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:

  1. 760Ω
  2. 260Ω
  3. 1060Ω
  4. 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:

  1. 10Ω
  2. 20Ω
  3. 5Ω
  4. 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,

RD = (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:

  1. 50/15 mho
  2. 0.3 mho
  3. 2/3 mho
  4. 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?

  1. 184.7 m
  2. 200 m
  3. 190.5 m
  4. 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?

  1. 110 m
  2. 122.6 m
  3. 100 m
  4. 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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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:

  1. 32Ω and 4Ω
  2. 16Ω and 20Ω
  3. 4Ω and 32Ω
  4. 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?

  1. Decreases
  2. Increases
  3. Does not change
  4. 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:

  1. 500Ω and 2 A
  2. 200Ω and 4 A
  3. 806.67Ω and 0.27 A
  4. 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?

  1. 1133 K
  2. 854 K
  3. 1217 K
  4. 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:

  1. 10Ω
  2. 40Ω
  3. 20Ω
  4. 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,

Req = R1 + R2 = 30 + 10 = 40Ω

Ques 11. Specific resistance of a wire depends on the

  1. Length of the wire
  2. Area of cross-section of the wire
  3. Resistance of the wire
  4. 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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CBSE CLASS XII Related Questions

  • 1.

    Consider a cylindrical conductor of length $l$ and area of cross-section $A$. Current $I$ is maintained in the conductor and electrons drift with velocity $\vec{v}_d$ ($|\vec{v}_d| = \frac{e|\vec{E}|}{m}\tau$), (where symbols have their usual meanings). Show that the conductivity $\sigma$ of the material of the conductor is given by

    $$\sigma = \frac{ne^2}{m}\tau.$$


      • 2.
        The resistance of a metal wire at \(20^\circ\text{C}\) is \(1.05\,\Omega\) and at \(100^\circ\text{C}\) is \(1.38\,\Omega\). Determine the temperature coefficient of resistivity of this metal.


          • 3.
            Consider the nuclear reaction \( X \to Y + Z \). Let \( M_x \), \( M_y \), and \( M_z \) be the masses of the three nuclei X, Y, and Z respectively. Then which of the following relations hold true?

              • \( (M_x - M_z)<M_y \)
              • \( (M_x - M_y)<M_z \)
              • \( M_x>(M_y + M_z) \)
              • \( M_x<(M_y + M_z) \)

            • 4.

              A ray of light MN is incident normally on the face corresponding with side AB of a prism with an isosceles right-angled triangular base ABC. Trace the path of the ray as it passes through the prism when the refractive index of the prism material is (i) $\sqrt{2}$, and (ii) $\sqrt{3}$.


                • 5.

                  A parallel beam of monochromatic light falls normally on a single slit of width $a$ and a diffraction pattern is observed on a screen placed at distance $D$ from the slit. Explain:

                  (I) the formation of maxima and minima in the diffraction pattern, and

                  (II) why the maxima go on becoming weaker and weaker with its increasing number $n$.


                    • 6.

                      A rectangular loop of size 5 cm $\times$ 8 cm is lying in the $x$-$y$ plane in a uniform magnetic field $\vec{B} = (2.0\ \text{T})\hat{k}$. The total magnetic flux linked with the loop is:

                        • 80 Wb
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                        • \(8 \times 10^{-2}\) Wb
                        • \(8 \times 10^{-3}\) Wb
                      CBSE CLASS XII Previous Year Papers

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