MCQ Questions for Class 10 Physics Electricity Quiz 11

Power dissipated across the

$8\Omega$ resistor in the circuit shown here is

$2W$ . The power dissipated in watt units across the

$3\Omega$ resistor is

0%
$2$
0%
$1$
0%
$0.5$
0%
$3$

Equivalent resistance between

$A$ and

$B$ is

0%
$3 \Omega$
0%
$6 \Omega$
0%
$1.5 \Omega$
0%
$4.5 \Omega$

The equivalent resistance of fig shown between 'a' and 'b' is

0%
$\dfrac {2r} {8}$
0%
$\dfrac {5r} {8}$
0%
$\dfrac {r} {8}$
0%
$\dfrac {3r} {8}$

Explanation

$\therefore$ Option

$B$ is correct.

1381088_1272365_ans_b53eb448eb9d4228943da0e818ec7eea.png

The resistance across A and B is

0%
$\dfrac {R} {2}$
0%
$R$
0%
$\dfrac {6R} {5}$
0%
$4R$

Explanation

The resistances in upper branch are a series combination of

$R$ and

$2R$

So their series resistance is

$R_1=R+2R=3R$

The resistances in lower branch are a series combination of

$R$ and

$R$

So their series resistance is

$R_2=R+R=2R$

Now

$3R$ and

$2R$ are in parallel.

$\dfrac{1}{R_{eq}}=\dfrac{1}{2R}+\dfrac{1}{3R}=\dfrac{5}{6R}$

$\therefore R_{eq}=\dfrac{6R}{5}$ .

1213192_1288848_ans_0e29e01f158848a8956b34bb2e836744.png

Find a TRUE statement

0%
Ohm's law is applicable to all conductors of electricity
0%
The resistance of an incandescent lamp is lesser when the lamp is switched on
0%
Specific resistance of a wire depends upon its dimensions
0%
The resistance of carbon decreases with an increase in temperature

A wire of resistance 'R' is bent into a circular ring of radius r. Equivalent resistance between two points 'x' and 'y' on its circumference, when angle XOT is A, can be given by

0%
$\dfrac { Ra }{ { 4\pi }^{ 2 } } \left( 2\pi -a \right)$
0%
$\dfrac { R }{ { 2\pi } } \left( 2\pi -a \right)$
0%
$R\left( 2\pi -a \right)$
0%
$\dfrac { 4\pi }{ { Ra } } \left( 2\pi -a \right)$

When two resistors are connected in parallel, then the equivalent resistance is

0%
${R_1} + {R_2}$
0%
$\dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$
0%
$\dfrac{{{R_1} \times {R_2}}}{{{R_1} + {R_2}}}$
0%
$\dfrac{{{R_1}}}{{{R_2}}}$

Potential difference across

$AB$ shown in the figure is:

0%
10 V
0%
12 V
0%
8 V
0%
0 V

Explanation

In the given figure, $A$ and $B$ are on the same potential as there is a direct connection between $A$ and $B$ without any other element (via $D$ )

$V_A=V_B$

Potential difference

$=V_B-V_A=0$

Alternatively,

we can see that

Voltage between

$A$ and

$C$ i.e.

$V_{AC}=V_A-V_C= 10 \ V$ . . . (i)

And,

Voltage between

$C$ and

$B$ i.e.

$V_{CB}=V_C-V_B=-10 \ V$ . . . (ii)

Adding

$(i)$ and

$(ii)$ ,

$(V_A-V_C)+(V_C-V_B)=10+(-10)$

$\Rightarrow V_A-V_B=0$

So Voltage between

$A$ and

$B$ i.e.

$V_{AB}=V_A-V_B= 0 \ V$

Thus, option D is correct.

2050624_1309413_ans_0e25510e8ebc4b058cab8662cb0cbf9c.png

A wire of resistance R is cut into n equal parts these parts are then connected in parallel with each other. The equivalent resistance of the combination is :

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

Explanation

when wire of resistance R is cut into n equal parts then Resistance of each part will be

$\dfrac Rn$ ;

such

$n$ parts are joined in parallel

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

$\dfrac{1}{(R/n)}+$

$\dfrac{1}{(R/n)}+$ ...

$R_{eq}=\dfrac {R}{n^2}$ .

A conductor of resistance

$3 \Omega$ is stretched uniformly till its length is doubled. The wire is now bent in the form of an equilateral triangle. The effective resistance between the ends of any side of the triangle in

$\Omega$

0%
$\dfrac{9}{2}$
0%
$\dfrac{8}{3}$
0%
$2$
0%
$1$

Explanation

If we double the length of the conductor

$,$ then its cross

$-$ sectional area must become half of its original value (since

$,$

$volume = area \times length$ is constant).

Therefore

$,$ we can say that after stretching the new resistance is

$R' = \rho \dfrac{{2l}}{{A/2}} = 4\rho \dfrac{1}{A} = 4{R_0} = 4 \times 3\Omega = 12\Omega$

Now

$,$ the wire is bent such that each side has a resistance

$4\Omega ,$ (since the resistance is divided uniformly).

Now from the diagram, we can see that between two vertices

$.$ there is an arrangement of two

$4\Omega$ resistors in series connected in parallel with another

$4\Omega$ resistor

$.$

The net resistance becomes

$:-$

$\dfrac{{8 \times 4}}{{8 + 4}} = \dfrac{8}{3}\Omega$

Hence,

option

$(B)$ is correct answer.

1225928_1296969_ans_fc6c30f0ee534f64a88f51adbfb0bf4e.png

Electric current flows from the positive terminal of a battery to the negative terminal.

0%
True
0%
False

When resistors are connected in parallel, the current divides in the ...................

0%
Direct ratio of resistances
0%
Inverse ratio of resistances
0%
Inverse ratio of the potential differences
0%
None of these

In the circuit shown in the figure below, the magnitude and direction of the current will be

0%
$2.3 \ Amp$ from A to B via E
0%
$2.3 \ Amp$ from B to A via E
0%
$1.0 \ Amp$ from B to A via E
0%
$1.0 \ Amp$ from A to B via E

Explanation

The voltage of the battery

$V_1=10 \ V$ and that of the battery

$V_2=4 \ V$ get added as they are connected in series with same polarity.

Net emf

$V_{net}=10\,V+4\,V=14\,V$

The resistances

$1 \Omega, 3 \Omega$ and

$2 \Omega$ are connected in series.

Total resistance

$R_{series}=1+3+2=6\Omega$

By Ohm's law,

$V_{net}=IR_{series}$

$\Rightarrow I=\dfrac{V_{net}}{R_{series}}$

Hence, Current

$I=\dfrac {14}{6}=2.3\,A$

For the direction of current, we can see net emf is in the same sense of the individual batteries because

$10\,V$ battery and

$4\,V$ battery are connected back-to-back. So

$2.3\,A$ current will flow in the direction

$A$ to

$B$ via

$E$ .

A toaster operating at

$240V$ has a resistance of 120

$\Omega$ . The power is

0%
$400 W$
0%
$200 W$
0%
$480 W$
0%
$240 W$

What is the minimum resistance which can be made using five resistors each of

$\dfrac{1}{5} \Omega$ ?

0%
$\dfrac{1}{5}\Omega$
0%
$\dfrac{1}{25}\Omega$
0%
$\dfrac{1}{10}\Omega$
0%
$\dfrac{1}{50} \Omega$

Explanation

For minimum resistance resistors should be connected in parallel combination.

Suppose value of one resistance is

$R$ .

$R = \dfrac{1}{5} \Omega$

Equivalent resistance of

$5$ resistance-

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

$\Rightarrow \dfrac{1}{R_{eq}} = \dfrac{5}{R}$

$\Rightarrow {R_{eq}} = \dfrac{R}{5}$

put the value of

$R$ -

$\Rightarrow {R_{eq}} = \dfrac{1}{5 \times 5}$

$\Rightarrow {R_{eq}} = \dfrac{1}{25} \Omega$

Inside a cell in a closed circuit the conventional electric current

0%
always travels from - ve terminal to + ve terminal
0%
always travels from + ve terminal to - ve terminal
0%
may travel from +ve terminal to -ve terminal
0%
may travel from -ve terminal to +ve terminal

Ashokbhai goes to an electric shop to buy an electric bulb to obtain light in a small room of his home. Which of the following statement is scientifically correct

0%
Give bulb of $60\ V$
0%
Give bulb of $60\ J/s$
0%
Give bulb of $60\ A$
0%
Give bulb of $60\ W$

Explanation

Give bulb of

$60W$ is the right statement to say while buying a bulb because in market bulbs are sold according to their power ratings and with power ratings it is easy to identify which bulb will suit a particular requirement.

Find the current flowing through the resistance

${ R }_{ 1 }$ of the circuit shown in the figure if the resistances are equal to

${ R }_{ 1 }=10\Omega ,{ R }_{ 2 }=20\Omega$ and

${ R }_{ 3 }=30\Omega$ and the potential of points 1,2 and 3 are equal to

${ \varphi }_{ 2 }=6V,and{ \varphi }_{ 3 }$ 5 V.

0%
0.1 A
0%
0.2 A
0%
0.3 A
0%
0.4 A

A uniform wire is cut into

$10$ equal parts and all the equal parts are connected in parallel. The effective resistance

0%
decreases $10$ times
0%
increases $10$ times
0%
increases $100$ times
0%
decreases $100$ times

When three resistors of resistances

$3 \Omega, 4 \Omega$ and 5

$\Omega$ are connected in parallel, the currents
through them are in the ratio

0%
$5:4:3$
0%
$20:15:12$
0%
$3:4:5$
0%
$9:8:7$

Inside a cell in a closed circuit the conventional electric current:

0%
always travels from $- ve$ terminal to $+ ve$ terminal
0%
always travels from $+ ve$ terminal to $- ve$ terminal
0%
may travels from $- ve$ terminal to $+ ve$ terminal
0%
may travels from $+ ve$ terminal to $- ve$ terminal

Explanation

The electric current was considered to be the flow of positive charges and the direction of flow of positive charges was taken to be the direction of electric current. Conventionally, in an electric circuit, the direction of electric current is taken as opposite to the direction of the flow of electrons, which are negative charges.

Inside a cell, in a closed circuit the conventional electric current always travels from

$−ve$ terminal to

$+ve$ terminal.

A lamp of

$600 \mathrm{W}-240 \mathrm{V}$ is connected to 220

$\mathrm{V}$ mains. Its resistance is

0%
$96 \Omega$
0%
$84 \Omega$
0%
$90 \Omega$
0%
$64 \Omega$

Graphs between electric current and potential difference across two conductors A and B are shown in the figure. Which of the following conductor has more resistance?

0%
B
0%
A
0%
Both have equal resistance
0%
None of these

Explanation

For comman voltage

$V$ Current in conductor B is greter then currecnt in conductor A.

$i_B>i_A$ for same value of V

From ohms law,

$R=\dfrac VI\\R\propto\dfrac1I$

Hence,

$i_B>i_A$

$R_A>R_B$

Option B

For the circuit shown in figure given below, the equivalent resistance between points A and B is

0%
$10\Omega$
0%
$5\Omega$
0%
$\dfrac{10}{3} \Omega$
0%
$2\Omega$

$5$ resistors of

$2.5 \Omega$ each are connected in parallel then what would be its resultant resistance?

0%
$5\Omega$
0%
$0.5\Omega$
0%
$50\Omega$
0%
$6.25\Omega$

Explanation

Given:

Five resistances of

$R=2.5 \ \Omega$ each are connected in parallel.

By formula of parallel combination,

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

$\Rightarrow \dfrac{1}{R_{eq}}=5 \times \dfrac{1}{R}=5 \times \dfrac{1}{2.5}$

$\Rightarrow R_{eq}=\dfrac{2.5}{5}=0.5 \ \Omega$

So the resultant resistance is

$0.5 \ \Omega.$

If five equal pieces of

$25\ \Omega$ are connected in parallel, then their equivalent resistance will be _____

0%
$25\Omega$
0%
$1\Omega$
0%
$5\Omega$
0%
$\cfrac{1}{5}\Omega$

Explanation

In the parallel combination-

$\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2}+\dfrac{1}{R_3}+\dfrac{1}{R_4} + \dfrac{1}{R_5}$

Here all resistors are equal,

$R_{1}=R_2=R_3=R_4=R_5=R=25\ \Omega$

So,

$\Rightarrow \dfrac{1}{R_{eq}} = \dfrac{5}{R}$

$\Rightarrow \dfrac{1}{R_{eq}} = \dfrac{5}{25}$

$\Rightarrow \dfrac{1}{R_{eq}} = \dfrac{1}{5}$

$\Rightarrow R_{eq} = 5\ \Omega$

Three resistances are arranged as shown. The potential of the point O is :

0%
3 V
0%
1 V
0%
4 V
0%
5 V

Which unit is not used in either the definition of the coulomb or the definition of the volt?

0%
ampere
0%
joule
0%
ohm
0%
second

$220 \ V$ and

$100 \ W$ lamp is connected to

$220 \ V$ power supply. What is the power consumed?

0%
$100$ W
0%
$200$ W
0%
More than $100$ W
0%
None of these

Explanation

The power consumed would be obviously

$100 \ W$ .

It is because the lamp is operated at its rated voltage

$V =220 \ V$ . So the power consumed will be equal to its power rating, i.e.

$100 \ W.$

Let us solve it to prove it.

We know, power

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

Putting

$P=100 \ W$ and

$V =220 \ V$

$\Rightarrow R=\dfrac{(220)^2}{100}=484 \ \Omega$

By Ohm's law,

Current in the circuit

$I=\dfrac{\text{power supply voltage V}}{\text{resistance R}}$

$\Rightarrow I=\dfrac{220}{484} \ A$

Power consumed, P can be calculated as

$I^2R.$

$P={\left( \dfrac{220}{484}\right)}^2 \times 484=\dfrac{(220)^2}{484}=100 \ W$

Hence, option (A) is correct.

A wire has resistance

$12\Omega$ . It is bent in the form of a circle. The effective resistance between the two points on any diameter of the circle is?

0%
$12\Omega$
0%
$24\Omega$
0%
$6\Omega$
0%
$3\Omega$

Three

$2\Omega$ resistors are connected to form a triangle. The resistance between any two corners is?

0%
$6\Omega$
0%
$2\Omega$
0%
$\dfrac{3}{4}\Omega$
0%
$\dfrac{4}{3}\Omega$

In the circuit shown, the current in

$3\Omega$ resistance is?

0%
$1$ A
0%
$1/7$ A
0%
$5/7$ A
0%
$15/7$ A

Explanation

We know,

The voltage of ground is considered to be 0V,

Let the grounds be named as 1,2,3 respectively,

And let the current in

$1^{st}$

$2\Omega$ be i, and the

$3\Omega$ be

$i_1$ ,

Applying Kirchhoff Voltage law , from ground 1 to ground 2,

$0+10-2i-2(i-i_1)=0$

$2i-i_1=5$

$-(i)$

Applying KVL from ground 1 to ground 3,

$0+10-2i-3i_1-3i_1=0$

$2i+6i_1=10$

$-(ii)$

Subtracting

$(i)$ from

$(ii)$ ,

$7i_1=5$

$i_1=\dfrac{5}{7} A$

Option

$\textbf C$ is the correct answer.

1545759_1704589_ans_8fb5d7d56784411382ffd8880f48feef.PNG

The power consumed by

$4$ V battery in the circuit as shown is?

0%
$8$ W
0%
$7$ W
0%
$6$ W
0%
$5$ W

Explanation

$E_{\text{net}} = 10 - 4 = 6\ V$

$R = 3\ \Omega$

$i = \dfrac{V}{R} = \dfrac{6}{3} = 2A$

$\therefore$ Power consumed by

$4V$ battery

$= VI = 4 \times 2$

$= 8 VA = 8W$

2093315_1704921_ans_bb856ceb77f34ddcb8bfa8a247ef437b.PNG

An electrical conductor has a resistance of

$5.6k\Omega$ . A potential difference (p.d.) of

$9.0V$ is applied
across its ends.
How many electrons pass a point in the conductor in one minute?

0%
$6.0\times 10^{20}$
0%
$1.0\times 10^{19}$
0%
$6.0\times 10^{17}$
0%
$1.0\times 10^{16}$

Explanation

We know that

$V = IR$

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

$\implies I = \dfrac{9}{5.6 \times 10^3}$ A

This means that no.of electron passes through a point in one second =

$\dfrac{9}{5.6 \times 10^3 \times 1.6 \times 10^{-19}}$

$\implies$ no. of electron passes through a point in one minute =

$\dfrac{9}{5.6 \times 10^3 \times 1.6 \times 10^{-19}} \times 60$

$6.0 \times 10^{17 }$ electrons

Choose the correct option:
Three bulbs of

$100 \ W$ ,

$200 \ W$ and

$40 \ W$ are connected in series to the main supply of

$200$ V. The current will be?

0%
Maximum through $100$ W
0%
Maximum through $200$ W
0%
Maximum through $40$ W
0%
Same in all

If two identical heaters each rated as (

$1000$ W-

$220$ V) are connected in parallel to

$220$ V, then the total power consumed is?

0%
$200$ W
0%
$2500$ W
0%
$250$ W
0%
$2000$ W

The ratio

$R_{2}/R_{3}$ is

0%
$\dfrac{(k-1)^{2}}{k}$
0%
$k^{2}-\dfrac{1}{k}$
0%
$\dfrac{k}{k-1}$
0%
$k-\dfrac{1}{k^{2}}$

Explanation

Current in

$R_1$ and

$R_2$ will be the same:

$\dfrac {V_{n-1}-V_n}{R_1}=\dfrac {V_n}{R_3}$

$\dfrac {V_{n-1}-\dfrac {V_{n-1}}{k}}{R_1}=\dfrac {V_{n-1}}{k\ R_3}$

$\Rightarrow \ R_1 =R_3 (k-1)$

Put the value of

$R_1$ in the above expression, we get:

$\dfrac {R_2}{R_3}=\dfrac {k}{k-1}$

1627248_1753608_ans_658f690939bf4b5099c2723f0182a27b.png

The equivalent resistance between

$A$ and

$B$ ( of the circuit as shown ) is

0%
$4.5\Omega$
0%
$12\Omega$
0%
$5.4\Omega$
0%
$20\Omega$

Explanation

Clearly from the figure,

Resistors

$7\Omega$ and

$3\Omega$ are in parallel;

$6\Omega$

and

$4\Omega$ are in parallel and both in series.

$R_{eq}=\dfrac{7\times 3}{7+3}+\dfrac{4\times 6}{4+6}=4.5\Omega$

The equivalent resistance between

$A$ and

$B$ in the network in figure is

0%
$\dfrac{4}{3}\Omega$
0%
$\dfrac{3}{2}\Omega$
0%
$3\Omega$
0%
$2\Omega$

Explanation

Above circuit can be reduced as shown ,

Now,

The equivalent of the network is given in figure.
The equivalent of the above network is a parallel combination of

$3\Omega, 4\Omega$ and

$6\Omega$ , i.e,

$\dfrac{1}{R}=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{6}\Rightarrow R=\dfrac{4}{3}\Omega$

1627528_1753700_ans_7be7746d0f0d4a19b9fb1738fe188a73.png

In the network shown, each resistance is equal to R. The equivalent
resistance between diagonally opposite corners is:

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

The ratio

$R_{1}/R_{2}$ is

0%
$k^{2}-\dfrac{1}{k}$
0%
$\dfrac{k}{k-1}$
0%
$k-\dfrac{1}{k^{2}}$
0%
$\dfrac{(k-1)^{2}}{k}$

Explanation

Given ,

$V_1 =\dfrac {V_0}{k}, V_2 =\dfrac {V_1}{k}, V_3 =\dfrac {V_2}{k},$

$\ I=I_1 +I_2$

$\dfrac {V_0-V_0 /k}{R_1}=\dfrac {V_0/ k-V_0 /k^2}{R_1}+\dfrac {V_0 /k}{R_2}$

$\dfrac {R_1}{R_2}=\dfrac {(k-1)^2}{k}$

1627243_1753606_ans_03c576ab47424ab693a6c863213b6d3a.png

In Fig. when an ideal voltmeter is connected across 4000

$\Omega$ resistance, it reads 30 V. If the voltmeter is connected across 3000

$\Omega$ resistance, it will read

0%
$20 V$
0%
$22.5 V$
0%
$35 V$
0%
$40 V$

Explanation

Let

$I$ be the current in the circuit then,

$I\times R=V$

$\Rightarrow I\times 4000=30 V$

$\Rightarrow I=\dfrac{30}{4000}$

Reading of voltmeter across

$3000\Omega$

$V=I\times 3000=\dfrac{30}{4000}\times 3000=22.5 V$

A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is doubled if

0%
both the length and the radius of the wire are halved
0%
both the length and the radius of the wire are doubled
0%
the radius of the wire is doubled
0%
the length of the wire is doubled

Explanation

The heat produced in the metallic wire is given using

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

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

$H=\dfrac{V^{2}t\pi r^{2}}{\rho L}$

Therefore, heat produced will be doubled if both length and the radius of the wire are doubled.

The equivalent resistance in series combination is :

0%
smaller than the largest resistance.
0%
larger than the largest resistance
0%
smaller than the smallest resistance
0%
larger than the smallest resistance.

An Insulator like wood and rubber has a very low electrical resistance.

0%
True
0%
False

Write true or false for the following statements :
The resistance of a piece of rubber is more than that of piece copper.

0%
True
0%
False

Write true or false for the following statements :
Heat dissipated by a conductor when a current is passed through it is directly proportional to the resistance of the conductor.

0%
True
0%
False

Explanation

According to joule's law of heating

$H = i^2Rt$

Hence, heat dissipated by a conductor when a current is passed through it is directly proportional to the resistance

$R$ of the conductor.

Variation of current and voltage in a conductor has been shown in the diagram below. The resistance of the conductor is.

0%
$4$ ohm
0%
$2$ ohm
0%
$3$ ohm
0%
$1$ ohm

Explanation

from ohm's law

$R=\dfrac VI$

also, slope of given graph

$\dfrac VI$

$R=$ slope of graph

$s=\dfrac{y_2-y_1}{x_2-x_1}$

$s=\dfrac{4-1}{4-1}=1$

Resistance = 1 ohm

Two wires of the same material are given. The first wire is twice as long as the second and has twice the diameter of the second. The resistance of the first will be.

0%
Twice of the second
0%
Half of the second
0%
Equal of the second
0%
Four times of the second

Explanation

we know

$R=\rho l /A$

$R\propto \dfrac{l}{r^2}$

According to the question:

$l_1=2l_2$ and

$r_1=2r_2$

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

The resistance of the first will be Half of the second.

Two resistors of resistance

$R_{1}$ and

$R_{2}$ having

$R_{1} > R_{2}$ are connected in parallel. For equivalent resistance

$R$ , the correct statement is

0%
$R > R_{1}+R_{2}$
0%
$R_{1} < R < R_{2}$
0%
$R_{2} < R < (R_{1}+R_{2})$
0%
$R < R_{1}$

Explanation

For parallel combination:

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

$\dfrac{1}{R_1}+$

$\dfrac{1}{R_2}$

Equivalent resistance of parallel resistors is always less than any of the member of the resistance system.

$R < R_{1}$

CBSE Questions for Class 10 Physics Electricity Quiz 11 - MCQExams.com

0:0:3

Practice Class 10 Physics Quiz Questions and Answers

CBSE Questions for Class 10 Physics Electricity Quiz 11 - MCQExams.com

0:0:3

Practice Class 10 Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.