Home Subjects Electrical Engg (EEE) Electrical Measurements

Electrical Engg (EEE)
Electrical Measurements

Electrical machines, power systems, circuits

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Topics in Electrical Engg (EEE)
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?
A Becomes 250 lux
B Becomes 125 lux
C Becomes 1000 lux
D Remains 500 lux
Correct Answer:  B. Becomes 125 lux
EXPLANATION

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

Test
In a three-phase power measurement using two-wattmeter method, the power factor can be determined from which relationship?
A tan(φ) = √3(W₁ - W₂)/(W₁ + W₂)
B cos(φ) = (W₁ + W₂)/√3VIL
C sin(φ) = (W₁ - W₂)/(W₁ + W₂)
D Power factor = (W₁ + W₂)/(W₁ - W₂)
Correct Answer:  A. tan(φ) = √3(W₁ - W₂)/(W₁ + W₂)
EXPLANATION

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

Test
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?
A 2300 Ω
B 2381 Ω
C 2500 Ω
D 2000 Ω
Correct Answer:  B. 2381 Ω
EXPLANATION

Gauge factor = (ΔR/R₀)/strain; 2.1 = 5/(R₀ × 0.001); R₀ = 5/(2.1 × 0.001) ≈ 2381 Ω.

Test
In a Kelvin's double bridge used for measuring low resistances (< 0.1 Ω), which configuration is adopted?
A Series arrangement of two arms
B Cross-connection between potential and current terminals
C Parallel-series hybrid configuration
D Bridge with four identical arms
Correct Answer:  B. Cross-connection between potential and current terminals
EXPLANATION

Kelvin's double bridge uses cross-connection to eliminate lead resistance errors when measuring very low resistances in industrial applications.

Test
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:
A 100 Ω
B 50 Ω
C 200 Ω
D 400 Ω
Correct Answer:  A. 100 Ω
EXPLANATION

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

Test
For accurate power measurement in a three-phase system with unbalanced load, the minimum number of wattmeters required is:
A 1
B 2
C 3
D 4
Correct Answer:  C. 3
EXPLANATION

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

Test
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?
A 45.45 V
B 50 V
C 54.55 V
D 60 V
Correct Answer:  C. 54.55 V
EXPLANATION

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

Test
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