Unit of Voltage MCQs with Solutions

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Voltage is defined as the electric potential between two points in a conductor. The SI unit of voltage is volt. It is represented by the symbol ‘V’. A volt can be expressed as the electric potential in a wire when an electric current of 1 Ampere dissipates power of 1 watt. This can be written as;

V = W/A

In terms of energy, volt can be defined as the potential difference between two points in an electrical circuit that gives 1 Joule of energy per coulomb of charge. This can be expressed as;

V = Potential Energy / Charge

V = J/C

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MCQs on Unit of Voltage

Ques 1. In the ladder network shown, current through the resistor 3Ω is 0.25 A. The input voltage 'V' is equal to:

In the ladder network shown, current through the resistor 3Ω is 0.25 A.

10 V
20 V
5 V
15/2 V

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Ans. (b). 20 V

Explanation: Current through the 3Ω resistor is 0.25 A. Total resistance of this branch is 6, same as the parallel branch since both have the same resistance so current will also be the same.

Total current from 5Ω resistance is 0.25 + 0.25 = 0.5 A

Total resistance of 5Ω branch = 5 + (6 6) / 6 + 6 = 8Ω

If current through resistor 8Ω is 0.5 A, total resistance of that branch is 8.

So, total current from 7Ω resistance is 0.5 + 0.5 = 1 A

Total resistance of the circuit = 7 + (8 8) / (8 + 8) + 9 = 20Ω

V = IR

V = 1 20

V = 20 Volts

Ques 2. If two bulbs of power 25 W and 100 W respectively each rated at 220 V are connected in series with the supply of 440 V, which bulb will fuse?

25 W bulb
40 Watts
Both a and b
None of these

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Ans. (a). 25 W bulb

Explanation: P₁ = 25 W ; P₂ = 100 W ; V = 220 V and V’ = 440 V

P = V₂/R

R₁ = V₂/P₁

R₁ = (220)²/25 = 1600 Ω

R₂ = V₂/P₂

R₂ = (220)²/100 = 400 Ω

They are connected in series, so effective resistance is;

R₁ = R₁ / R₁ + R₂

R₂ = R₂ / R₁ + R₂

∴ V₁ = V’ R₁ / R₁ + R₂

V₂ = V’ R₂ / R₁ + R₂

V₁ = 440 1600/2000 = 352 V

V₂ = 440 440/2000 = 96.8 V

So, Voltage across 25 W is greater than the range specified and thus this bulb will fuse.

Ques 3. A charge is moved from point A to point B. The work done to move unit charge during this process is called:

Potential at A
Potential at B
Potential Difference between A and B
Current from A to B

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Ans. (c). Potential Difference between A and B

Explanation: The work done per unit charge in moving it from point A to B is called potential difference between A and B.

Ques 4. The potential difference (V_A - V_B) between the points A and B in the given figure is:

The potential difference (VA - VB) between the points A and B

9 V
- 3 V
3 V
6 V

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Ans. (a). 9 V

Explanation: I = 2 A

Using Kirchhoff’s laws,

V_A - I(2) - 3 - I(1) - V_B = 0

∴ V_A - 2 2 - 3 - 2 1 - V_B = 0

V_A - 9 - V_B = 0

V_A - V_B = 9 V

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Ques 5. In the below circuit, the potential difference between A and B is:

In the below circuit, the potential difference between A and B

20 V
30 V
10 V
40 V

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Ans. (c). 10 V

Explanation: Let q₁ and q₂ be the charge flowing in the loops I and II respectively.

Let q1 and q2 be the charge flowing in the loops I and II respectively

Applying Kirchhoff’s rule in loop I;

q₁/1 + (q₁ + q₂ / 3) - 190 = 0

4q₁ + q₂ = 190 x 3

Applying Kirchhoff’s rule in loop II;

(q₁ + q₂ / 3) + q₂/3 + q₂/1 = 0

q₁ = – 5q₂

Now, put this value of q₁ in the above equation,

4 (- 5q₂) + q₂ = 190 x 3

q₂ = 30 μC

Thus, potential difference V_AB = q₂ / 3μF = 30μ C / 3μF

V_AB = 10 V

Ques 6. In the given circuit, the potential difference across 3 F capacitor will be:

In the given circuit, the potential difference across 3 F capacitor

12 V
10 V
6 V
4 V

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Ans. (b). 10 V

Explanation: The capacitor and battery are in series, so

V_eq = 20 - 4 = 16 V

C_eq = 3 x 5 / 5 + 3 = 15/8

q = C_eq V_eq = 15/8 x 16

q = 30

V₃ = q/c = 30/3 = 10 V

Ques 7. A current of 2 A flows in an electric circuit as shown in the figure. The potential difference (V_R - V_s) in volts (V_R and V_s are potentials at R and S respectively) is:

A current of 2 A flows in an electric circuit as shown in the figure

- 4 V
2 V
4 V
- 2 V

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Ans. (c). 4 V

Explanation: Current through R = 1 A and through S is also 1 A.

V_R - V_s = (V_p - V_s) - (V_p - V_R)

V_R - V_s = (7 x 1) - (3 x 1)

V_R - V_s = 4 V

Ques 8. In the circuit shown, the internal resistance of the battery is 1.5 Ω and V_p and V_Q are potentials at P and Q respectively. What is the potential difference between points P and Q?

In the circuit shown, the internal resistance of the battery is 1.5 Ω and Vp and VQ are potentials at P and Q respectively

Zero
4 V (V_p > V_Q)
4 V (V_Q > V_p)
2.5 V (V_Q > V_p)

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Ans. (d). 2.5 V (V_Q > V_P)

Explanation: R_eq = 5/2 Ω

I = 20 / (5/2 + 1.5) = 5 A

Potential difference between X and P

V_X - V_P = (5/2) 3 = 7.5 V …..(i)

Potential difference between X and Q

V_X - V_Q = (5/2) 2 = 5 V …..(ii)

Solving (i) and (ii),

V_P - V_Q = - 2.5 V

Minus sign indicates that V_Q > V_P

Ques 9. The unit of electric potential difference is:

J
J/C
JC
C/J

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Ans. (b). J/C

Explanation: The potential difference is given by;

ΔV = Q/C

Thus, the unit of electric potential difference is joule per coulomb denoted by J/C.

Ques 10. The potential difference between the points B and A is:

The potential difference between the points B and A

- 9.1 V
7.9 V
8.9 V
9.1 V

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Ans. (b). 7.9 V

Explanation: V_B - V_A = 5 - 1 9 - 6 - 4 - 29 + 2 - 49 - 39

V_B - V_A = -1 - 4 + 2 - I (1 + 2 + 4 + 3)

V_B - V_A = -5 + 2 - I (10)

V_B - V_A = -7 - 1/10

V_B - V_A = -7 - 0.9

V_B - V_A = -7.9 V

|V_B - V_A| = 7.9 V

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Unit of Voltage MCQs with Solutions

MCQs on Unit of Voltage

CBSE CLASS XII Related Questions

1.
Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

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A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

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Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

5.
Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

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Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

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Unit of Voltage MCQs with Solutions

MCQs on Unit of Voltage

CBSE CLASS XII Related Questions

1.Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

2.A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

3.What is displacement current (\( i_d \))? Considering the case of charging of a capacitor, show that \( i_d = \varepsilon_0 \frac{d\Phi_E}{dt} \). What is the value of \( i_d \) for a conductor across which a constant voltage is applied?

4.Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

5.Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

6.Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

2.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

3.
What is displacement current (\( i_d \))? Considering the case of charging of a capacitor, show that \( i_d = \varepsilon_0 \frac{d\Phi_E}{dt} \). What is the value of \( i_d \) for a conductor across which a constant voltage is applied?

4.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

5.
Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

6.
Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –