MCQs on Electrical Resistance: Introduction & Explanation

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Electrical resistance is the property of a material to resist the flow of the electric current through it. When electric current flows through a conductor, some amount of it is obstructed from flowing. This obstruction to flow of current is measured as electrical resistance. It is denoted by ‘R’. Electrical resistance is a major concept covered in the chapter Current Electricity.

The formula for calculating electrical resistance of a conductor is;

V = IR

Where, V = Potential difference

I = Current

R = Resistance

The electrical resistance of a conductor depends on the following factors:

The area of cross-section of the conductor
Length of the conductor
Temperature of the conductor
Material of the conductor

Electrical resistance is directly proportional to the length of the conductor and inversely proportional to the area of its cross-section. This relation can be expressed as;

R = ρ(L/A)

where, is the resistivity of the material

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MCQs on Electrical Resistance

Ques 1. A potentiometer wire 10 m long has a resistance of 40Ω . It is connected in series with a resistance box and a 2 V storage cell. If the potential gradient along the wire is 0.1 mV/cm, the resistance in the box is:

760Ω
260Ω
1060Ω
960Ω

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Ans. (a). 760Ω

Explanation: Potential gradient = 0.1 mV/cm

ΔV/1 = (0.1 x 10^-3) / (1 x 10^-2)

ΔV = 1 x 0.01 = 0.1

0.1 / (2 - 0.1) = 40/R

R = 760Ω

Ques 2. A wire of resistance 20Ω is bent in the form of a square. The resistance between the ends of diagonal is:

10Ω
20Ω
5Ω
15Ω

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Ans. (c). 5Ω

Explanation: The wire is bent in the form of a square. In a square, all the sides are equal, therefore, resistance will be same for each side i.e. 5Ω.

When a diagonal is created, the two faces on each side of the diagonal will have two sides and the resistance will be 10.

R = 20Ω

R/4 = 5Ω

R₁ = 10Ω and R₂ = 10Ω

So, resistance between the ends of the diagonal is,

R_D = (10 x 10) / (2 x 10) = 5Ω

Ques 3. The equivalent conductance of two wires of resistance 10 and 5 joined together in parallel is:

50/15 mho
0.3 mho
2/3 mho
35/50 mho

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Ans. (b). 0.3 mho

Explanation: Effective resistance in parallel combination is given by,

1/R = 1/5 + 1/10

1/R = 3/10

So, conductance = 3/10 = 0.3 mho

Ques 4. What length of copper wire of resistivity 1.7 x 10^-8 Ωm and radius 1 mm is required so that its resistance is 1Ω?

184.7 m
200 m
190.5 m
150 m

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Ans. (a). 184.7 m

Explanation: We know, R = ρL/A …….(i)

R = 1Ω

Area of cross-section, A = πr² = 3.14 x 0.0012 = 3.14 x 10^-6

Resistivity = 1.7 x 10-8 m

Putting these values in eq (i),

1 = (1.7 x 10^-8 L) / (3.14 x 10^-6)

L = (1 x 3.14 x 10^-6) / (1.7 x 10^-8)

L = 3.14 x 10^-6 / 1.7 x 10^-8

L = 1.847 x 10² m

Ques 5. A copper wire has diameter 0.5 mm and resistivity of 1.6 x 10^-8 Ωm. What will be the length of this wire to make its resistance 10Ω?

110 m
122.6 m
100 m
150 m

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Ans. (b). 122.6 m

Explanation: d = 0.5 mm = 0.5 x 10^-3 m

R = d/2 = (0.5 x 10^-3) / 2

R = 0.25 x 10^-3 m

A = πr² = 3.14 x (0.25 x 10^-3)²

ρ= 1.6 x 10^-8 Ωm

R = 10Ω

We know, R = ρL/A

I = RA/ρ

I = 10 x 3.14 x (0.25)² x 10^-6 / 1.6 x 10^-8

I = 1.226 x 10²

I = 122.6 m

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Resistivity Temperature Dependence	Unit of Resistance	Relation between Power and Resistance
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Ques 6. An aluminium ( α = 4 x 10^-3 K^-1) resistance R₁ and a carbon ( α = 0.5 x 10^-3 K^-1) resistance R₂ are connected in series to have a resultant resistance of 36 Ω at all temperatures. The values of R₁ and R₂ respectively are:

32Ω and 4Ω
16Ω and 20Ω
4Ω and 32Ω
20Ω and 16Ω

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Ans. (c). 4Ω and 32Ω

Explanation: Here, R₁ + R₂ = 36 …..(i)

R₁’ = R₁ [1 + α₁ΔT]

R₂’ = R₂ [1 + α₂ΔT]

R₁’ + R₂’ is independent of T

R₁α₁ + R₂α₂ = 0

Also, R₁/R₂ = α₂/α1

R₁/R₂ = 0.5 x 10^-3 / 4 x 10^-3

R₁/R₂ = ⅛

R₂ = 8R₁

Put this value in eq (i),

R₁ + 8R₁ = 36

9R₁ = 36

R₁ = 4Ω

And R₂ = 32Ω

Ques 7. If the temperature is increased, what will be the effect on the resistance of a conductor?

Decreases
Increases
Does not change
May increase or decrease depending on other factors

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Ans. (b). Increases

Explanation: Since, Resistance is directly proportional to temperature, hence when temperature is increased resistance will also increase.

Ques 8. On a bulb it is written 220 V and 60 W. The resistance of the bulb and the value of the current flowing through it is:

500Ω and 2 A
200Ω and 4 A
806.67Ω and 0.27 A
100Ω and 1 A

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Ans. (c). 806.67Ω and 0.27 A

Explanation: Power (P) = V²/R

R = V²/P

R = (220)²/60

R = 806.67Ω

Now, according to Ohm’s law,

V = IR

I = 220/806.67

I = 0.27 A

Ques 9. The temperature coefficient of resistance of a wire is 0.0012/°C. Its resistance is 1Ω at 300 K. At what temperature, its resistance will be 2Ω?

1133 K
854 K
1217 K
1154 K

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Ans. (a). 1133K

Explanation: Given, = 0.0012/°C

R₃₀₀ = 1Ω

T₁ = 300 K

R’ = 2Ω

Resistance at a temperature, R’ = R₃₀₀ (1 + αΔT)

2 = 1 (1 + 0.0012 x ΔT)

ΔT = T’ - 300

ΔT = 833 + 300 = 1133 K

Ques 10. Two wires of the same metal have the same length but their cross-sections are in the ratio 3:1. They are joined in series. The resistance of thick wire is 10Ω . The total resistance of combination is:

10Ω
40Ω
20Ω
100Ω

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Ans. (b). 40Ω

Explanation: R₁ = 10Ω , A₁ : A₂= 3:1

R = ρL/A

R ∝ 1/A

So, R₁/R₂ = A₂/A₁ = ⅓

R₂ = 3R₁ = 3 x 10 = 30Ω

The resistance are in series, so,

R_eq= R₁ + R₂ = 30 + 10 = 40Ω

Ques 11. Specific resistance of a wire depends on the

Length of the wire
Area of cross-section of the wire
Resistance of the wire
Material of the wire

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Ans. (d). Material of the wire

Explanation: The resistivity or specific resistance of a material is an intensive property and depends only on the material and temperature.

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MCQs on Electrical Resistance: Introduction & Explanation

MCQs on Electrical Resistance

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MCQs on Electrical Resistance: Introduction & Explanation

MCQs on Electrical Resistance

CBSE CLASS XII Related Questions

1. If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

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3.A light copper ring is freely suspended by a light string. A bar magnet is held horizontally with its length along the axis of the ring. The magnet is moved towards the ring with its N pole facing the loop. What will happen to the ring and its position? Explain.

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. If Bohr’s quantization postulate (angular momentum \( = \frac{nh}{2\pi} \)) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun? Explain.

6.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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If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

2.
Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

3.
A light copper ring is freely suspended by a light string. A bar magnet is held horizontally with its length along the axis of the ring. The magnet is moved towards the ring with its N pole facing the loop. What will happen to the ring and its position? Explain.

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.
If Bohr’s quantization postulate (angular momentum \( = \frac{nh}{2\pi} \)) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun? Explain.

6.
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