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Class XII · Physics

Current Electricity: NEET MCQ practice

204 practice questions on Current Electricity, each with an explanation. Practise them chapter by chapter in the GyaanLoop app.

204

Practice questions

107

Easy

96

Medium

1

Hard

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What Current Electricity covers

Topics in the NEET (UG) 2026–27 syllabus for this chapter.

  • Electric current, drift velocity, mobility
  • Ohm's law, V-I characteristics, electrical resistance and resistivity
  • Series and parallel combination of resistors, temperature dependence of resistance
  • Internal resistance, potential difference and emf of a cell, combination of cells
  • Kirchhoff's laws and applications, Wheatstone bridge, metre bridge

Sample questions on Current Electricity

Try each one, then open the answer and explanation.

Question 1 · Practice question

The resistance of a heating element is 99Ω at room temperature. What is the temperature of the element if the resistance is found to be 116Ω? (Temperature coefficient of the material of the resistor is 1.7 × 10 −4∘ C −1 )

  1. A.999.9 ∘ C
  2. B.1005.3 ∘ C
  3. C.1020.2 ∘ C
  4. D.1037.1 ∘ C
Show answer

Answer: D. 1037.1 ∘ C

Use the temperature dependence formula: R = R₀(1 + αΔT), with R₀ = 99 Ω, R = 116 Ω, α = 1.7 × 10⁻⁴ °C⁻¹. Solve for ΔT = (116/99 − 1)/α ≈ 1010.1°C. Since room temperature is taken as 27°C, final temperature = 27 + 1010.1 = 1037.1°C. Option (a) 999.9°C incorrectly assumes initial temperature 0°C; option (b) 1005.3°C and (c) 1020.2°C come from arithmetic mistakes.

Question 2 · Practice question

The resistance of wire in a heater at room temperature is 65Ω. When the heater is connected to a 220V supply the current settles after a few seconds to 2.8A. What is the steady temperature of the wire. (Temperature coefficient of resistance α = 1.70 × 10 −4∘ C −1 )

  1. A.955 ∘ C
  2. B.1055 ∘ C
  3. C.1155 ∘ C
  4. D.1258 ∘ C
Show answer

Answer: D. 1258 ∘ C

First, use Ohm’s law to find the steady-state resistance: R = V/I = 220 V / 2.8 A ≈ 78.57 Ω. Then, use the temperature dependence formula R = R₀[1 + α(T−T₀)], solving for T. Assuming room temperature T₀ = 30°C (a common assumption when not specified), substituting gives T ≈ 1258°C. The tricky part is realizing that 220/2.8 is not a round number, so careful calculation is needed, and one must recall the formula linking resistance and temperature.

Question 3 · Practice question

A wire has a resistance of 2.5Ω at 28 ∘ C and a resistance of 2.9Ω at 100 ∘ C. The temperature coefficient of resistivity of material of the wire is

  1. A.1.06 × 10 −3∘ C −1
  2. B.3.5 × 10 −2∘ C −1
  3. C.2.22 × 10 −3∘ C −1
  4. D.3.95 × 10 −2∘ C −1
Show answer

Answer: C. 2.22 × 10 −3∘ C −1

Use the linear approximation formula R₂ = R₁[1 + α(T₂ – T₁)] with R₁ = 2.5 Ω at 28°C, R₂ = 2.9 Ω at 100°C. Solving gives α ≈ 2.22 × 10⁻³ °C⁻¹. Distraction (a) could come from misplacing the decimal in the arithmetic, but careful step-by-step calculation yields the correct value.

Question 4 · Practice question

If voltage across a bulb rated 220V, 100W drops by 2.5% of its rated value, the percentage of the rated value by which the power would decrease is

  1. A.20%
  2. B.2.5%
  3. C.5%
  4. D.10%
Show answer

Answer: C. 5%

Power in a resistive load is proportional to the square of the voltage (P = V²/R). Here, voltage decreases by 2.5%, so V(new) = 0.975 V(rated). Squaring gives 0.975² ≈ 0.9506, meaning power decreases by about 4.94% ≈ 5%. Option (b) 2.5% is a trap if you assume linear relationship, while option (d) 10% would be if voltage dropped by 5%.

Question 5 · Practice question

Four wires of the same diameter are connected, in turn, between two points maintained at a constant potential difference. Their resistivities and lengths are; ρ and L (wire 1), 1.2ρ and 1.2L (wire 2), 0.9ρ and 0.9L (wire 3) and ρ and 1.5L (wire 4). Rank the wires according to the rates at which energy is dissipated as heat, greatest first,

  1. A.4 > 3 > 1 > 2
  2. B.4 > 2 > 1 > 3
  3. C.1 > 2 > 3 > 4
  4. D.3 > 1 > 2 > 4
Show answer

Answer: D. 3 > 1 > 2 > 4

Since all wires have the same cross-section and are connected to the same potential difference, power dissipated is inversely proportional to resistance R. R is proportional to ρL. Ranking ρL from smallest to largest gives wire 3 (0.81ρL), wire 1 (ρL), wire 2 (1.44ρL), wire 4 (1.5ρL). Thus, power is greatest for wire 3 and least for wire 4, matching option D. Don't mistakenly think longer or higher resistivity alone means more heating—under constant voltage, higher resistance means less power.

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