Current Electricity

13 NEET-UG practice questions on Current Electricity, part of Physics. The first 13 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

1. Question 1 · Medium

   A uniform wire of resistance 25 Ω is stretched until its length is 20% greater. Its volume and resistivity do not change. Find its new resistance.

   1. A20.8 Ω
   2. B30 Ω
   3. C43.2 Ω
   4. D36 Ω

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   Answer: D · 36 Ω

   Constant volume means the area falls by the factor 1.2 as the length rises by 1.2. Since R = ρl/A, R becomes 1.2 × 1.2 = 1.44 times. New R = 25 × 1.44 = 36 Ω.

   • A: R was divided by 1.2 as if it fell on stretching.
   • B: Only the length rise was counted; the fall in area was ignored.
   • C: The ratio 1.2 was cubed instead of squared.

   Remember this

   At constant volume R is proportional to l², so a fractional stretch multiplies R by the square of the length ratio.

2. Question 2 · Easy

   An electric heater is rated 1200 W when it runs on a 240 V supply. What is the resistance of its element while it works at this rating?

   1. A48 Ω
   2. B96 Ω
   3. C0.2 Ω
   4. D5 Ω

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   Answer: A · 48 Ω

   Power and voltage give the resistance through P = V²/R. So R = V²/P = (240 × 240)/1200 = 48 Ω.

   • B: An extra factor of 2 has been put into V²/P.
   • C: V/P is the reciprocal of the current and has no meaning as resistance.
   • D: P/V gives the current of 5 A, not the resistance.

   Remember this

   For a rated appliance use R = V²/P; a higher rated power at the same voltage means a smaller resistance.

3. Question 3 · Medium

   A conductor of cross-section 2.0 mm² carries a steady current of 3.2 A. It has 1.0 × 10²⁹ free electrons per m³. Taking e = 1.6 × 10⁻¹⁹ C, find the drift speed of the electrons.

   1. A5.0 × 10⁻⁵ m/s
   2. B1.0 × 10⁻⁷ m/s
   3. C2.0 × 10⁻⁴ m/s
   4. D1.0 × 10⁻⁴ m/s

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   Answer: D · 1.0 × 10⁻⁴ m/s

   From I = neAv, v = I/(neA). Here neA = (1.0 × 10²⁹)(1.6 × 10⁻¹⁹)(2.0 × 10⁻⁶) = 3.2 × 10⁴ C/m. So v = 3.2/(3.2 × 10⁴) = 1.0 × 10⁻⁴ m/s.

   • A: The electron charge was doubled by mistake.
   • B: 1 mm² was wrongly converted to 10⁻³ m².
   • C: The factor 2.0 in the area was left out.

   Remember this

   Drift speed is v = I/(neA), very small, and 1 mm² must be converted to 10⁻⁶ m².

4. Question 4 · Hard

   An ideal 12 V battery feeds two parallel branches. Branch P has 2 Ω and 4 Ω in series, and branch Q has 3 Ω and 1 Ω in series, with the 2 Ω and 3 Ω resistors joined to the positive terminal. The junction between the two resistors of P is now connected to the junction in Q by a wire of zero resistance. Find the current in this wire.

   1. A3.6 A
   2. B2.4 A
   3. C1.2 A
   4. D6.0 A

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   Answer: B · 2.4 A

   The wire makes the two junctions one node, so the network is (2 Ω || 3 Ω) in series with (4 Ω || 1 Ω), which is 1.2 + 0.8 = 2 Ω, and the battery current is 6 A. The junction potential is 6 × 0.8 = 4.8 V, so the 2 Ω carries 3.6 A and the 4 Ω carries 1.2 A. By Kirchhoff's junction rule the wire carries 3.6 - 1.2 = 2.4 A.

   • A: This is the current in the 2 Ω resistor.
   • C: This is the current in the 4 Ω resistor.
   • D: This is the total battery current.

   Remember this

   With an unbalanced bridge, a shorted link carries the difference between the currents entering and leaving its junction.

5. Question 5 · Easy

   The temperature of a copper wire is raised. How do the average relaxation time of its free electrons and its resistivity change?

   1. ARelaxation time decreases and resistivity decreases
   2. BRelaxation time increases and resistivity increases
   3. CRelaxation time decreases and resistivity increases
   4. DRelaxation time increases and resistivity decreases

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   Answer: C · Relaxation time decreases and resistivity increases

   Resistivity is ρ = m/(ne²τ). In a metal n hardly changes with temperature, but collisions with the ions become more frequent, so τ falls and ρ rises.

   • A: A shorter relaxation time raises resistivity; this is the behaviour wrongly reversed.
   • B: ρ varies inversely with τ, so both cannot rise together.
   • D: This describes the wrong trend for τ and for ρ in a metal.

   Remember this

   In metals resistivity rises with temperature because the relaxation time falls, while carrier density stays nearly constant.

6. Question 6 · Medium

   Match List-I with List-II. List-I (quantity) A. Resistivity B. Conductivity C. Mobility D. Current density List-II (SI unit) I. A m⁻² II. Ω m III. m² V⁻¹ s⁻¹ IV. S m⁻¹

   1. AA-IV, B-II, C-III, D-I
   2. BA-II, B-III, C-IV, D-I
   3. CA-II, B-IV, C-I, D-III
   4. DA-II, B-IV, C-III, D-I

   Answer and explanation in LearnFree account

7. Question 7 · Hard

   Two cells are joined in parallel, positive terminal to positive terminal: one of emf 10 V and internal resistance 2 Ω, the other of emf 4 V and internal resistance 1 Ω. A 2 Ω resistor is connected across this combination. What happens in the 4 V cell?

   1. A2.0 A flows and the cell is being charged
   2. B0.5 A flows and the cell is discharging
   3. C4.0 A flows and the cell is discharging
   4. D0.5 A flows and the cell is being charged

   Answer and explanation in LearnFree account

8. Question 8 · Medium

   In a Wheatstone bridge ABCD, the arm resistances are AB = 4 Ω, BC = 6 Ω, AD = 8 Ω and DC = 12 Ω. A galvanometer of resistance 5 Ω joins B and D, and a 20 V battery of negligible internal resistance joins A and C. What current does the battery supply?

   1. A1.0 A
   2. B3.0 A
   3. C0.67 A
   4. D2.0 A

   Answer and explanation in LearnFree account

9. Question 9 · Medium

   In this meter bridge experiment the galvanometer shows no deflection with the jockey at J, on a uniform wire AB of length 100 cm. What is the unknown resistance X?

   

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   1. A20 Ω
   2. B18 Ω
   3. C45 Ω
   4. D12 Ω

   Answer and explanation in LearnFree account

10. Question 10 · Medium

    A cell of constant emf and internal resistance r is connected to a variable external resistance R. Which graph shows how the power P delivered to R varies with R?

    

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    1. AGraph B
    2. BGraph A
    3. CGraph C
    4. DGraph D

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11. Question 11 · Hard

    Both cells in this network have negligible internal resistance. Find the current through the 4 V cell.

    

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    1. A1 A, entering the cell at its positive terminal
    2. B3 A, entering the cell at its positive terminal
    3. C1 A, leaving the cell at its positive terminal
    4. D2 A, leaving the cell at its positive terminal

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12. Question 12 · Easy

    Lines P and Q in the graph are for the same metal wire at two different temperatures. Assertion A: Line Q belongs to the higher temperature. Reason R: The resistivity of a metal increases as its temperature rises. Choose the correct answer.

    

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    1. ABoth A and R are correct but R is not the correct explanation of A
    2. BA is not correct but R is correct
    3. CA is correct but R is not correct
    4. DBoth A and R are correct and R is the correct explanation of A

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13. Question 13 · Medium

    How much power is dissipated in the 6 Ω resistor of this circuit? The cell has negligible internal resistance.

    

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    1. A6 W
    2. B54 W
    3. C12 W
    4. D24 W

    Answer and explanation in LearnFree account