MCQ Questions for Class 10 Physics Electricity Quiz 6

The figure shows two bulbs with a resistance of

$10\ \Omega$ with switches and a fuse connected to mains through a socket. How are the two bulbs joined and what is net resistance?

0%
In parallel, $20\ \Omega$
0%
in series, $20\ \Omega$
0%
In parallel, $5\ \Omega$
0%
In series , $5\ \Omega$

Explanation

The two bulbs are connected in parallel.

Given,

Resistance of each bulb,

$R_1=R_2=R=10\ \Omega$

For two resistors connected in parallel

Equivalent resistance,

$R_{eq}$

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

$\dfrac{1}{R_{eq}}=\dfrac{1}{10}+\dfrac{1}{10}$

$\dfrac{1}{R_{eq}}=\dfrac{2}{10}=\dfrac 15$

$R_{eq}=5\ \Omega$

The power supplied by the battery will be :

0%
15 W
0%
24 W
0%
3.6 W
0%
20 W

Explanation

In this case, the resistances 4 ohms and 6 ohms are connected in parallel.
The total resistance is

$\dfrac { 1 }{ 4 } +\dfrac { 1 }{ 6 } =\dfrac { 1 }{ 0.416 } =2.4\quad ohms$ . The voltage is given as 6 V.
The power is calculated as follows.

$P=VI\quad =\quad \dfrac { { V }^{ 2 } }{ R } (as\quad I=\dfrac { V }{ R } )$ .
Therefore,

$P=\quad \dfrac { { 6 }^{ 2 } }{ 2.4 } \quad =\quad 15W$ .
Hence, the power supplied by the battery is 15 W.

What is the current through an electrical appliance of rating 5 kW, 200 V and can you use a fuse which is rated 8 A ?

0%
25 A, yes
0%
2.5 A, yes
0%
25 A, no
0%
2.5 A, no

Explanation

Fuses are safety devices that are to be built into our electrical system. If there were no fuses and we operated too many appliances on a single circuit, the cable carrying the power for that circuit would get extremely hot, short circuit, and possibly start a fire. To prevent electrical overloads, fuses are designed to trip or blow, stopping the flow of current to the overloaded cable.
The fuse must always be connected to the mains and it must be of correct value. For example, a 15-ampere fuse should trip when the current through it exceeds 15 amperes. A 20-ampere fuse should blow when the current through it exceeds 20 amps.
In this case, let us consider a device of power 5 kW, that is, 5000 W and 200 V.

The current through the appliance is given as

$I=P/V.$

That is,

$I=\dfrac { P }{ V } =\dfrac { 5000 }{ 200 } =25A$ .

The fuse of 8 A cannot be used because the current through the appliance is

greater the fuse rating. When the 25 A current flows through the fuse it will

blow off and it will not be able to fulfill it purpose.

Hence, a fuse is rated 8 A cannot be used with this appliance.

The power dissipated in each resistor will be :

0%
$P_A= 6 W, P_B=9 W$
0%
$P_A= 9 W, P_B=6 W$
0%
$P_A= 9 W, P_B=3 W$
0%
$P_A= 3 W, P_B=6 W$

Explanation

The voltage across the circuit is given as 6 V.
So the power dissipated by the 4 ohms resistor is given as

$P=\dfrac { { V }^{ 2 } }{ R } =\dfrac { { 6 }^{ 2 } }{ 4 } =\quad 9W$ . And,
the power dissipated by the 6 ohms resistor is given as

$P=\dfrac { { V }^{ 2 } }{ R } =\dfrac { { 6 }^{ 2 } }{ 6 } =\quad 6W$ .
Hence, the power dissipated by the 4 ohms and 6 ohms resistors are 9 W and 6 W.

A particular resistance wire has a resistance of 3.0 ohm per metre. The total resistance of three lengths of this wire each 1.5 m long, joined in parallel will be :

0%
$2.5\, \Omega$
0%
$3\, \Omega$
0%
$4.5\, \Omega$
0%
$1.5\, \Omega$

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 but 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,

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

In this case, the resistance of one meter wire is 3 ohms. So the resistance of 1.5 meters of the same wire is given as

$\dfrac { 3\times 1.5 }{ 1 } =4.5\quad ohms$ .

When three such wires of 1.5 meters length and resistance of 4.5 ohms are joined in parallel the resistance is given as

$\dfrac { 1 }{ { R }_{ total } } =\dfrac { 1 }{ 4.5 } +\dfrac { 1 }{ 4.5 } +\dfrac { 1 }{ 4.5 } \quad =\dfrac { 1 }{ 0.666 } =1.5\quad ohms$ .

Hence, the total resistance is given as 1.5 ohms.

What we used to protect the electric circuits from over loading and short circuit ?

0%
capacitance
0%
inductance
0%
fuse wire
0%
filament wire.

The rating of a fuse connected in the general household lighting circuit is:

0%
15 A
0%
5 A
0%
10 A
0%
zero

Explanation

General household wire has a current supply of 5 ampere. Hence fuse also have that. hence The rating of a fuse connected in the general household lighting circuit is

$5A$

Three resistors of

$6.0 \Omega, \, 2.0 \Omega$ and

$4.0 \Omega$ are joined to an ammeter A and a cell of e.m.f. 6.0 V as shown in figure. The effective resistance of the circuit is :

0%
$6.0\, \Omega$
0%
$2.0\, \Omega$
0%
$3.0\, \Omega$
0%
$12.0\, \Omega$

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,

${ R }_{ total }={ R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }$ .
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 but 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,

$\dfrac { 1 }{ { R }_{ total } } =\dfrac { 1 }{ { R }_{ 1 } } +\dfrac { 1 }{ { R }_{ 2 } } +\dfrac { 1 }{ { R }_{ 3 } } \quad that\quad is,\quad R={ { R }_{ total } }^{ -1 }$ .
In this case, the resistance

$R_2$ and

$R_3$ are connected in series and the combined resistance is connected in parallel with the resistance

$R_1.$
Therefore, the combined resistance of

$R_2$ and

$R_3$ is given as

$4+2 = 6\Omega$ . The total resistance in the circuit is given as

$\dfrac{ 1 }{ 6 } +\dfrac{ 1 }{ 6 } =\dfrac{ 1 }{ 0.33 } \ =\ 3\ \Omega$ .

Hence, the effective resistance of the circuit is 3 Ohms.

If a bulb of 60 W is connected across a source of 220 V, the current drawn by it is:

0%
0.3 A
0%
0.27 A
0%
3.67 A
0%
0.5 A

Explanation

The electric power in watts associated with a complete electric circuit or a circuit component represents the rate at which energy is converted from the electrical energy of the moving charges to some other form, e.g., heat, mechanical energy, or energy stored in electric fields or magnetic fields. The power is given by the product of applied voltage and the electric current. That is, P=VI. The power of the bulb is given as 60 W and the voltage is 220 V.
The current drawn in calculated as follows.

${ I=\dfrac { P }{ V } =\dfrac { 60W }{ 220V } }=0.2727A$
Hence, the current drawn by the bulb is 0.2727 A.

How many electrons constitute a current of one microampere?

0%
$6.25 \times 10^6$
0%
$6.25 \times 10^{12}$
0%
$6.25 \times 10^9$
0%
$6.25 \times 10^{15}$

Explanation

1 microampere=

$10^{-6}A=10^{-6}C/s$

Hence, charge flowing per second=

$q=10^{-6}C$

Hence, number of electrons=

$N=\dfrac{q}{e}=\dfrac{10^{-6}}{1.6\times 10^{-19}}$

$\implies N=6.25\times 10^{12}$

Answer-(B)

Two wires of same metal have the same length but their cross-sections are in the ratio 3:They are joined in series. The resistance of the thicker wire is

$10\Omega$ . The total resistance of the combination is:

0%
$\displaystyle \frac {5}{2} \Omega$
0%
$\displaystyle \frac {40}{3} \Omega$
0%
$40\Omega$
0%
$100\Omega$

Explanation

Resistance of a wire $R=\dfrac{\rho l}{A}$ where $\rho=$ resistivity, $l=$ length, $A=$ cross section of the wire.
As both have same material and length so $R \propto \dfrac{1}{A}$
Thus, $\dfrac{R_2}{R_1}=\dfrac{A_1}{A_2}=\dfrac{3}{1} \Rightarrow R_2=3R_1$ .
here $R_1$ is the resistance of thicker wire so its resistance $R_1=10 \Omega$ (given)
so, $R_2=3(10)=30 \Omega$
As they are connected in series so the equivalent resistance is
$R_{eq}=R_1+R_2=10+30=40 \Omega$

There are two wires, A and B, made of same material. Both the wires have the same length. It is observed that the resistance of wire A is four times the resistance of wire B, then find the ratio of their cross sectional areas.

0%
$4:1$
0%
$1:4$
0%
$1:2$
0%
$2:1$

Explanation

We know that

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

Given the material of both wires A and B is same and their lengths are also given to be same.

$\rho and l$ are constant.

So, we have the relation:

$\dfrac{R_A}{R_B} = \dfrac{A_B}{A_A} = \dfrac{4}{1}$

So,

$\dfrac{A_A}{A_B} = \dfrac{1}{4}$

The equivalent resistance for the '

$n$ ' resistors with resistance

$R$ connected in parallel is

0%
$nR$
0%
$R$
0%
$\dfrac{R}{n}$
0%
$nR^2$

Explanation

Series circuit:
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,

${ R }_{ total }={ R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }$ .
Parallel circuit:
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 but 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,

$\dfrac { 1 }{ { R }_{ total } } =\dfrac { 1 }{ { R }_{ 1 } } +\dfrac { 1 }{ { R }_{ 2 } } +\dfrac { 1 }{ { R }_{ 3 } } \quad that\quad is,\quad R={ { R }_{ total } }^{ -1 }$ .

In an electric circuit containing a battery, the charge (assumed positive) inside the battery

0%
always goes from the positive terminal to the negative terminal.
0%
may go from the positive terminal to the negative terminal.
0%
always goes from the negative terminal to the positive terminal.
0%
does not move.

A technician has

$10$ resistors each of resistance

$0.1\ \Omega$ . The largest and smallest resistance that he can obtain by combining these resistors are:

0%
$10\ \Omega \ \text{and}\ 1\ \Omega$
0%
$1\ \Omega \ \text{and}\ 0.1\ \Omega$
0%
$1\ \Omega \ \text{and}\ 0.01\ \Omega$
0%
$0.1\ \Omega \ \text{and}\ 0.01\ \Omega$

Explanation

Given,

Number of resistors,

$n=10$

Resistance of each resistors,

$R=0.1\ \Omega$

For the largest resistance, all ten resistors should be connected in series.

The effective resistance of

$n$ resistors having equal resistances

$R$ in series,

$R_s=n\times R$

$R_s=10\times 0.1$

$R_s=1\ \Omega$

For the smallest resistance, all ten resistors should be connected in parallel.

The effective resistance of

$n$ resistors having equal resistances

$R$ in parallel,

$R_p=\dfrac{R}{n}$

$R_p=\dfrac{0.1}{10}$

$R_p=0.01\ \Omega$

An electric toaster draws

$8\ A$ current in a

$220\ V$ circuit. It is used for

$2\ hr$ . What is the cost of operating the toaster if the cost of electrical energy is

$Rs.\ 4.5 / kWh$ ?

0%
$Rs.\ 9.50$
0%
$Rs.\ 25.50$
0%
$Rs.\ 14.84$
0%
$Rs.\ 15.84$

Explanation

Given,

$i=8\ A\\E=220\ v\\t=2\ hr$

Energy dissipated is given by:

$E = E\ i\ t$

$=220 \times 8 \times 2\ Wh$

$=3.52\ kWh$

The cost per kWh is

$Rs.\ 4.50$

Hence, the cost of running the kettle for two hours is:

$3.52 \times 4.50 = Rs.\ 15.84$

Fill in the blank.

The heat generated in calorie is equal to _________.

0%
$VIt$
0%
$(VIt)/4.2$
0%
$(V^2Rt)/4.2$
0%
$VI^2t$

Explanation

The right matches are given below.
Heat Generated

$\rightarrow$ Propotional to the square of current

$\rightarrow$

$\frac { VIt }{ 4.18 } cal$
Resistance in parallel

$\rightarrow$ Is used to reduce effective resistance in a circuit

$\rightarrow$

$\frac { 1 }{ { R }_{ p } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } }$
Resistivity

$\rightarrow$ Depends on the material of the conductor

$\rightarrow$

$\rho =\frac { RA }{ l }$
Ohm's law

$\rightarrow$ Gives relation between V and I

$\rightarrow$ V=IR

An electric heater of power

$3 \ kW$ is used for

$10 \ h$ . How much energy does it consume? Express your answer in joule.

0%
$1.08\times 10^5$
0%
$1.08\times 10^8$
0%
$1.08\times 10^7$
0%
$1.08\times 10^3$

Explanation

We know that

$Power = work/time$

$Work \ or\ energy = Power\times time$

$Work = 3 kW \times 10 h$

$Energy = 30 KWh$ .

$Energy = 30\times 10^{3}\times 60\times 60\times =1.08\times 10^{8}\ J$ .

A bulb is connected to a cell. How is the resistance of circuit affected if another identical bulb is connected in series ?

0%
Resistance is doubled
0%
Resistance is same.
0%
Resistance is tripled
0%
Resistance is one half.

Explanation

${ R }_{ total }={ R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }$ .
When a bulb of the same resistance is connected in the series, the total resistance in the circuit is given as sum of both resistances connected. Hence the resistance is doubled.

Substances which allow electricity to pass through them are known as

0%
Electrical conductors
0%
Electrical insulators
0%
Both conductors & insulators
0%
None of these

State whether true or false.

Bulbs C and D are connected in series.

0%
True
0%
False

The chemical reaction due to passage of electric current depends on:

0%
electrodes.
0%
magnitude of current.
0%
density of liquid.
0%
all of the above.

Two wires of equal length, one of copper and the other of manganin (an alloy) have the same thickness. Which one can be used for electrical heating devices? Why?

0%
Manganin because being an alloy it has more resistivity than copper
0%
Copper because it has more conductivity.
0%
Both because both are good conductors of heat.
0%
None of them can be used as heating device.

The given figure is a 'material tester'.
Which of the following objects will make the bulb glow when put on the gap shown?

If we touch a naked current carrying wire, we get a shock. This is because our body is a :

0%
conductor of electricity
0%
insulator of electricity
0%
source of electricity
0%
both B&C

In a cell, by convention, the current is taken to be flowing from :

0%
positive electrode to the negative electrode
0%
negative electrode to the positive electrode
0%
both A and B
0%
none of the above

Three unequal resistors in parallel are equivalent to a resistant 1 ohm. If two of them are in the ratio 1 :2 and if no resistance value is fractional, the largest of the three resistance in ohm is :-

0%
4
0%
6
0%
8
0%
12

Explanation

$1={\dfrac {1}{R_1}}+{\dfrac {1}{R_2}}+{\dfrac {1}{R_3}}$
${\dfrac {R_1}{R_2}}={\dfrac {1}{2}}$ (Given)
$R_2=2R_1$
$1={\dfrac {2}{R_2}}+{\dfrac {1}{R_2}}+{\dfrac {1}{R_3}}$
$1={\dfrac {3}{R_2}}+{\dfrac {1}{R_3}}$
$1-{\dfrac {3}{R_2}}={\dfrac {1}{R_3}}$
${\dfrac {{R_2}-3}{R_2}}={\dfrac {1}{R_3}}$
${R_3}={\dfrac {R_2}{{R_2}-3}}$
Now only $R_2=6 \Omega$ satisfies this above equation $\Rightarrow R_3=2 \Omega$ & $R_1=3 \Omega$

Two 220 volt, 100 watt bulbs are connected first in series and then in parallel. Each time the combination is connected to a 220 volt a.c.supply line. The power drawn by the combination in each case respectively will be

0%
50 watt, 100 watt
0%
100 watt, 50 watt
0%
200 watt, 150 watt
0%
50 watt, 200 watt

Explanation

Resistance of the 220V, 100W bulb is given by

$R=\quad \frac { { V }^{ 2 } }{ P } =\frac { 220^{ 2 } }{ 100 } \\ \quad =484\Omega \quad$
Thus, when connected in series, total resistance

$R_{ s }=484+484\quad\Omega \quad\\=968\Omega \quad$
Power drawn in case of series combination,

${ P }_{ s }=\quad \frac { { V }^{ 2 } }{ R_{ s } } =\frac { 220^{ 2 } }{ 968 } \\ \quad =50W$
When connected in parallel, total resistance

$R_{ p }=\frac { 484 }{ 2 }\Omega \quad\\=242\Omega \quad$
Power drawn in case of series combination,

${ P }_{ p }=\quad \frac { { V }^{ 2 } }{ R_{ p } } =\frac { 220^{ 2 } }{ 242 } \\ \quad =200W$

In how many parts (equal) a wire of

$100\Omega$ be cut so that a resistance of

$1\Omega$ is obtained by connecting them in parallel?

0%
$10$
0%
$5$
0%
$100$
0%
$50$

Explanation

The resistance of the conductor is proportional to length of the conductor.
That is, $R=\rho \dfrac { l }{ A }$ .
The resistance of one wire of length l, R1 = 100 ohms.
If we cut it in a half: $\dfrac { { R }_{ 1 } }{ l } =\dfrac { { R }_{ 2 } }{ \dfrac { l }{ 2 } } \quad \rightarrow \quad { R }_{ 2 }=\dfrac { { R }_{ 1 } }{ 2 }$ .
If we cut it in n parts the resistance will be $R=\dfrac { { R }_{ 1 } }{ n }$ .
Because we shorten the length n times so it became $\dfrac { l }{ n }$ .
When we connect these parts in parallel the output resistance must be 1 ohm.
So, $1=\dfrac { 1 }{ R } +\dfrac { 1 }{ R } +...=n\dfrac { 1 }{ R } =n\dfrac { 1 }{ \dfrac { { R }_{ 1 } }{ n } } =\dfrac { { n }^{ 2 } }{ { R }_{ 1 } } =\dfrac { { n }^{ 2 } }{ 100 } \rightarrow n=10$ .
Hence, a resistance of 1 ohm is obtained by connecting 10 equal parts of the wire and connecting them in parallel.

You are given two fuse wires

$A$ and

$B$ with current rating

$2.5 A$ and

$6 A$ respectively. Which of the two wires would you select for to use with a

$1100 W$ ,

$220 V$ room heater?

0%
$A$
0%
$B$
0%
$A$ and $B$
0%
None of these

Explanation

Answer is B.
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 a short circuit, the wires get heated and melts. As a result, the circuit is broken and the current stops flowing.
The fuse must always be connected to the mains and it must be of the correct value. For example, a 15-ampere fuse should trip when the current through it exceeds 15 amperes. A 20-ampere fuse should blow when the current through it exceeds 20 amps.
In this case, the electrical appliance is of 1100 W power and 220 V. So the current that passes through the appliance is calculated from

$P=VI$ . That is,

$I=P/V$ .

So,

$I = 1100\ W/220\ V = 5\ A$ .

Hence, wire B with a current rating of 6 A should be used for the fuse. so B is the answer

SI unit of current is :

0%
Ampere
0%
Volt
0%
Watt
0%
Hertz
0%
None of these

Which one of the following will not conduct electricity?

0%
solid NaCl
0%
$CuSO_4$ solution
0%
graphite
0%
acidified water

In the following figure, the equivalent resistance between the points

$A$ and

$B$ in ohm will be :

0%
$1$
0%
$2$
0%
$3$
0%
$9$

Explanation

Resistance of each resistor

$R=3\Omega$

According to above fig. Resistors are in parallel combination.

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

${ R }_{ eq. }=\dfrac { R }{ 3 } =\dfrac { 3 }{ 3 } =1\Omega$

The diagram shows two resistors connected in parallel across a voltage source. Which statement is true for this circuit?

0%
The combined resistance is equal to the sum of the resistance
0%
The current at every point in the circuit is the same
0%
The current from the source is equal to the sum of the currents in X and in Y
0%
The sum of the potential difference across X and across Y is equal to the potential difference across the voltage source

Explanation

Both

$X$ and

$Y$ are in parallel so the total current delivered by source(battery) is divided into

$X$ and

$Y$ . So total current from the source is equal to the sum of the currents in

$X$ and in

$Y$ .

$R_1$ and

$R_2$ are the filament resistances of a 200 W bulb and a 100 W bulb respectively designed to operate on the same voltage, then

0%
$R_1 \,=\, 2R_2$
0%
$R_2 \,=\, 2R_1$
0%
$R_2 \,=\, 4R_1$
0%
$R_1 \,=\, 4R_2$

Explanation

Given that,

Power of bulbs

$P_1=200\ W\\P_2=100\ W$

The power

$P = VI$ , and

$I=V/R$

On substituting we get,

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

Here, the power

$P_1$ used in case of

$R_1$ is

$P_1=\dfrac { { V }^{ 2 } }{ R_1 } =200W$ ..........(1)

The power

$P_2$ used in case of

$R_2$ is

$P_2=\dfrac { { V }^{ 2 } }{ R_2 } =100W$ .........(2)

On taking ration,

$\dfrac { P_1 }{ P_2 } =\dfrac { R_2 }{ R_1 } =\dfrac { 200}{ 100 }$ .

Hence, $R_2 = 2R_1$ .

Option B

In how many parts (equal) a wire of 100

$\Omega$ be cut so that a resistance of 1

$\Omega$ is obtained by connecting them in parallel ?

0%
10
0%
5
0%
100
0%
50

Explanation

The resistance of the conductor is proportional to length of the conductor.
That is,

$R=\rho \dfrac { l }{ A }$ .
The resistance of one wire of length l, R1 = 100 ohms.

If we cut it in a half:

$\dfrac { { R }_{ 1 } }{ l } =\dfrac { { R }_{ 2 } }{ \dfrac { l }{ 2 } } \quad \rightarrow \quad { R }_{ 2 }=\dfrac { { R }_{ 1 } }{ 2 }$ .

If we cut it in n parts the resistance will be

$R=\dfrac { { R }_{ 1 } }{ n }$ .
Because we shorten the length n times so it became

$\dfrac { l }{ n }$ .

When we connect these parts in parallel the output resistance must be 1 ohm.
So,

$\dfrac{1}{R_{eq}}=\dfrac11=\dfrac { 1 }{ R } +\dfrac { 1 }{ R } +...=n\dfrac { 1 }{ R } =n\dfrac { 1 }{ \dfrac { { R }_{ 1 } }{ n } } =\dfrac { { n }^{ 2 } }{ { R }_{ 1 } } =\dfrac { { n }^{ 2 } }{ 100 }$

$\rightarrow n=10$ .
Hence, a resistance of 1 ohm is obtained by connecting 10 equal parts of the wire and connecting them in parallel.

A car headlamp of 48 W works on the car battery of 12 V. The correct fuse for the circuit is

0%
5A
0%
2A
0%
3A
0%
15A

Explanation

Answer is A.

Fuses are safety devices that are to be built into our electrical system.
In this case, the car headlamp is

$48\ W$ and the voltage is

$12\ V$ .
We know that power

$P = V\times I$

$\therefore I =\dfrac{P}{V} = \dfrac{48}{12} = 4\ A$
This is the maximum current that should pass through the device. If the current is more than this then the fuse must disconnect the circuit. Hence, the fuse of 5 A should be connected to the circuit.

The temperature of a metal wire rises when an electric current is passed through it because:

0%
collision of conduction electrons with the atoms of metal gives them energy which appears as heat
0%
.when electrons fall from higher energy level to lower energy level, heat energy is released
0%
collisions of metal atoms with each other releases heat energy.
0%
collisions of conduction electrons with each other releases heat energy.

Three identical bulbs are connected in parallel with a battery. The current drawn from the battery is 6 A. If one of the bulbs gets fused, what will be the total current drawn from the battery ?

0%
6A
0%
2A
0%
4A
0%
5A

An electric iron is connected to a

$200 V$ mains supply and draws a current of

$4.0 A$ What is the power rating of the iron ?

0%
$800W$
0%
$50W$
0%
$106W$
0%
$112W$

Explanation

Operating voltage

$V=200 V$

Current drawn

$I=4 A$

Power rating

$P=V\times I=200\times 4=800 W$

In the given figure below, the current passing through

$6\Omega$ resistor is

0%
$0.40A$
0%
$0.48A$
0%
$0.72A$
0%
$0.80A$

Explanation

$P.d$ across the circuit

$=1.2\times \dfrac {6\times 4}{6+4}=2.88$ volt

Current through

$6$ ohm resistance

$=\dfrac {2.88}{6}=0.48\ A$

A potential difference of 20V is needed to make a current of 0.05A flow through a resistor. What potential difference is needed to make a current of 300 mA flow through the same resistor?

0%
60V
0%
120V
0%
40V
0%
150V

Explanation

Answer is B.

According to Ohm's law,

$V = IR$ . That is,

$R = \dfrac VI$ .
In this case, the voltage is

$20\ V$ and the current is

$0.05\ A$ .
So, the resistance

$R = 20/0.05 = 400$ ohms.
Now, to make a current of 300 mA flow through the same resistor of 400 ohms, the potential difference to be applied is calculated as follows.
Here,

$I = 300 mA = 0.3\ A$ and

$R = 400$ ohms.
Therefore,

$V = 0.3\times 400 = 120\ V$ .
Hence, to make a current of 300 mA flow through the same resistor of 400 ohms a potential difference of 120 V is applied.

$8 \Omega$ resistance wire is doubled up by folding. What is the new resistance of the wire?

0%
$0.5 \Omega$
0%
$1.5 \Omega$
0%
$2\Omega$
0%
$4\Omega$

Explanation

Resistance of the conductor depends on the following factors:

1. Length of the conductor

2. Area of the cross-section of the conductor

3. Material used as a conductor.

Resistance is directly proportional to the length of the wire, and inversely proportional to the cross sectional area of the wire.

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

$\rho$ is the material's resistance in ohms,

$l$ is the length, and

$A$ is the cross sectional area in

${m}^2$ .

Which one of the following statements is not correct?

0%
The two main organs of the human body where the magnetic field produced is quite significant are the heart and the brain.
0%
In a house circuit, lamps are used in parallel
0%
Switches, fuses and circuit breakers should be placed in the neutral wire
0%
A generator with commutator produces direct current

The value of equivalent resistance between the points

$A\;and\;B$ in the given circuit, will be

0%
$6\;R$
0%
$\displaystyle\frac{4\;R}{11}$
0%
$\displaystyle\frac{11\;R}{4}$
0%
$\displaystyle\frac{R}{6}$

Explanation

$\text{Refer image-1}$

In the given circuit resistor cd, de and ef are in series.

$\therefore R_{eq}=cd+de+ef=R+R+R=3R$

Now, the circuit will be like

$\text{Refer image-2}$

In this circuit cf and de are parallel

$\therefore R_{eq}=\dfrac{R_{cf}\times R_{de}}{R_{cf}+R_{de}}$

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

$=\dfrac{3R}{4}$

Now, the circuit will be like

$\text{Refer image-3}$

In this circuit,

$Ac,cf,fB$ are in series

$\therefore R_{AB}=R+\dfrac{3R}{4}+R=2R+\dfrac{3R}{4}=\dfrac{11R}{4}$

$\therefore$ Correct option is C.

2057909_321414_ans_8e0e98c173694ea4ae2bd723813beff4.png

The minimum resistance that can be obtained by connecting 5 resistance of

$\displaystyle \frac{1}{4} \Omega$ each is

0%
$\displaystyle \frac{4}{5} \Omega$
0%
$\displaystyle \frac{4} \Omega$
0%
$20 \Omega$
0%
$0.05 \Omega$

Explanation

Minimum resistance will be obtained from the combination if they are connected in, parallel

$\therefore$ When 5 resistance of

$\displaystyle \frac{1}{4} \Omega$ each are connected in parallel then

$\displaystyle \frac{1}{R_p} =\frac{5}{\frac{1}{4}} =20$

$\Rightarrow R_p = \displaystyle \frac{1}{20} = 0.05 \Omega$

Two electric bulbs whose resistance are in the ratio of

$1 : 2$ are connected in parallel to a constant voltage source. The ratio of the power dissipated in them will be:

0%
$1 : 4$
0%
$1 : 2$
0%
$1 : 1$
0%
$2 : 1$

Explanation

Given :

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

And, the voltage difference across the resistors is the same in parallel connection.

And,

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

Ratio of power dissipated

$\dfrac{P_1}{P_2} = \dfrac{V^2/R_1}{V^2/R_2} = \dfrac{R_2}{R_1} = 2$

What is the effective resistance between points P and Q?

0%
$\displaystyle \frac{r}{2}$
0%
$\displaystyle \frac{r}{3}$
0%
$\displaystyle \frac{2r}{3}$
0%
$\displaystyle \frac{4r}{3}$

Explanation

$\textbf{Hint: Resistances are connected in parallel}$

Step 1: Find the the type of connection for each resistance

After moving from first 2r resistance, it has two paths one is with one more 2r resistance and other is pass through the conductor. So, these two 2r resistances are in parallel connection. Similarly, there is one more way that goes from above and connects directly to r which is other option with these 2r path. So, it is also in parallel connection. Hence, all resistance are parallel to each other.

Step2: Find the net resistance of the circuit

Assume that net resistance is R_{net}. So, from parallel resistance formula,

$\dfrac{1}{R_{net}} = \dfrac{1}{2r} + \dfrac{1}{2r} + \dfrac{1}{r} = \dfrac{1}{r} (\dfrac{1}{2} + \dfrac{1}{2} + 1) = \dfrac{2}{r}$

$\Rightarrow R_{net} = \dfrac{r}{2}$

$Answer:$

Hence, option A is the correct answer.

What is the effective resistance between points A and B?

0%
$\displaystyle \frac{R}{3}$
0%
$\displaystyle \frac{4R}{3}$
0%
$\displaystyle \frac{2R}{3}$
0%
$\displaystyle \frac{R}{2}$

Explanation

$\textbf{Hint: Here all the resistors are connected in parallel}$

Step – 1: Determine the types of connection for each resistance

Here, there are 3 possible ways to reach from point A to B and in all the ways there is only resistance R. First is from

$R_1$ , second is from

$R_2$ and third is from

$R_3$ .

As there are 3 ways, so all these 3 connections are in parallel to each other.

Step – 2: Determine net resistance

Assume that net resistance is

$R_{net}$ . So, from the formula of parallel connections,

$\dfrac{1}{R_{net}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} = \dfrac{1}{R} + \frac{1}{R} + \dfrac{1}{R} = \dfrac{3}{R}$

$=> R_{net} = \dfrac{R}{3}$

$Answer:$

Hence, option A is the correct answer.

A wire of resistance 12 ohm is bent in the form of a circular ring. The effective resistance between the two points on any diameter of the circle is:

0%
24 ohm
0%
12 ohm
0%
6 ohm
0%
3 ohm

Explanation

Resistance of wire is 12 ohm. Therefore, resistance of each semicircle

= 6 ohm=6Ω
Two resistances are in parallel.

Their equivalent resistance =

$\dfrac{6 \times 6}{6+6}=3$ ohm

CBSE Questions for Class 10 Physics Electricity Quiz 6 - MCQExams.com

0:0:2

Practice Class 10 Physics Quiz Questions and Answers

CBSE Questions for Class 10 Physics Electricity Quiz 6 - MCQExams.com

0:0:2

Practice Class 10 Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.