects of electric current and energy change.

a. An electric bulb converts the electrical energy into light energy
b. An electric fan converts the electrical energy into a mechanical effect and hence the fan rotates as a result.
c. An electric iron box converts electrical energy into heat energy.
d. A battery exhibits the chemical effect of electric current.

A 800 W electrical device is designed to work in 200 V. What will be its power if the device is working in 100 V?

Resistance of the appliance

$R= \dfrac{V^2}{P} = \dfrac{200 \times 200}{800}= 50 \ \Omega$

Power when connected in 100 V

$P= \dfrac{V^2}{R}= \dfrac{100 \times 100}{50}$

$= 200 \ W$

If we apply 220 V for a device having 230 V and 400 W, what will happen to the power of the device?
(increases, decreases, doesn't change)

If we apply

$220\ V$ for a device having

$230\ V$ and

$400\ W$ , the power of the device will decrease.

$\text{Power} = \dfrac{V^2}{R}$

Now,

$V$ is reduced so power will decrease.

Electric current intensity is made three times Heat produced becomes ....(1)....
In a circuit, fuse wire is connected in ......(2).....
Electric current intensity is reduced to $1/4$ Heat produced becomes ....(3)....
Analyse the table and fill it suitably.

(1)As we know that

$H= I^2Rt \Rightarrow H \propto I^2$ Hence, if

$I$ is made to 3 times , heat produced becomes 9 times,

(2) In a circuit, fuse wire is connected in series.

(3) As we know that

$H= I^2Rt \Rightarrow H \propto I^2$ Hence, if

$I$ is made to

$\dfrac{1}{4}$ times , heat produced becomes

$\dfrac{1}{16}$ times,

Four resistors are connected in parallel across a $9.20-V$ battery. They carry currents of $150 \,mA, \,45.0 \,mA, \,14.0 \,mA,$ and $4.00 \,mA$ . If the resistor with the largest resistance is replaced with one having twice the resistance,
(a) what is the ratio of the new current in the battery to the original current?
(b) What If? If instead the resistor with the smallest resistance is replaced with one having twice the resistance, what is the ratio of the new total current to the original current?

1978383_1864757_ans_8dad4288a5d7415ab3d76d4eccfc0a6e.jpg

Resistors in Series and Parallel(19)
Calculate the power delivered to each resistor in the circuit shown in Figure

1895912_1863985_ans_c89795e3976e4b699402e634ad2dea13.jpg

A V-I graph for a nichrome wire is given below. What do you infer from this graph? Draw a labelled circuit diagram to obtain such a graph.
Refer above image.

The V-I graph of the nichrome wire shows a straight line. It means that the resistance of the wire remains constant when the current supply is changed. It follows the ohm's law and therefore, it is considered as an ohmic conductor. It follows ohm's law

$\rightarrow V = IR$

$R = \dfrac{V}{I}$ which shows that the slope of graph

$\dfrac{V}{I}$ is a constant.

1949522_1943171_ans_2816a397841d4d97a12bfd470787124a.png

Conceptual Questions
When the potential difference across a certain conductor is doubled, the current is observed to increase by a factor of $3$ . What can you conclude about the conductor?

According to the Ohm's law,

$V=iR$

In the given case, the conductor does not follow Ohm's law. According to the given data, the current does not depend linearly on the voltage, so the conductor is not ohmic. The resistivity of the conductor may be dependent on current or temperature.

A 400 W electrical device is designed to work in 100 V. What will be its power if the device is working in 100 V?

Rated power

$P=400\ W$

Rated voltage

$V=100\ V$

From the above information, it is clear that when 100 V potential is applied across the device it will consume 400 W power.

Here, the working voltage is 100 V, so the power consumed by the device will be equal to 400 W.

Resistors in Series and Parallel
Three 100-V resistors are connected as shown in Figure. The maximum power that can safely be delivered to anyone resistor is $25.0 \, W$ .
(a) What is the maximum potential difference that can be applied to the terminals $a$ and $b$ ?
(b) For the voltage determined in part (a), what is the power delivered to each resistor?
(c) What is the total power delivered to the combination of resistors?

(a) The equivalent resistance is 150 Ohms. All of the current flows through the first resistor, and the current divides in half and flows through both of the resistors in parallel. Thus the resistor with the highest power load will be the first one and that load will be I² R, and I will be the voltage applied divided by 150 Ohms. Thus we want 25 (maximum power) = (V/150)² * 100. That gives V = 75 Volts.

(b) The current delivered is V/R = 75/150 = 0.5 Amps. In the first resistor, using I² R, the power is 0.5² 100 = 25 Watts, in the other two it is (0.5/2)² 100 = 6.25 Watts. The sum is the total power, 25 + 6.25 + 6.25 = 37.5 Watts. This is also I*V = 0.5 Amps * 75 Volts.

Resistors in Series and Parallel(15)
Two resistors connected in series have an equivalent resistance of $690 \Omega$ . When they are connected in parallel, their equivalent resistance is $150 \,\Omega$ . Find the resistance of each resistor.

1895903_1863974_ans_e7da5f5d28704dffa7334d6c4a458457.jpg

An electric iron has a rating of $750\ W, 220\ V$ . Calculate the (i) current flowing through it, and (ii) its resistance when in use.

Given :

Power rating

$P = 750\ W$

Rated voltage

$V = 220\ V$

$(i)$ We know

$P = VI$

$750 = 220 \times I$

$\Rightarrow$

$I = \dfrac{750}{220} = 3.40\ A$

$(ii)$ Also, power can also be wriiten as

$P = \dfrac{V^2}{R}$

$\Rightarrow$

$R = \dfrac{V^2}{P} = \dfrac{220^2}{750}$

So resistance of electric iron

$R = 64.53\ \Omega$

While studying the dependence of potential difference (V) across a resistor on the current (I) passing through it, in order to determine the resistance of the resistor, a student took $5$ readings for different values of current and plotted a graph between V and I. He got a straight line graph passing through the origin. What does the straight line signify? Write the method of determining resistance of the resistor using this graph.

Straight line signify that the potential difference applied across the resistor is directly proportional to the current flowing through it.
To determine the resistance from the graph, read the current value, in amperes corresponding to a given voltmeter reading and take the ratio.
Thus the resistance of conductor is determined in ohms.

In the experiment to study the dependence of current (I) on the potential difference (V) across a resistor, a student obtained a graph as shown,
(I) What does the graph depict about the dependence of current on the potential difference?
(ii) Find the current that flows through the resistor when the potential difference across it is $2.5V$ .

$\textbf{Hint}$

$\textbf{Step1:Conclusion from graph}$

$\textbf{Step2:Find current when V=2.5 volt}$

From graph slope is

$\dfrac{I}{V}=\dfrac{0.2}{2}=0.1$

So to find current at 2.5 V suppose current is I.

Graph is straight line so slope is constant.

Now, slope

$0.1=\dfrac{I}{2.5}$

$\Rightarrow I=0.25 A$

Calculate the value of resistor $X$ .

In the circuit, one

$4\Omega$ resistor, one

$12\Omega$ resistor and a

$6\Omega$ resistor are in parallel.

Let the equivalent resistance of these three resistors is

$R_p$

$\therefore \dfrac{1}{R_p}=\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{4}$

$\therefore R_p=2\Omega$

Now, the

$4\Omega$ ,

$R_p$ and

$X$ are in series.

Thus, the total equivalent resistance of the circuit is

$2+4+X=6+X$

From Ohn's law, $$V=IR$$

Given that

$V=8\ volt$ ,

$I=0.5\ A$

$\therefore 8=0.5\times (6+X)$

$\Rightarrow X=10\Omega$

Find the value of equivalent resistance and total current flowing into the circuit.

Equivalent resistance

$R = 4 + \dfrac{{12 \times 12}}{{12 + 12}} = 10\;\Omega$

Current $I = \dfrac{V}{R}$

$I = \dfrac{5}{{10}} = 0.5\;{\rm{A}}$

Thus equivalent resistance is $10\Omega$ and current is $0.5{\rm{A}}$ .

With the help of neat circuit, derive the expression for the eqivalent resistance of 3 resistances connected in series.

When three resistances ${R_1}$ , ${R_2}$ and ${R_3}$ are connected in series, then

Current passing through each resistor will be same and potential difference is distributed among them.

Now $V = {V_1} + {V_2} + {V_3}$

$I{R_{eq}} = I\left( {{R_1} + {R_2} + {R_3}} \right)$

Thus,

${R_{eq}} = \left( {{R_1} + {R_2} + {R_3}} \right)$

995988_1073114_ans_aa4e79d8771449068f9876bf6ec238a0.JPG

Out of joule, calorie, kilowatt and electron volt, which one is not the unit of energy?

Kilowatt is not a unit of energy. It is the unit of power. When 1000 J of energy is delivered in a second,

The power delivered is said to be 1 kilowatt (kW).

A wire of given material of length $l$ and area of cross section $A$ , has a resistance of $2 \Omega$ .Find the resistance of another wire of the same material having length $2l$ and area of cross-section $A/2$ .

1966721_1406259_ans_0ec82e88605a4df98de95a6413a7b10f.jpg

Express 1KWh in joules.

We know that -

$1\ kW = 1000\ W$ and

$1\ hr = 3600\ s$

$\therefore 1\ kWh = 1000\ W\times 1\ hr = 1000\ W\times 3600\ s = 3.6\times 10^5\ J$

Compute the heat generating while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.

Q=96000 C

t=1 hr=3600seconds

$\dfrac{Q}{t}$ =

$\dfrac{96000}{3600}$ =

$\dfrac{80}{3}$ A

V=50 V

Amount of heat generated is

H=VIt

$50\times\dfrac{80}{3}\times3600$

=4800000 J

Power input by the source is

P=VI

$50\times\dfrac{80}{3}$

=1333.3 W

Calculate (i) effective resistance (ii) current (iii) Potential difference across 10 resistor of a circuit shown in the figure.

All resistance are connected in parallel across 2 V batteries,

So,

(i) Effective resistance

$= 5 \Omega + 10\Omega$

$R_{eff}=15\Omega$

(ii) Current

$\Rightarrow i = \dfrac{V}{R_{eff}} = \dfrac{2}{15}amp$

(iii) Potential across

$10\Omega \Rightarrow V = IR = \dfrac{2}{15}\times 10 = \dfrac{4V}{3}$

2086774_1367832_ans_42d836ccb6e647e5b158a8a70b999336.png

In the circuit shown find the equivalent resistance between $A$ and $B$ .

From the above figure,

$R_{eq}=\dfrac{E}{I}$

loop

$CFAC:E=rI_{1}+R(I_{1}-I_{2})\Rightarrow E=(r+R)I_{1}-RI_{2}$

loop

$ACBDEA:rI_{1}+rI_{2}=R(I-I_{1})\Rightarrow (r+P)I_{1}+rI_{2}=RI$

$rI_{2}+r(I-I_{1}+I_{2})=R(I_{1}-I_{2})$

$\Rightarrow (r+R)I_{3}-(R+2r)I_{2}=rI$

From Eqs (ii)-(iii), we get

$(3r+R)I_{2}=(R-r)I\Rightarrow I_{2}=\dfrac{(R-r)I}{3r+R}$

Putting the above value in eq (ii), we get

$(r+R)I_{1}+\dfrac{r(R-r)}{3r+R}I=RI\Rightarrow I_{1}=\dfrac{(R-r)I}{3r+R}$

Putting

$I_{1}$ and

$I_{2}$ in Eq (i) we get

$R_{eq}=\dfrac{E}{I}=\dfrac{(R+r)^{2}}{3r+R}-\dfrac{r(R-r)}{3r+R}=\dfrac{r(r+3R)}{3r+R}$

1626688_1539959_ans_e3df63b40e7d4ecabb384ccccda5087c.png

Figure 6.32 shows a meter bridge (which is nothing but a practical wheatstone bridge) , consisting of two resistors X and Y toghether in parallel with a meter long constant wire of uniform corss section.
With the help of movable contact D, one can change the ratio of resistance of the two segments of the wire until a sensitive galvanometer G connected across B and D shows no deflection. The null point is found to be at a distance of 33.7 cm. The resistor Y is shunted by a resistance of 12 $\Omega$ and the null point is found to sift by a distance of 18.2 cm. Determine the resistance of X and Y.

1883298_1753298_ans_0dd6f363c25e4e599c2d813764007e9d.jpg

Five electric fans of 120 watts each are used for 4 hours. Calculate the electrical energy consumed in kilowatt hours.

Power of 5 electric fans,

$P = 5 \times 120 = 600 W= 0.6\ kW$

$t =4\ hr$

$E = p \times t =2.4\ kWh$

Two heating coils,one of fine wire and other of thick wire, made of the same material and of the same length are connected one by one to a source of electricity.Which coil will produce heat at a greater rate ?

Given,

2 heating coils of same length,

$l$

Let resistance of thin wire be

$R_1$

resistance of thick wire be

$R_2$

We know,

for same material, resistivity

$\rho$ is same

Cross section area of thin wire =

$A_1$

Cross section area of thick wire =

$A_2$

$A_2 > A_1$

$R = \dfrac{\rho l}{A}\implies R\alpha\dfrac{1}{A}$

$\implies R_1 > R_2\ (\because A_2 > A_1)$

Rate of heat generation,

$P = \dfrac{V^2}{R}$

Both are connected to the same source V

$P\alpha\dfrac{1}{R}$

$\implies P_2 > P_1\ (\because R_1 > R_2)$

A thick wire coil will produce heat at a rate of heat generation,

$P = \dfrac{V^2}{R}$

Both are connected to the same source V

$P\alpha\dfrac{1}{R}$

$P_1$ = Rate of heat generation by thin wire

$P_2$ = Rate of heat generation by thick wire

$R_1 > R_2$

$\implies P_2 > P_1$

The thick wire coil will generate heat at a greater rate.

Define one watt hour and one kilowatt hour.

One watt-hour is the energy consumed when

$1$ watt of power is used for

$1$ hour. One kilowatt-hour is the energy consumed when

$1$ kilowatt (

$1000\ watts$ ) of power is used for 1 hour.

One kilowatt-hour is known as one unit to calculate the electricity bill.

An electric heater of power $3\, kW$ is used for $10\, h$ . How much energy does it consume? Express your answer in (i) kWh. (ii) joule.

Fatergy consumed

$= power \times time$

(i) Energy

$= 3\, kW \times 10\, h = 30\, kWh$

(ii)

$1$ kilowatt has (kWh)

$= 3.6 \times 10^6\, J$

$30\,kWh= 30 \times 3.6 \times 10^6\, J$

$=1.08 \times 10^8\,J$

A heating element in a stove is designed to receive $3000 \,W$ when connected to $240 \,V$ . Assuming the resistance is constant, calculate the current in the heating element.

Given:

The power rating of the stove

$P=3000\ W=3\ kW$

Voltage

$V=240\ V$

$\underline{\text{Calculating current}}:$

We know,

$P=VI$ , where

$I$ is the current.

$\therefore I=\dfrac{P}{V}=\dfrac{3000}{240}=12.5\ A$

A light bulb has a resistance of $240\Omega$ when operating with a potential difference of $120 V$ across it. What is the current in the light bulb?

1974715_1863475_ans_85a1f1b5e2724743beb81b43be3eea1b.jpg

Additional Problems(51)
Four $1.50-V \,AA$ batteries in series are used to power a small radio. If the batteries can move a charge of $240 \,C$ , how long will they last if the radio has a resistance of $200 \,\Omega$ ?

1896257_1864754_ans_54c0eb036afa43bdb07434841be0c515.jpg

Operating voltage	Resistance of the device (R)	Currents flowing in the device $I = V/R$	Heat generated per second, Heat $H = V \times I \times t$	Power given by the device $P = V \times I$ or $P = H/t$	Reason for the change in power
230V	57.5 $\Omega$	4A	920J	920W	Difference in resistance
230V	11.5 $\Omega$	2A	460J	460W	Difference in resistance
230V	230 $\Omega$	1A	230J	230W	Difference in resistance
115V	57.5 $\Omega$	2A	230J	230W	Difference in resistance
460V	57.5 $\Omega$	8A	3680J	3680W	Difference in resistance

Electric appliance	Energy change (electrical energy $\to$ )	Effect of electric current
Bulb	.......a.......	Lighting effect
Electric Fan	Mechanical Energy	........b.........
Electric Iron box	........c.......	Heating effect
Strong battery (While charging)	Chemical Energy	.........d.........

Electric current intensity is made three times	Heat produced becomes ....(1)....
In a circuit, fuse wire is connected in	......(2).....
Electric current intensity is reduced to $1/4$	Heat produced becomes ....(3)....

Electricity - Class 10 Physics - Extra Questions

The electronic current flows opposite to the conventional current. True/False?

What are insulators? Define on the basis of atomic structure. Give four examples of insulators.

State three characteristics of a series circuit?

A heating element is marked 210V210 V, 630W630 W. Find the resistance of the element when connected to a 210V210 V dc source.

The wattage of an electric heater which draws 5 A current when connected to a 220 V supply is _____________.

...................... does not conduct electricity.

A 1000 W heater has more resistance than a 100 W bulb.

The current drawn by a bulb rated 40 W, 220 V when it is connected to a 220 V supply is ___________.

State the characteristics required in a material to be used as an effective fuse wire.

Two copper wires are of the same length, but one is thicker than the other.Which wire will have more resistance?

Match the following:

Write the differences between static electricity and electric current.

(i)(i) What happens to the resistivity of semi-conductors with the increase of temperature?(ii)(ii) For a fuse, higher the current rating ________ is the fuse wire. (Thicker / Thinner)

Two resistors of 4Ω4\Omega and 6Ω6\Omega are connected in parallel to a cell to draw 0.5 A0.5\ A current from the cell.Draw a labelled circuit diagram showing the above arrangement

The SI units of current is .........

(i) Which particles are responsible for current in conductors?(ii) To which wire of a cable in a power circuit should the metal case of a geyser be connected? (iii) To which wire should the fuse be connected ?

Write three safety measures in using electricity.

An electric bulb is rated 200V−100W200V - 100W. Find the resistance of the bulb.

Murugan measured the electric current. What unit should he use?

Write the S.I. unit of electric power.

Equivalent resistance when 22 and 33 ohm resistances are connected in series is _______.

Give two examples each of conductors and insulators of heat.

Explore and answer :Does pure water conduct electricity? If not, what can we do to make it conduct?

Write equivalent resistance of given circuit.

Though same current flows the wire and the filament of a bulb, yet only the filament glows.Why?

Calculate the resistance across points AD.

Define electric current. What is its SISI unit?

Power of a lamp is 60 W60\ W. Find the energy in joules consumed by it in 1 s1\ s.?

Define 11 ampere.

Name and define the SI unit of current

Name and define unit for electric current.

6Ω,8Ω6\Omega,8\Omega and 10Ω10\Omega resistance are connected in series. Find total resistance of the circuit.

Two resistance 3 ohm and 2 ohm are in parallel connection and a potential difference of 12 V is applied across them.Finda) The equivalent resistance of the parallel combinations.b) The circuit current and c) The branch currents.

What are conductors of electricity ?

How much current flows through a 10 Ω10\ \Omega resistor when it is connected to a battery of 3 v3\ v ?

Define electric current.

Name the quantity whose unit is (i) kilowatt , and (ii) kilowatt - hour.

What is the unit of electric charge ?

Find the total resistance in the given circuit.

By what name is the physical quantity coulomb / second called ?

What is the unit of electric current ?

What is meant by electrical resistivity of a material?

A wire of length L and resistance R is stretched so that the length is doubled and area of cross section halved. How will (i) resistance change and (ii) resistivity change.

Why does the cord of an electric heater not glow while the heating element does ?

What is the total resistance of 5 resistors each of resistance 5Ω\Omega connected in parallel?

The voltage - current (V−I)(V-I) graph of a metallic conductor at two different temperature T1T_{1} and T2T_{2} is shown below. At which temperature is the resistance higher?

Match the entries in column A with appropriate ones from column B.

A current of 8 A flows through a 25 V car headlight bulb. Calculate the electric power (in Watts).

Two resistors of resistances 10Ω10 \Omega and 20Ω20 \Omega are connected in parallel. A battery supplies 6A6 A of current to the combination. Calculate the current in each resistor.

A certain household has consumed 250 units250\ units of energy during a month. How much energy is this in Joules?

How many 176 Ω176\ \Omega resistors (in parallel) are required to carry 5A on a 220V line?

A toaster produces more heat than an electric lamp when they are connected in parallel. Which of the two has the greater resistance?

Which of the following bulbs offers more resistance?(A) 60 W - 220 V(B) 100 W - 250 V

A resistance of 3 ohms3\ ohms is connected in series with another resistance of 5 ohms5\ ohms. A potential difference of 66 volts is applied across the combination. Calculate the current through the circuit and potential difference across the 3 ohm3\ ohm resistance.

Give examples for good conductors solid and liquid.

The potential difference between the ends of a conductor is 40V. If the conductor draws a current of 4A from the source, what current will the conductor draw if the potential difference is increased to 100V ?

The figure shows a circuitWhen the circuit is switched on, the ammeter reads 0.5A0.5 A.(i) Calculate the value of the unknown resistor RR.(ii) Calculate the charge passing through the 3Ω3 \Omega resistor in 120s120 s.(iii) Calculate the power dissipated in the 3Ω3 \Omega resistor.

Deduce the expression for the the equivalent resistance of three resistors connected in series.

How do you verify that resistance of a conductor is proportional to the length of the conductor for constant cross section area and temperature?

(i) Name the device used to protect the electric circuits from overloading and short circuits.(ii) On what effect of electricity does the above device work?

Deduce the expression for the the equivalent resistance of three resistors connected in Parallel.

An unknown circuit draws a current of 2A from a 12V battery equivalent resistance is.......

Three resistors of values 2Ω,4Ω,6Ω2 \Omega, 4\Omega, 6 \Omega are connected is series. The equivalent resistance of combination of resistors is............

Difference between resistances in series and parallel.

Find the expression for the resistors connected in series. Write any two characteristics of series combination of resistors.

With a neat labeled diagram, derive the equation for three resistance connected in parallel.

Find the expression for the resistors connected in series and state its two characteristics. (Draw figure).

When do you say that the resistors are connected in series? Find the expression for the resistors connected in this way. (Draw circuit diagram).

Three resistances having the values 5Ω\Omega, 10Ω\Omega, 30Ω\Omega are connected in parallel with each other. Calculate the total circuit resistance.

A domestic electric circuit has a fuse of 5A5A. What is the maximum number of 100W(220V)100 W ( 220 V ) bulbs that can be safely used in the circuit?

Two lamps one rated 60 W60\ W, 220 V220\ V and the other 40 W40\ W, 220 V220\ V are connected in parallel to a 220 V220\ V electric supply mains. What is the total current drawn from the electric mains if the voltage of the electric supply is 220 V220\ V?

Three resistors are connected in series with 10V10 V supply as shown in the figure. Find the voltage drop across each resistor, and effective resistance of series combination.

My mother was operating the washing machine on Sunday morning. Suddenly, she saw a spark coming out of the electric board and the electric current in the house failed. An electrician was called to look into the matter. What should he have noticed?

Show that the effective resistance of series combination in a circuit is equal to the sum of their resistances.

A wire of resistance 4R4 R is bent in the form of a circle. What is the effective resistance between the ends of the diameter?

Explain series connections of Resistors and drive the formula of equivalent resistance.

With the help of a circuit diagram, obtain the expression for equivalent resistance of two resistors connected in parallel.

Nichrome and copper wire of same length and same radius are connected in series. Current II is passed through them. Which wire gets heated up more? Justify your answer.

Name the device used to protect the electric circuits from overloading and short circuits.

A heating element is marked $210 V$ , $630 W$ . Find the resistance of the element when connected to a $210 V$ dc source.

Two copper wires are of the same length, but one is thicker than the other.
Which wire will have more resistance?

$(i)$ What happens to the resistivity of semi-conductors with the increase of temperature?
$(ii)$ For a fuse, higher the current rating ________ is the fuse wire. (Thicker / Thinner)

Two resistors of $4\Omega$ and $6\Omega$ are connected in parallel to a cell to draw $0.5\ A$ current from the cell.
Draw a labelled circuit diagram showing the above arrangement

(i) Which particles are responsible for current in conductors?
(ii) To which wire of a cable in a power circuit should the metal case of a geyser be connected?
(iii) To which wire should the fuse be connected ?

An electric bulb is rated $200V - 100W$ . Find the resistance of the bulb.

Equivalent resistance when $2$ and $3$ ohm resistances are connected in series is _______.

Explore and answer :
Does pure water conduct electricity? If not, what can we do to make it conduct?

Define electric current. What is its $SI$ unit?

Power of a lamp is $60\ W$ . Find the energy in joules consumed by it in $1\ s$ .?

Define $1$ ampere.

$6\Omega,8\Omega$ and $10\Omega$ resistance are connected in series. Find total resistance of the circuit.

Two resistance 3 ohm and 2 ohm are in parallel connection and a potential difference of 12 V is applied across them.Find
a) The equivalent resistance of the parallel combinations.
b) The circuit current and
c) The branch currents.

How much current flows through a $10\ \Omega$ resistor when it is connected to a battery of $3\ v$ ?

What is the total resistance of 5 resistors each of resistance 5 $\Omega$ connected in parallel?

The voltage - current $(V-I)$ graph of a metallic conductor at two different temperature $T_{1}$ and $T_{2}$ is shown below. At which temperature is the resistance higher?

Two resistors of resistances $10 \Omega$ and $20 \Omega$ are connected in parallel. A battery supplies $6 A$ of current to the combination. Calculate the current in each resistor.

A certain household has consumed $250\ units$ of energy during a month. How much energy is this in Joules?

How many $176\ \Omega$ resistors (in parallel) are required to carry 5A on a 220V line?

Which of the following bulbs offers more resistance?
(A) 60 W - 220 V(B) 100 W - 250 V

A resistance of $3\ ohms$ is connected in series with another resistance of $5\ ohms$ . A potential difference of $6$ volts is applied across the combination. Calculate the current through the circuit and potential difference across the $3\ ohm$ resistance.

The figure shows a circuit
When the circuit is switched on, the ammeter reads $0.5 A$ .
(i) Calculate the value of the unknown resistor $R$ .
(ii) Calculate the charge passing through the $3 \Omega$ resistor in $120 s$ .
(iii) Calculate the power dissipated in the $3 \Omega$ resistor.

(i) Name the device used to protect the electric circuits from overloading and short circuits.
(ii) On what effect of electricity does the above device work?

Three resistors of values $2 \Omega, 4\Omega, 6 \Omega$ are connected is series. The equivalent resistance of combination of resistors is............

Three resistances having the values 5 $\Omega$ , 10 $\Omega$ , 30 $\Omega$ are connected in parallel with each other. Calculate the total circuit resistance.

A domestic electric circuit has a fuse of $5A$ . What is the maximum number of $100 W ( 220 V )$ bulbs that can be safely used in the circuit?

Two lamps one rated $60\ W$ , $220\ V$ and the other $40\ W$ , $220\ V$ are connected in parallel to a $220\ V$ electric supply mains. What is the total current drawn from the electric mains if the voltage of the electric supply is $220\ V$ ?

Three resistors are connected in series with $10 V$ supply as shown in the figure. Find the voltage drop across each resistor, and effective resistance of series combination.

A wire of resistance $4 R$ is bent in the form of a circle. What is the effective resistance between the ends of the diameter?

Nichrome and copper wire of same length and same radius are connected in series. Current $I$ is passed through them. Which wire gets heated up more? Justify your answer.

Vani's hairdryer has a resistance of $50 \Omega$ when it is first turned on.
i) How much current does the hairdryer draw from the $230\ V$ line in Vani's house?
ii) What happens to the resistance of the hairdryer when it runs for a long time?

Three resistors $R_1, R_2, R_3$ are joined in series. Draw labelled diagram and obtain expression of its equivalent resistance.

Resistors are sometimes joined together and they have several applications in electronics. Obtain the expression for the effective resistance in the combination of two resistors ${R}_{1}$ and ${R}_{2}$ in parallel.

Calculate the resultant resistance when three resistances of $3\Omega$ are connected in parallel?

Resistors are sometimes joined together and they have several applications in electronics. Draw a series combination of three resistors ${R}_{1}$ , ${R}_{2}$ and ${R}_{3}$ .

You have three resistors of values $2\Omega, 3\Omega$ and $5\Omega$ . How will you join them so that the total resistance is more than $7\Omega$ ? Draw a diagram for the arrangement.

You have three resistors of values $2\Omega, 3\Omega$ and $5\Omega$ . How will you join them so that the total resistance is more than $7\Omega$ ? Calculate the equivalent resistance.

A resistance coil is made by joining in parallel two resistances each of 10 $\Omega$ . An emf of 1V is applied between the two ends of coil for 5 minutes. Calculate the heat produced in calories.

If the value of resistance is $10.845 \Omega$ and the value current is $3.23$ A, the potential difference is $35.02935\,\,V$ . Find its value in significant number.

A copper wire has diameter $0.5 \ mm$ and resistivity of $1.6\times { 10 }^{ -8 }\Omega m$ . What will be the length of this wire to make its resistance $10\ \Omega$ ?

Heat energy is being produced in a resistance in a circuit at the rate of $100\ W$ . The current of $2\ A$ is flowing in the circuit. what must be the value of the resistance (in $\Omega$ )?

Two resistances when connected in parallel give resultant value of $2\Omega$ . When connected in series the value becomes $9\Omega$ . Calculate the value of each resistance.

Find the reading of ammeter. Is this the current thought $6\Omega$ ?

Calculate combined resistance of resistances $3 \Omega$ and $6 \Omega$ connected in parallel.

Calculate the power of soldering iron with a hot resistance $4$ $K\Omega$ and operating voltage of $200$ $V$ .

In the above question, Find individual power consumed.
Given $R_1=1 \ k \Omega$ and $R_2=0.5\ k\Omega$

Judge the equivalent resistance when the following are connected in parallel
(i) $1 \Omega$ and $10^6 \Omega$
(ii) $1 \Omega, 10^3 \Omega$ and $10^6 \Omega.$

$B_1, B_2$ and $B_3$ are three identical bulbs connected as shown in Figure 12.When all the three bulbs glow, a current of 3A is recorded by the ammeter A.
What happens to the reading of $A_1, A_2, A_3$ and A when the bulb $B_2$ gets fused?

Resistors having resistances of $15\Omega$ , $20\Omega$ , and $10\Omega$ are connected in parallel. What is the effective resistance in the circuit.

$B_1, B_2$ and $B_3$ are three identical bulbs connected as shown in Figure 12.When all the three bulbs glow, a current of 3A is recorded by the ammeter A.
How much power is dissipated in the circuit when all the three bulbs glow together?

A combination consists of three resistors in series. Four similar sets are connected in parallel. If the resistance of each resistor is $2\ \Omega$ . Find the equivalent resistance of the combination.

Find the resistor each $3 \Omega$ are connected as shown, calculate the resistance
a) between the paint p and q
2) between the paint x and y

How many bulbs of power $60\ W$ can be connected in parallel across a $240\ V$ source if current rating of the circuit is $5A$ ? What is the resistance of each bulb?

Two lamps of rating $100 W$ , $220V$ and $60 W$ , $220V$ are joined in parallel with $220 V$ line. How much current will flow through the circuit?

What is the ratio of heat generation in two resistors having resistances $R$ and $2R$ in same time $t$ .

A lamp consumes $1000\ J$ of electrical energy in $10\ s$ . What is its power?

A piece of wire of resistance R is cut into five equal parts. When these parts are connected in parallel. Then the equivalent resistance of this combination is R' .The ratio $\frac{R}{R'}$ is

An electric bulb consumes $7.2\ kJ$ of electrical energy in $2\ min$ . What is the power of the electric bulb?

Find the followings -
a) total resistance of the circuit,
b) the value of resistor R
c) the current flowing through R.

A constantan wire of length $50\ cm$ and $0.4\ mm$ diameter is used in making a resistor. If the resistivity of constantan is $5\times 10^{-7} \ \Omega-m$ , calculate the value of the resistance.

Classify the following into good conductors, resistors and insulators :
Rubber, Mercury, Nichrome, Polythene, Aluminium, Wood, Manganin, Bakelite, Iron, Paper, Thermocol, Metal coin.

A uniform wire with a resistance of $27 \Omega$ is divided into three equal pieces and then they are joined in parallel . find the equivalent resistance of the parallel combination.

If two resistors with resistance 5 $\Omega$ and 10 $\Omega$ respectively , are connected to a battery of emf 6V, then determine the strength of the total resistance and current in the circuit when connected in
(a) parallel combination and (b) series combination

An electric bulb is rated $220\ V$ and $100\ W$ . When it is operated on $110\ V$ , what will be the power consumed?

(i)What is an electric fuse?
(ii)An electric fuse of rating 3 ampere is connected in a circuit in which an electric iron of power 1 kw is connect which operates at 220 V. What would happen, Explain?

How does the heat $H$ produced by a current passing through a fixed resistance wire depend on the magnitude of current $I$ ?

With the help of labelled circuit diagram derive a formula to find combined resistance (R) when two or more resistance $(R_1,R_2,R_3.....)$ are connected in parallel taking symbols potential difference $(V)$ and current $(I)$ .

(i)Draw a circuit diagram to show three resistors connected in parallel with a cell and a key. (ii)Write formula for the equivalent resistance $R$ of these combination if the individual resistance of the resistance are $R_1, R_2$ and $R_3$ .

Three bulbs, each having a resistance of $180 \Omega$ , are connected in parallel to an ideal battery of emf $60 V$ . Find the current delivered by the battery when
(a) all the bulbs are switched on,
(b) two of the bulbs are switched on and
(c) only one bulb is switched on.

Consider the circuit shown in figure. Find the current through the $10 \Omega$ resistor when the switch $S$ is (a) open (b) closed.