All Free Electrical Engineering MCQs with Answers

• A. 100 N
• B. 400 N
• C. 600 N
• D. 1000 N

Explanation: For a conductor perpendicular to the field, F = BIl = 0.67 × 50 × 3 = 100.5 N, approximately 100 N.

• A. 20 Wb
• B. 2 Wb
• C. 0.2 Wb
• D. 0.02 Wb

Explanation: Faraday’s law gives e = N(ΔΦ/Δt), so ΔΦ = eΔt/N = 10 × 0.5/250 = 0.02 Wb. The total flux change is therefore option d.

• A. Air core
• B. Laminated iron core
• C. Iron core
• D. Powdered iron core

Explanation: An air-core inductor has no conducting magnetic core in which circulating eddy currents can form, so its eddy-current loss is least.

• A. 0.01 s
• B. 0.1 s
• C. 0.5 s
• D. 5 s

Explanation: Using e = N(ΔΦ/Δt), the required time is t = NΔΦ/e = 500 × 0.06/60 = 0.5 s. Thus the flux change must occur over half a second.

• A. 50 ohms
• B. 500 ohms
• C. 1000 ohms
• D. 5000 ohms

Explanation: For a coil with negligible resistance, the applied voltage is across its inductive reactance, so Xₗ = V/I = 50/0.01 = 5000 Ω.

• A. 50 N
• B. 120 N
• C. 240 N
• D. 480 N

Explanation: The force is F = BIL sin θ = 1.1 × 125 × sin 60° ≈ 119 N, which rounds to 120 N.

• A. 10 V
• B. 15 V
• C. 25V
• D. 50V

Explanation: For a conductor moving perpendicular to a magnetic field, the induced emf is e = Blv = 1 × 2 × 12.5 = 25 V.

• A. 6 H
• B. 8 H
• C. 12 H
• D. 24 H

Explanation: With zero mutual inductance, series inductances simply add: L = L₁ + L₂ = 10 + 2 = 12 H.

• A. induced current
• B. charge
• C. energy
• D. induced e.m.f.

Explanation: Lenz's law gives the induced emf a direction that opposes the change producing it, preventing energy from appearing without an input.

• A. 2.0
• B. 1.0
• C. 0.5
• D. zero

Explanation: The coefficient of coupling is k = mutual flux divided by the total flux produced by the first coil that could link the second.

• A. 6 mH
• B. 14 mH
• C. 24 mH
• D. 48 mH

Explanation: For the same core geometry, inductance varies as the square of the number of turns, L ∝ N².

• A. 4 mH
• B. 5 mH
• C. 6 mH
• D. 12 mH

Explanation: Mutual inductance is M = k√(L₁L₂) = 0.5√(8 × 18) mH = 0.5 × 12 = 6 mH. The coupling coefficient alone cannot be multiplied directly by…

• A. voltage across the inductance
• B. magnetic field
• C. both A. and B.
• D. neither (A) nor (B)

Explanation: In a linear magnetic circuit, current is proportional to flux and hence to magnetic field strength, while voltage is proportional to the…

• A. Apparent power is zero
• B. Relative power is.zero
• C. Actual power of the circuit is zero
• D. Any capacitance even if present in the circuit will not be charged

Explanation: A pure inductor consumes no average or actual power because current is 90° out of phase with voltage, although its apparent power is not…

• A. 64 A/s
• B. 32 A/s
• C. 16 A/s
• D. 4 A/s

Explanation: Using e = L(di/dt), the rate of change of current is di/dt = e/L = 16/4 = 4 A/s.

• A. 3 mH
• B. 12 mH
• C. 24mH
• D. 48mH

Explanation: For a coil on the same core, inductance is inversely proportional to magnetic-path length, L ∝ 1/l.

• A. Ohm
• B. Henry
• C. Ampere turns
• D. Webers/metre

Explanation: The henry is the SI unit of inductance, equal to one volt-second per ampere.

• A. Capacitance
• B. Inductance
• C. Resistance
• D. All of the above

Explanation: An inductor opposes any change in current by producing a back emf, according to v = L(di/dt).

• A. flux density of the magnetic field
• B. amount of flux cut
• C. amount of flux linkages
• D. rate of change of flux-linkages

Explanation: Faraday’s law gives induced emf as e = −dλ/dt, where λ is flux linkage.

• A. Henry
• B. Coulomb/volt ampere
• C. Volt second per ampere
• D. All of the above

Explanation: Henry, volt-second per ampere, and weber per ampere are units of inductance.