Electrical Engg (EEE) — MCQ Practice Questions & Answers

Question 1

A voltmeter has an internal resistance of 100 kΩ and shows a reading of 50 V across a 1 MΩ resistor. What is the true voltage across the resistor?

Accepted Answer

Correct answer: C. 54.55 V — The voltmeter and resistor form a parallel circuit. Using voltage divider: V_true = 50 × (1000 + 100) / 1000 ≈ 54.55 V

Answer

A. 45.45 V

Answer

B. 50 V

Answer

D. 60 V

Question 2

For accurate power measurement in a three-phase system with unbalanced load, the minimum number of wattmeters required is:

Accepted Answer

Correct answer: C. 3 — For unbalanced three-phase loads, three wattmeters are required—one in each phase—to measure total power accurately.

Answer

A. 1

Answer

B. 2

Answer

D. 4

Question 3

In an AC bridge at balance condition, if R₁ = 100 Ω, R₂ = 50 Ω, and R₃ (inductive) = 200 Ω with series R, then R₄ is:

Accepted Answer

Correct answer: A. 100 Ω — Using AC bridge balance: Z₁/Z₂ = Z₃/Z₄; therefore 100/50 = 200/Z₄, which gives Z₄ = 100 Ω

Answer

B. 50 Ω

Answer

C. 200 Ω

Answer

D. 400 Ω

Question 4

In a Kelvin's double bridge used for measuring low resistances (< 0.1 Ω), which configuration is adopted?

Accepted Answer

Correct answer: B. Cross-connection between potential and current terminals — Kelvin's double bridge uses cross-connection to eliminate lead resistance errors when measuring very low resistances in industrial applications.

Answer

A. Series arrangement of two arms

Answer

C. Parallel-series hybrid configuration

Answer

D. Bridge with four identical arms

Question 5

A strain gauge has gauge factor of 2.1 and resistance change of 5 Ω for a strain of 0.001. What is the original resistance of the gauge?

Accepted Answer

Correct answer: B. 2381 Ω — Gauge factor = (ΔR/R₀)/strain; 2.1 = 5/(R₀ × 0.001); R₀ = 5/(2.1 × 0.001) ≈ 2381 Ω.

Answer

A. 2300 Ω

Answer

C. 2500 Ω

Answer

D. 2000 Ω

Question 6

In a three-phase power measurement using two-wattmeter method, the power factor can be determined from which relationship?

Accepted Answer

Correct answer: A. tan(φ) = √3(W₁ - W₂)/(W₁ + W₂) — For two-wattmeter method in three-phase systems: tan(φ) = √3(W₁ - W₂)/(W₁ + W₂), used to calculate power factor.

Answer

B. cos(φ) = (W₁ + W₂)/√3VIL

Answer

C. sin(φ) = (W₁ - W₂)/(W₁ + W₂)

Answer

D. Power factor = (W₁ + W₂)/(W₁ - W₂)

Question 7

A light intensity meter using a photocell shows reading of 500 lux. If the luminous intensity of the source is constant, what happens to the reading when distance is doubled?

Accepted Answer

Correct answer: B. Becomes 125 lux — Illuminance follows inverse square law: E ∝ 1/d²; when distance doubles, illuminance becomes 1/4th: 500/4 = 125 lux.

Answer

A. Becomes 250 lux

Answer

C. Becomes 1000 lux

Answer

D. Remains 500 lux

Electrical Engg (EEE) - MCQ Practice Questions