Current Electricity — Class 12 MCQs with Answers

Current Electricity — important questions & MCQs with answers (Class 12 Physics)

14 solved questions from this chapter's difficulty levels, each with its answer and explanation. The other 76 are timed and scored when you sign in.

1. Q1Easy

   The SI unit of resistivity (ρ) is:

   A.ohm

   B.ohm/metre

   C.ohm-metre✓ Correct

   D.siemens

   Answer: C. ohm-metre

   Explanation: Resistivity is defined by R = ρL/A, so ρ = RA/L carries units Ω·m² ÷ m = Ω·m. A common slip is picking plain ohm, confusing resistance (one wire's property) with resistivity, a material constant.

2. Q2Easy

   Kirchhoff's current law (KCL) is based on conservation of:

   A.Energy

   B.Charge✓ Correct

   C.Mass

   D.Momentum

   Answer: B. Charge

   Explanation: Kirchhoff's current law says ΣI_in = ΣI_out at any junction — charge can't pile up or vanish there, so it's a statement of charge conservation, not energy or momentum (that's KVL's job).

3. Q3Easy

   Equivalent resistance of n equal R in series:

   A.R/n

   B.R

   C.nR✓ Correct

   D.R²

   Answer: C. nR

   Explanation: In series, the same current flows through every resistor, so their voltage drops (and hence resistances) simply add: R_eq = R+R+...+R = nR — not R/n, which is the parallel result.

4. Q4Easy

   When the temperature of a metallic conductor increases, its resistance:

   A.Increases✓ Correct

   B.Decreases

   C.Becomes zero

   D.Remains the same

   Answer: A. Increases

   Explanation: Higher temperature increases lattice-ion vibrations, so conduction electrons collide more often — relaxation time τ falls, and since R ∝ 1/τ, resistance rises. This is why metals have a positive temperature coefficient α.

5. Q5Easy

   KVL is based on conservation of:

   A.Energy✓ Correct

   B.Charge

   C.Current

   D.Mass

   Answer: A. Energy

   Explanation: Kirchhoff's voltage law says the sum of EMFs around a closed loop equals the sum of IR drops — this is energy conservation, since moving a charge round a loop and back does zero net work.

6. Q6Easy

   Equivalent resistance of n equal R in parallel:

   A.nR

   B.n/R

   C.R

   D.R/n✓ Correct

   Answer: D. R/n

   Explanation: In parallel, conductances add: 1/R_eq = n/R, giving R_eq = R/n. Confusing this with the series rule (nR) is the classic mix-up.

7. Q7Medium

   A wire of length 2 m and cross-section 1 mm² has resistivity 1.7 × 10⁻⁸ Ωm. Its resistance is approximately:

   A.0.34 Ω

   B.0.034 Ω✓ Correct

   C.3.4 Ω

   D.34 Ω

   Answer: B. 0.034 Ω

   Explanation: R = ρL/A = (1.7×10⁻⁸ Ω·m × 2 m)/(1×10⁻⁶ m²) = 3.4×10⁻² Ω = 0.034 Ω — forgetting to convert 1 mm² to 1×10⁻⁶ m² is what makes 3.4 Ω or 34 Ω tempting but wrong.

8. Q8Medium

   In a loop, sum of IR drops = sum of EMFs is essentially:

   A.KCL

   B.Ohm's law

   C.KVL✓ Correct

   D.Joule's law

   Answer: C. KVL

   Explanation: Summing IR drops around a loop and setting that equal to the sum of EMFs is exactly Kirchhoff's voltage law (KVL) — a direct statement that energy is conserved as charge goes around and returns to its start.

9. Q9Medium

   A cell of emf ε and internal resistance r drives current I through external resistance R. The terminal voltage is:

   A.V = ε

   B.V = ε + Ir

   C.V = IR + Ir

   D.V = ε − Ir✓ Correct

   Answer: D. V = ε − Ir

   Explanation: As current flows out of a cell, some EMF is used up driving charge through its own internal resistance, so terminal voltage V = ε − Ir is always less than ε on discharge — V = ε + Ir only applies while charging.

10. Q10Medium

    Current in conductor I = neAv_d. If n doubles, v_d (for same I, A):

    A.Doubles

    B.Quadruples

    C.Unchanged

    D.Halves✓ Correct

    Answer: D. Halves

    Explanation: From I = neAv_d, for fixed I and A, v_d ∝ 1/n — doubling the carrier density n means each electron need only drift half as fast to carry the same current.

11. Q11Medium

    Three resistors of 2 Ω, 3 Ω and 6 Ω are connected in parallel. The equivalent resistance is:

    A.1/11 Ω

    B.11 Ω

    C.1 Ω✓ Correct

    D.6 Ω

    Answer: C. 1 Ω

    Explanation: For parallel resistors, 1/R_eq = 1/2+1/3+1/6 = 3/6+2/6+1/6 = 1, so R_eq = 1 Ω — remember to invert at the end; forgetting that step is why 6 Ω looks tempting.

12. Q12Hard

    If the cross-section of a wire is halved while keeping current and material constant, the drift velocity:

    A.Halves

    B.Doubles✓ Correct

    C.Quadruples

    D.Stays the same

    Answer: B. Doubles

    Explanation: From I = neAv_d, v_d ∝ 1/A for fixed current and carrier density, so halving the cross-section doubles the drift velocity — the same current now squeezes through half the area, so each electron must move faster.

13. Q13Hard

    In a Wheatstone bridge, the bridge is balanced when:

    A.P + Q = R + S

    B.P × Q = R × S

    C.P − R = Q − S

    D.P/Q = R/S✓ Correct

    Answer: D. P/Q = R/S

    Explanation: Balance means zero galvanometer current, which by Kirchhoff's laws forces the ratio condition P/Q = R/S — not a sum or product relation, which is why those distractors look tempting but aren't dimensionally the balance rule.

14. Q14Hard

    Galvanometer (G = 50 Ω, full-scale 1 mA) is to read 1 A. Shunt resistance:

    A.5 Ω

    B.50 Ω

    C.≈ 0.05 Ω✓ Correct

    D.500 Ω

    Answer: C. ≈ 0.05 Ω

    Explanation: A shunt diverts the bulk of the current around the galvanometer: S = I_g·G/(I−I_g) = (10⁻³×50)/(1−10⁻³) ≈ 0.05 Ω — it must be tiny since it carries nearly all of the 1 A while the galvanometer takes only 1 mA.

Current Electricity — FAQs

More Class 12 Physics chapters

• 01Electric Charges and Fields
• 02Electrostatic Potential and Capacitance
• 04Moving Charges and Magnetism
• 05Magnetism and Matter
• 06Electromagnetic Induction
• 07Alternating Current

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