MCQ Questions for Class 10 Physics Electricity Quiz 5

When a fuse is rated 8A, it means :

0%
It will not work if current is less than 8A
0%
It has a resistance of 8 ohm
0%
It will work only if current is 8A
0%
It will burn if current exceeds 8A

Fuse wire is made of :

0%
Platinum
0%
Copper
0%
Aluminium
0%
Alloy in tin and lead

Explanation

Electric fuse wire is a wire which allows only upto a limit of current through it, if a particular amount of value of current is exceeded then it melts down.

so for achieving the melting property the fuse wire is made up of an alloy made of tin

$63 \ \%$ & lead

$37\ \%$ . which can melt down over low temperatures.

An air conditioner is rated

$240\ V, 1.5\ kW$ . The air conditioner is switched on

$8\ hrs$ each day. What is electrical energy consumed in

$30\ days$ ?

0%
$2.88\ kWh$
0%
$360\ kWh$
0%
$120\ kWh$
0%
$240\ kWh$

Explanation

Energy consumed is given by:

$E= Pt$

$E$ : Energy consumed

$P$ : Power

$t$ : Time

$\Rightarrow E = 1.5\ kW \times 30 \times 8\ hrs$

$\Rightarrow E= 360\ kWh$

The resistors of resistances

$2 \Omega, 4 \Omega, 5 \Omega$ are connected in parallel. The total resistance of the combination will be :

0%
$\displaystyle \frac{29}{10} \Omega$
0%
$\displaystyle \frac{19}{20} \Omega$
0%
$\displaystyle \frac{10}{20} \Omega$
0%
$\displaystyle \frac{20}{19} \Omega$

Explanation

According to parallel combination.

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

$=\displaystyle \frac{1}{2} + \frac{1}{4} + \frac{1}{5}$

$=0.5 + 0.25 + 0.2$

$=0.95 \Omega$

$\therefore R_{eq} \displaystyle =\frac{1}{0.95} =\frac{20}{19} \Omega$

Five resistors each of 1 ohm are connected in parallel. The resultant resistance of the combination is :

0%
$5 \Omega$
0%
$1 \Omega$
0%
$0.2 \Omega$
0%
$2 \Omega$

Explanation

$\displaystyle R_{eq} = \frac{R}{5}$
Given

$R= 1 \Omega$

$R_{eq} =\frac{1}{5} = 0.2 \Omega$

$1kWh=$ _________?

0%
$3600000\ J$
0%
$10000\ J$
0%
$4.2\ J$
0%
$25000\ J$

Explanation

Kilowatt hour is the unit of energy to measure the amount of electricity used in an hour

$1\ kW$ in

$1\ hr$ .

Hence,

$1\ kWh = 1 kW \times 1\ hr$

$= 10^3 W \times 3600\ s$

$= 3600000\ J$

Kilowatt is the unit of electrical _______ but kilowatt-hour is the unit of electrical _______.

0%
Power, Energy
0%
Power, Electric potential
0%
Power, Current
0%
Energy, Power

Explanation

In SI system, unit of power is Watt (W). Kilowatt is equal to one thousand (

$10^3$ ) watts. The kilowatt-hour is a unit of energy and is equal to 1,000 watt-hours. If the energy is being transmitted or used at a constant rate (power) over a period of time, the total energy in kilowatt-hours is the product of the power in kilowatts and the time in hours.

What is one unit of household electrical energy in J?

0%
$1000\ J$
0%
$3600\ J$
0%
$0.6 \times 10^3 J$
0%
$3.6 \times 10^6 J$

Explanation

One unit of household is

$1\ kwh$

$kwh\ =$ kilo watt hour

$1\ \text{unit}$

$= 1\ kWh$

$\Rightarrow 1\ \text{unit} =10^3 W \times 3600\ s$

$\Rightarrow 1\ \text{unit} = 3.6 \times 10^6 J$

What is the maximum resistance one can make with ten

$1 \Omega$ resistors?

0%
$1 \Omega$
0%
$2 \Omega$
0%
$5 \Omega$
0%
$10 \Omega$

Explanation

Given ten

$1 \Omega$ resistors.
If all these resistors are connected in series a maximum value of resistance is obtained then

$R_{eq} = (1+1+1+1+1+1+1+1+1+1) \Omega = 10 \Omega$

Which of the following is false about fuse ?

0%
A fuse controls the current that flows to the appliance.
0%
A fuse increases the resistance in the circuit to minimize the current flow.
0%
A fuse allows high current to flow through the circuit without damaging the appliance.
0%
A fuse protects the wiring and the appliance in the circuit from overheating.

In which of the following situations will a fuse possibly melt?

0%
The earth wire is broken
0%
There is a short circuit in the electrical circuit
0%
The fuse is fixed along the neutral wire instead of the live wire
0%
The fuse wire is connected to earth wire

Two fuse wires of the same length are rated 5 A and 20 A.Which of the fuse wires is thicker and why ?

0%
5 A
0%
20 A
0%
Both have same thickness
0%
Cannot be same

An electric kettle is rated as

$2.5\ kW, \ 250\ V$ . Find the cost of running the kettle for two hours at

$60\ \text{paisa}/\text{unit}$ .

0%
$Rs\ 6$
0%
$Rs\ 3$
0%
$Rs\ 9$
0%
$Rs\ 1$

Explanation

$Rate=0.60 Rs/kWh$

$Energy=Power \times time$

Energy consumed,

$E =2.5 \times 2 = 5\ kWh$

Cost,

$C = 5 \times 0.60 = Rs\ 3$

Energy of the electric current is

0%
Thermal
0%
Nuclear
0%
Electrical
0%
Chemical

Explanation

$\textbf{Given:}$ Different energy forms.

$\textbf{To find:}$ Which form of energy is the energy of electric current.

We know electric current flows in a circuit due to potential differences. Electric current flows due to the work done on each charge to move it across the potential difference. The charges contain energy which gives them the capacity to do this work. The nature of this energy is electrical.

So option (C) is correct.

A current of

$3\ A$ flows through a resistance when it is connected across a 12V source. What should the current be if the resistance is increased by

$4\ \Omega$ and connected across the same voltage source?

0%
$1.5\ A$
0%
4 A
0%
1.6 A
0%
24 A

Explanation

$\underline{\text{First Case:}}$

Voltage

$V_1=12\ V$

Current

$I=3\ A$

Resistance =

$R$

From Ohm's law:

$12=3\times R$

$\therefore R=\dfrac{12}{3} = 4\ \Omega$

$\underline{\text{Second Case:}}$

Voltage

$V_2=12\ V$

Resistance =

$R+4=8\ \Omega$

From Ohm's law:

$12=I'\times 8$

$\therefore I'=\dfrac{12}{8}=1.5\ A$

The least resistance that can be obtained by combining resistance of

$6 \Omega, 3 \Omega$ and

$X \Omega$ is

$1 \Omega$ .
Then

$X$ is

0%
$3 \Omega$
0%
$2 \Omega$
0%
$4 \Omega$
0%
$6 \Omega$

Explanation

Here, we are given two resistance say

$R_{1} = 6\Omega$

$R_{2} = 3\Omega$

$R_{3} = x\Omega$

Equivalent Resistance (least)

$= 1\Omega$

We know that equivalent resistance of a combination is minimum when all the resistances are connected in parallel.

So,

$\dfrac {1}{R_{eq}} = \dfrac {1}{R_{1}} + \dfrac {1}{R_{2}} + \dfrac {1}{R_{2}}$

$\Rightarrow \dfrac {1}{1} = \dfrac {1}{6} + \dfrac {1}{3} + \dfrac {1}{x}$

$\Rightarrow \dfrac {1}{x} = 1-\dfrac {1}{6} - \dfrac{1}{3}$

$\Rightarrow \dfrac {1}{x} = \dfrac {1}{2}$

$\Rightarrow x = 2\Omega$

Hence value of

$x$ is

$2\Omega$ , option B is correct.

Which one of the following is different from the others?

0%
Joule
0%
Kilowatt hour
0%
Erg
0%
Watt

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct

Conventionally, the direction of the current is taken as the direction of flow

0%
of negative charges
0%
the direction of flow of atoms
0%
of molecules
0%
of positive charges

In the network given below, the current flowing through the resistance

$2 R$ is

0%
$\dfrac { E }{ R }$
0%
$\dfrac { E }{ 7R }$
0%
$\dfrac { 2E }{ 7R }$
0%
$\dfrac { 2E }{ R }$

Explanation

The combination of given resistors is a series combination of

$R$ and parallel combination of

$4R, 2R$ .

Resultant resistance of $4R$ and $2R$ :

$\dfrac{1}{R_p}=\dfrac{1}{4R}+\dfrac{1}{2R}$

$\Rightarrow R_p=\dfrac{(4R)(2R)}{2R+4R}=\dfrac{8R^2}{6R}=\dfrac{4R}{3}$

Resultant resistance of $R+(4R||2R)$

$R_{eq}=R+R_p$

$\Rightarrow R_{eq}= R+ \dfrac{4R}{3}= \dfrac{7R}{3}$

By Ohm's law,

Current supplied by the battery

$I=\dfrac{E}{R_{eq}}=\dfrac{3E}{7R}$

Voltage across the parallel branch is

$V_p = I \times R_p$

$\Rightarrow V_p=\left( \dfrac{3E}{7R} \right) \left(\dfrac{4R}{3}\right)=\dfrac{4E}{7}$

The voltage

$V_p$ would be same across

$4R$ as well as

$2R$ as they are connected across the same two points.

Hence, current across

$2R$ is

$I_{2R} = \dfrac{V_p}{2R} = \dfrac{\left( \dfrac{4E}{7}\right)}{2R}$

$I_{2R} = \dfrac{2E}{7R}$

Option C is correct.

State whether given statement is True or False
Conventional current always flows from a body having high electron density to body having low electron density.

0%
True
0%
False

A resistance of

$0.01\ \Omega$ is connected in parallel with a resistance of

$1\ k\Omega$ The resistance of the combination will be:

0%
$1000\ \Omega$
0%
$10\ \Omega$
0%
$1\ \Omega$
0%
$0.01\ \Omega$

Explanation

initial velocity,

$u = 10\ m/s$

Resistances,

$R_1 = 0.01\ \Omega$

$R_2 = 1K\ \Omega = 1000\ \Omega$

for 2 resistances, connected in parallel equivalent resistance,

$R_{eq}$

$\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} = \dfrac{1}{0.01} + \dfrac{1}{1000}$

$\dfrac{1}{R_{eq}} = \dfrac{100000 + 1}{1000}$

$R_{eq} = \dfrac{1000}{100001} = 0.009999$

$R_{eq} \sim 0.01\ \Omega$

Three resistors each of

$10 \Omega$ are connected in series to a battery of potential difference 150 v. The current flowing through it is ..... A.

0%
45
0%
5
0%
15
0%
20

Explanation

$V=IR$

Current is given by,

$V=I\times{R_{eq}}$

$I=\frac{V}{R_{eq}}$

The three resistance are in series. So, the equivalent resistance will be:

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

$R_{eq}=10+10+10$

$=30\ \Omega$

Substitute the values in the given formula:

$I=150/30$

$I=5$

$A$

If a man has five resistors each of value

$\dfrac { 1 }{ 5 } \Omega$ then is the maximum resistance he can obtain by connecting them, is :

0%
$1\Omega$
0%
$5\Omega$
0%
$\dfrac { 1 }{ 2 } \Omega$
0%
$\dfrac { 2 }{ 5 } \Omega$

Explanation

Given,

$R=\dfrac15\ Omega$ total number of resistance, n=5

For maximum effective resistance, all resistors should connect in series.

For series combination

$R_{eq}=R_1+R_2+ ..........+R_n\\R_{eq}=nR$ (All resistor have same resistance)

On putting value of

$n$ and

$R$

$R_{eq}=5\times \dfrac15=1\ \Omega$

Hence maximum resistance is 1 ohm.

Fuse should be connected to ..... wire of the circuit

0%
Earth
0%
Neutral
0%
Phase
0%
Any of the above

How much energy in kilowatt hour is consumed in operating ten 50 watt bulbs for 10 hours per day in a month (30 days)

0%
1500
0%
5000
0%
15
0%
150

Explanation

Energy consumed

$=10\times 50\times 10\times 30\times 3600 J$

$[1 Kwh=3600\times 1000J]$

$=\frac {10\times 50\times 10\times 30\times 3600}{3600\times 1000}kWh=150$

An 1800 W toaster, a 1.3 kW frying pan are plugged into the same

$20 A-120V$ line then

0%
fuse will not blow
0%
fuse will blow
0%
supply will spark
0%
fuse is just saved

Explanation

$I_{toaster}=\dfrac{1800}{120}= 15 A,$

$I_{frying pan}=\dfrac{1300}{120}= 10.83 A$

$I_{net}= 25.83 A > 20 A$

Because the applied current is more than the possible value of current which is given in the question, the fuse will blow.

An electric heater, operating at

$220\ V$ , boils

$5\ l$ of water in

$5\ \text{minutes}$ . If it is used on a

$110\ V$ line, it will boil the same amount of water in:

0%
$10\ \text{minutes}$
0%
$20\ \text{minutes}$
0%
$5\ \text{minutes}$
0%
$1\ \text{minute}$

Explanation

Heat required to boil the water will be same in the two cases.

Heat produced by heater is:

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

$V^2 = \cfrac{HR}{t}$
Hence, other things remaining constant,

$V^2 \propto \dfrac{1}{t}$

Let the first case be labeled by subscript 1 and the second by subscript 2. Then,

$\cfrac{t_2}{t_1} = \cfrac{V_1^2}{V_2^2}$

$t_2 = \dfrac{220^2 \times 5}{110^2}$

$t_2 = 20\ min$

An electric bulb is rated '100 W, 250 V' . What is the resistance of bulb ?

0%
2.5 ohm
0%
625 ohm
0%
25000 ohm
0%
1500 ohm

Explanation

Generally electrical appliances such as an electric bulb, geyser, heater, etc. are rated with power and voltage. If a bulb is rated 100W - 250V, it means that when a bulb is lit on a 250V supply, it consumes 100 W of electric power. This means that, 100 Joules of electrical energy gets converted into heat and light energy in 1 second.
From this rating we can calculate the resistance of the filament of the appliance while it is glowing and also the current which flows through it.

$Since\quad P=\frac { { V }^{ 2 } }{ R } ,\quad R=\frac { { V }^{ 2 } }{ P } .$
For the bulb rated as 100 W - 250 V, the resistance of its filament

$R=\frac { { V }^{ 2 } }{ P } =\frac { 250^{ 2 } }{ 100 } =625\Omega .$
The resistance of the filament of a bulb is much less than this value when it is not lit because the resistance of a filament increases with the increasing temperature.

Choose the correct option.

0%
$kW=$ unit of power and $kWh =$ unit of energy.
0%
$kW=$ unit of energy and $kWh =$ unit of power.
0%
$kW$ and $kWh$ are both of unit of energy.
0%
$kW$ and $kWh$ are both of unit of power.

Explanation

Electrical Power is measured in watts (

$W$ ), kilo-watts (

$kW$ ), mega-watts (

$MW$ ), and giga-watts (

$GW$ ).

Electrical Energy is measured in kilowatt-hours (

$kWh$ ), megawatt-hours, and sometimes (less commonly) in watt-hours or watt-seconds.

A given wire of resistance

$1\,\Omega$ is stretched to double its length. What will be its new resistance?

0%
$1\, \Omega$ .
0%
$8\, \Omega$ .
0%
$2\, \Omega$ .
0%
$4\, \Omega$ .

Explanation

When we stretch a wire to make its length double or whatever the size, to conserve volume, its cross-sectional area must decrease. For a common observation, when we stretch something its length increases but thickness decreases. Now the tricky part is in the question we are being told the new length but not the cross-sectional area, so we have to find it first.

Let's say a wire having length L cross-sectional area A has resistance R. If it is so stretched that its length becomes 2L so new length L' = 2L and when the length is doubled by stretching the cross-sectional area decreases by half.

The resistance of the wire is expressed as

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

rho is the resistivity of the wire,

L is the length of the wire,

A is the area of the cross-section of the wire.

The new resistance is calculated as follows.

$R=\dfrac { \rho 2L }{ \dfrac { A }{ 2 } } \quad =\dfrac { 4\rho L }{ A } \quad$ .

That is, the new resistance is four times the old resistance.

Hence, the new resistance is

$1\times 4=4\quad ohms$ .

The filament of a bulb takes a current 0.1 A when potential difference across it is 0.2 V. When the potential difference across it becomes 1.0 V, the current becomes 0.4 A. The resistance of filament in each case will be

0%
$2.0\,\Omega$ , $2.5\,\Omega$
0%
$2.5\,\Omega$ , $2\,\Omega$
0%
$1.0\,\Omega$ , $2.5\,\Omega$
0%
$2.0\,\Omega$ , $2\,\Omega$

Explanation

Ohm's law states that the current through a conductor between two points is directly proportional to the potential difference across the two points. Introducing the constant of proportionality, the resistance, we arrive at the usual mathematical equation that describes this relationship:

$I = \dfrac{V}{R}$ , where, I is the current through the conductor in units of amperes, V is the potential difference measured across the conductor in units of volts, and R is the resistance of the conductor in units of ohms.

Therefore,

$R=V/I.$

In the first case, the voltage is o.2 V and the current is 0.1 A. So, the resistance is 2 ohms. In the second case, the voltage is 1 V and the current is 0.4 A. So, the resistance is 2.5 ohms.

The current in the wire is due to the drifting of electrons inside a wire in a direction opposite to the flow of electrons. During their drifting, they collide with their atoms vibrating about their mean position and lose some of the kinetic energy to the vibrating atoms which increase the amplitude of the vibration of the atoms, and the thermal energy of the metal increases and hence we observe the corresponding rise in temperature of the wire.

An electric heater kept in a vacuum is heated continuously by passing an electric current. Its temperature:

0%
It will go on rising with time
0%
It will stop after some time as it will lose heat to the surroundings by conduction
0%
It will rise for sometime and thereafter will starts falling
0%
It will become constant after sometime because of loss of heat due to radiation

Explanation

Treating initially that there is no radiation and on applying a voltage to the heater, the temperature of the heater rises. As the temperature rises, the radiation also increases.
As the input power(

$i^2r$ ) is constant, the output power should also be constant.
Initially, the temperature will be increasing and radiation as well. When the heater reaches a certain temperature, the radiation power equals the input power and the temperature of the heater stops rising.
Thus temperature becomes constant after some time.

The equivalent resistance R of three resistors

$R_1$ ,

$R_2$ and

$R_3$ joined in parallel is

0%
$\displaystyle R= R_1+R_2+R_3$
0%
$\displaystyle R= 1/R_1+1/R_2+1/R_3$
0%
$\displaystyle R= \frac{R_1R_2R_3}{R_1R_2+R_2R_3+R_3R_1}$
0%
$\displaystyle R=1/(R_1+R_2+R_3)$

Explanation

Equivalent resistance

$R$ is given by:

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

$\Rightarrow\ R=\dfrac{R_1R_2R_3}{R_1R_2+R_2R_3+R_3R_1}$

A uniform wire of resistance

$50\Omega$ is cut into 5 equal parts. These parts are now connected in parallel. The equivalent resistance of the combination is

0%
$2\Omega$
0%
$10\Omega$
0%
$250\Omega$
0%
$6250\Omega$

Explanation

Here, a uniform wire of resistance $50 \Omega$ is cut into $5$ equal parts. Hence, resistance of each part is
$r = \dfrac{R}{n}$

$\Rightarrow r = \dfrac{50}{5}$

$\Rightarrow r= 10\ \Omega$
Now, the parts are connected in parallel, the equivalent resistance of the combination is
$\dfrac{1}{R_{eq}} = \dfrac{1}{10} + \dfrac{1}{10} + \dfrac{1}{10} + \dfrac{1}{10} + \dfrac{1}{10}$
$\dfrac{1}{R_{eq}} = \dfrac{5}{10} = \dfrac{1}{2}$
$R_{eq} = 2\Omega$

In our household applications, commercial unit of electricity is used. One unit is equal to:

0%
$1\ kW$
0%
$1\ kWh$
0%
$1\ kW/h$
0%
$1\ Wh$

What is the approximate resistance setting of a rheostat in which

$650 \ mA$ of current flows with a

$150 \ V$ source?

0%
$230 \ \Omega$
0%
$56 \ \Omega$
0%
$15 \ \Omega$
0%
$13 \ \Omega$

Explanation

Given:

Current

$I=650 \ mA=650 \times 10^{-3} \ A$

Voltage

$V=150 \ V$

By Ohm's Law,

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

$\implies R=\dfrac{150}{650\times 10^{-3}}$

$\implies R\approx 230 \ \Omega$

Answer: Option (A)

In the circuit shown below in Fig., the value of x if the equivalent resistance between A and B is

$4\, \Omega$ will be:

0%
$2\, \Omega$
0%
$1\, \Omega$
0%
$1.5\, \Omega$
0%
$3\, \Omega$

Explanation

In this case, the 2 resistors of 4 ohms and 8 ohms are connected in series and the combined resistance is given as 4+8 = 12 ohms.
The 2 resistors of x ohms and 5 ohms are connected in series and the combined resistance is given as x+5 ohms.
The above 2 sets of resistors are connected in parallel. Thus, the total resistance is expressed as

$\frac { 1 }{ 12 } +\frac { 1 }{ x+5 } =\frac { 1 }{ 4 } as\quad the\quad total\quad resistance\quad is\quad given\quad as\quad 4\quad ohms\quad in\quad the\quad question.$
Solving for x in the above equation we get x=0.99. That is. 1 ohm approximately.
Hence, the value of the unknown resistance in the circuit is 1 ohm.

When two resistors with resistances

$R_1$ and

$R_2$ are connected in series, then the total equivalent resistance is

0%
$\dfrac{R_1R_2}{R_1+R_2}$
0%
$R_1+R_2$
0%
$1/R_1+1/R_2$
0%
$R_2-R_1$

Explanation

Components connected in series are connected along a single path, so the same current flows through all of the components. The current through each of the components is the same, and the voltage across the circuit is the sum of the voltages across each component. In a series circuit, every device must function for the circuit to be complete. One bulb burning out in a series circuit breaks the circuit. A circuit composed solely of components connected in series is known as a series circuit.
The total resistance of resistors in series is equal to the sum of their individual resistances. That is,

${ R }_{ total }={ R }_{ 1 }+{ R }_{ 2 }$ . The current is given as

$I={ I }_{ 1 }={ I }_{ 2 }$ .
Below is the diagrammatic representation of 2 resistors

${ R }_{ 1 }\quad and\quad { R }_{ 2 }$ connected in series.

Equivalent resistance is more than either of the two resistances when they are in:

0%
Series
0%
Series and parallel
0%
Parallel
0%
none of these.

Explanation

Equivalent resistance is more than either of the two resistances when they are connected in series.Components connected in series are connected along a single path, so the same current flows through all of the components. The current through each of the components is the same, and the voltage across the circuit is the sum of the voltages across each component. In a series circuit, every device must function for the circuit to be complete. One bulb burning out in a series circuit breaks the circuit. A circuit composed solely of components connected in series is known as a series circuit.The total resistance of resistors in series is equal to the sum of their individual resistances.

Hence, the equivalent resistance is more than the individual resistances because a sum is taken of all the individual resistances.

That is,

${ R }_{ total }={ R }_{ 1 }+{ R }_{ 2 }$ . The current is given as

$I={ I }_{ 1 }={ I }_{ 2 }$ .
Diagrammatic representation of 2 resistors

${ R }_{ 1 } \ and \ { R }_{ 2 }$ connected in series is given.

In the figure the equivalent resistance of the following combination of resistors

$r_1,\, r_2,\, r_3$ and

$r_4$ , if

$r_1\, =\, r_2\, =\,r_3\, =\, r_4\, =\, 2.0\, \Omega$ , between the points A and B is

0%
$2.0\, \Omega$
0%
$4.0\, \Omega$
0%
$6.0\, \Omega$
0%
$5.0\, \Omega$

Explanation

In this case, the first two resistors of 2 ohms each are connected in series and the third and fourth resistors of 2 ohms each are connected in parallel. The first set is connected in series with the second set of resistors.
The resistances R3 and R4 that are connected in parallel have a resistance given as

$\frac { 1 }{ 2 } +\frac { 1 }{ 2 } =\frac { 1 }{ 1 } =1\quad ohm$ . That is, these two resistances is reduced to a single resistance of 1 ohm.
The resistances R1 and R2 that are connected in series have a resistance given as 2+2 = 4 ohms.
This combined resistance when connected in series with reduced resistance of 1 ohm gives a total resistance of 4+1 = 5 ohms.
Hence, the total resistance in the circuit is 5 ohms.

Four resistors each of resistance

$2\, \Omega$ are connected in parallel. What is the effective resistance?

0%
$2\, \Omega$
0%
$5\, \Omega$
0%
$0.5\, \Omega$
0%
$8\, \Omega$

Explanation

$\textbf{Step 1: Calculation of net resistance}$

For parallel arrangement,

$\dfrac { 1 }{ { R }_{ eq } } =\dfrac { 1 }{ { R }_{ 1 } } +\dfrac { 1 }{ { R }_{ 2 } } +\dfrac { 1 }{ { R }_{ 3 } } +\dfrac { 1 }{ { R }_{ 4 } }$ .

Resistance of every resistor

$=2 \Omega$

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

$\Rightarrow R_{eq}=\dfrac { 1 }{ 2 } =0.5\quad \Omega$ .

Hence, Option C is correct.

2105515_173912_ans_a589aeac73124a57add5d5e8922602f9.png

A wire of uniform thickness with a resistance of

$27\, \Omega$ is cut into three equal pieces and they are joined in parallel. The equivalent resistance of the parallel combination will be :

0%
$3\, \Omega$
0%
$9\, \Omega$
0%
$27\, \Omega$
0%
$49\, \Omega$

Explanation

In this case, a wire of uniform thickness with a resistance of 27 is cut into three equal pieces and they are joined in parallel. As we know that the resistance is directly proportional to the length of the wire,

$R\propto L$ , so when the wire is cut into 3, the resistance of the cut wire is reduced to one third of the total resistance. That is, the resistance of each cut wire becomes 27/3= 9 ohms.
Now, when these 3 wires are connected in a parallel connection, the total resistance is given as

$\frac { 1 }{ 9 } +\frac { 1 }{ 9 } +\frac { 1 }{ 9 } =\frac { 1 }{ 0.333 } =\quad 3\quad ohms$ .
Hence, the total resistance is now 3 ohms only.

The energy consumed by

$100 \ W$ electric bulb in

$5 \ hours$ is:

0%
$0.5 \ kWh$
0%
$1.5 \ kWh$
0%
$5 \ kWh$
0%
$2.5 \ kWh$

Explanation

Given, power of the electric bulb,

$P =100 \:W=\dfrac{100}{1000} \:kW=0.1 \:kW$ ;

time for which bulb is used,

$t =5 \:h$

$\displaystyle P =\frac{W}{t}$ or

$W =Pt$

$W =0.1 \:kW \times 5 \:h =0.5 \:kWh$

Which of above combinations have equivalent resistance

$1\Omega$ between X and Y?

0%
(a)
0%
(b)
0%
(c)
0%
(d)

Explanation

Answer is option A and D.
Components connected in parallel are connected so the same voltage is applied to each component. In a parallel circuit, the voltage across each of the components is the same, and the total current is the sum of the currents through each component.
To find the total resistance of all components, the reciprocals of the resistances of each component is added and the reciprocal of the sum is taken. The total resistance will always be less than the value of the smallest resistance. That is,

$\frac { 1 }{ { R }_{ total } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } \quad that\quad is,\quad R={ { R }_{ total } }^{ -1 }$ . The voltage is given as

$V={ V }_{ 1 }={ V }_{ 2 }$ . The total resistance is always lesser than the least resistance.
In the figure (a), the resistances two 2 ohms resistors are connected in parallel and the total resistance is

$\frac { 1 }{ R } =\frac { 1 }{ 2 } +\frac { 1 }{ 2 } =\frac { 1 }{ 1 } .\quad That\quad is\quad R=1\quad ohm.$
In the figure (b), the resistances two 1 ohm and a 2 ohms resistors are connected in parallel and the total resistance is

$\frac { 1 }{ R } =\frac { 1 }{ 1 } +\frac { 1 }{ 1 } +\frac { 1 }{ 2 } =\frac { 1 }{ 2.5 } .\quad That\quad is\quad R=0.4\quad ohm.$
In the figure (c), the resistances two 1 ohm resistors are connected in parallel and the total resistance is

$\frac { 1 }{ R } =\frac { 1 }{ 1 } +\frac { 1 }{ 1 } =\frac { 1 }{ 2 } .\quad That\quad is\quad R=0.5\quad ohm.$
In the figure (d), the resistances two pairs 1 ohm resistors are connected in parallel. a pair of 1 ohm resistors form 2 ohms resistors as they are serially connected. Hence, the total resistance is

$\frac { 1 }{ R } =\frac { 1 }{ 2 } +\frac { 1 }{ 2 } =\frac { 1 }{ 1 } .\quad That\quad is\quad R=1\quad ohm.$
Hence, the figures a and d have a total resistance of 1 ohm each.

Two resistances are connected in two configurations- series or parallel. Which represents the parallel combination?

0%
A
0%
A and B both
0%
B
0%
None of these.

You have three resistors of values

$2\, \Omega,\, 3\, \Omega$ , and

$5\, \Omega$ . If they are in parallel ,then the total resistance is :

0%
$0.97\, \Omega$
0%
$2 \, \Omega$
0%
$10 \, \Omega$
0%
$3\, \Omega$

Explanation

Components connected in parallel are connected so the same voltage is applied to each component. In a parallel circuit, the voltage across each of the components is the same, and the total current is the sum of the currents through each component. In parallel circuits, each light has its own circuit, so all or one light could be burned out, and the last one will still function. A circuit composed solely of components connected in parallel is known as a parallel circuit.
To find the total resistance of all components, the reciprocals of the resistances of each component is added and the reciprocal of the sum is taken. Total resistance will always be less than the value of the smallest resistance That is,

$\frac { 1 }{ { R }_{ total } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } +\frac { 1 }{ { R }_{ 3 } } +\frac { 1 }{ { R }_{ 4 } } \quad that\quad is,\quad R={ { R }_{ total } }^{ -1 }$ .
In this case, we have three resistors of values 2 ohms, 3 ohms and 5 ohms. For the resistance to be less than 1 ohm, the resistors should be connected in parallel.
Therefore,

$\frac { 1 }{ { R }_{ total } } =\frac { 1 }{ 2 } +\frac { 1 }{ 3 } +\frac { 1 }{ 5 } =\frac { 1 }{ 1.03 } =0.97\quad \Omega$ .
Hence, the resistors are connected in parallel and their equivalent resistance is 0.97 ohms.

Two resistors of resistance

$4\, \Omega$ and

$6\, \Omega$ are connected in parallel to a cell to draw 0.5 A current from the cell. What is current through resistor

$4\, \Omega$ ?

0%
0.3 A
0%
0.5 A
0%
0.2 A
0%
1.2 A

Explanation

In a parallel resistance system current in 1st branch =

${ I }_{ 1 }=\left( \frac { { R }_{ 2 } }{ { R }_{ 1 }+{ R }_{ 2 } } \right) \times I\\$

${ I }_{ 1 }=\frac { 6 }{ 4+6 } \times 0.5\\$

${ I }_{ 1 }=\frac { 3 }{ 10 } =0.3A$

Find out the total power supplied by the 60-V source in the circuit shown in figure.

0%
360 W
0%
200 W
0%
400 W
0%
180 W

Explanation

First we will find the equivalent resistance of the circuit.

The

$7\Omega$ and

$5\Omega$ are in series so equivalent is

$7+5=12\Omega$ .

$12\Omega$ and

$6\Omega$ are in parallel, so equivalent is

$12\times6/(12+6)=4\Omega$ .

These

$4\Omega$ and

$12\Omega$ are in parallel, so their equivalent is

$12\times4/(12+4)=3\Omega$ .

This

$3\Omega$ and

$7\Omega$ are in series so equivalent is

$3+7=10\Omega$ .

So the equivalent resistance of the circuit is

$R_{eq}=10\Omega$ .

Thus the power supplied is

$\dfrac{V^2}{R_{eq}}=60^2/10=360\ W$

Two resistors of resistance

$4\, \Omega$ and

$6\, \Omega$ are connected in parallel to a cell to draw

$0.5 A$ current from the cell. What is the current through resistor

$6\, \Omega$ ?

0%
$0.2 \ A$
0%
$0.5 \ A$
0%
$2.4 \ A$
0%
$1.2 \ A$

Explanation

Two resistors of resistance

$4 \ Ω$ and

$6 \ Ω$ are connected in parallel.

Hence, equivalent resistance in parallel is given by,

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

$\Rightarrow \dfrac{1}{R_{eq}} = \dfrac{1}{4} + \dfrac{1}{6} = \dfrac{10}{24}$

$\Rightarrow R_{eq} = \dfrac{24}{10}= 2.4 \ \Omega$

Using Ohm's law,

$V=iR$

$V=0.5\times2.4=1.2 \ V$

Since, value of voltage is same for objects connected in parallel, Hence, voltage drop across

$6 \ \Omega$ resistor is also

$1.2\ V$

Hence, using Ohm's law , Current through resistor

$6 \ Ω$ :

$1.2=i\times6$

$i=0.2 \ A$

2106174_174091_ans_900459579ded4f159b9945852d36b55a.png

CBSE Questions for Class 10 Physics Electricity Quiz 5 - MCQExams.com

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Practice Class 10 Physics Quiz Questions and Answers

CBSE Questions for Class 10 Physics Electricity Quiz 5 - MCQExams.com

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Practice Class 10 Physics Quiz Questions and Answers

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