Q1. The approximate voltage drop in a transformer at lagging power factor is:
A. I₂R₀₂ sinφ + I₂X₀₂ cosφ
B. I₂R₀₂ cosφ + I₂X₀₂ sinφ
C. I₂R₀₂ cosφ − I₂X₀₂ sinφ
D. I₂R₀₂ − I₂X₀₂
Correct Option: B
Explanation:
At lagging power factor, current lags voltage. The resistive voltage drop depends on cosφ and the reactive drop depends on sinφ. Both drops add, giving I₂R₀₂ cosφ + I₂X₀₂ sinφ.
Q2. For leading power factor, approximate voltage drop becomes:
A. I₂R₀₂ cosφ + I₂X₀₂ sinφ
B. I₂R₀₂ cosφ − I₂X₀₂ sinφ
C. I₂R₀₂ sinφ − I₂X₀₂ cosφ
D. I₂R₀₂ − I₂X₀₂ sinφ
Correct Option: B
Explanation:
In leading power factor, current leads voltage. Reactive voltage drop reverses its sign, so the X-component is subtracted while resistive drop remains positive.
Q3. Exact voltage drop is equal to:
A. AN
B. AM
C. AC
D. MN
Correct Option: B
Explanation:
From the phasor diagram of transformer voltage regulation:
AN = approximate drop,
AM = exact drop,
NM = neglected component.
Hence exact voltage drop is AM.
Q4. Percentage regulation using approximate method is:
A. vr sinφ ± vx cosφ
B. vr cosφ ± vx sinφ
C. vr + vx
D. vr − vx
Correct Option: B
Explanation:
Approximate percentage regulation is given by:
%VR = vr cosφ ± vx sinφ.
“+” is used for lagging power factor and “−” for leading power factor.
Q5. In a transformer, voltage regulation is maximum when:
A. pf = 1
B. pf = 0.5 lag
C. pf = resistance / reactance ratio
D. pf = reactance / resistance ratio
Correct Option: C
Explanation:
Voltage regulation is maximum when tanφ = X/R.
This condition corresponds to power factor equal to R/Z, i.e., resistance-to-impedance ratio.
Q6. The equivalent circuit of transformer consists of:
A. R₀ only
B. X₀ only
C. R₀ and X₀ in parallel with series branch
D. Only ideal transformer
Correct Option: C
Explanation:
Practical transformer equivalent circuit has core-loss resistance R₀ and magnetising reactance X₀ in parallel, along with series winding resistance and leakage reactance.
Q7. Core-loss component of no-load current flows through:
A. Magnetising reactance
B. Core loss resistance
C. Leakage reactance
D. Equivalent impedance
Correct Option: B
Explanation:
The core-loss component is in phase with voltage and flows through core loss resistance R₀, representing hysteresis and eddy current losses.
Q8. In approximate equivalent circuit, exciting current is:
A. Neglected
B. Doubled
C. Halved
D. Made in phase with voltage
Correct Option: A
Explanation:
Exciting current is only about 2–5% of full-load current, so it is neglected for simplified approximate calculations.
Q9. Open-circuit test is conducted to determine:
A. Copper losses
B. Leakage reactance
C. Core loss
D. Efficiency only
Correct Option: C
Explanation:
In OC test, rated voltage is applied at no-load. Current is very small, so copper loss is negligible and wattmeter reading gives core loss.
Q10. In open-circuit test wattmeter reading gives:
A. Copper loss
B. Hysteresis loss only
C. Eddy current loss only
D. Core loss
Correct Option: D
Explanation:
Wattmeter measures total core loss, which includes both hysteresis loss and eddy current loss.
Q11. In open-circuit test normally energized winding is:
A. Low voltage
B. High voltage
C. Any winding
D. Secondary only
Correct Option: B
Explanation:
High-voltage winding is energized so that rated voltage can be applied with very small current, keeping instruments within safe limits.
Q12. Short-circuit test is used to determine:
A. Iron losses
B. Magnetising current
C. Full-load copper loss
D. Total impedance only
Correct Option: C
Explanation:
During SC test, applied voltage is very small, flux is negligible and iron loss is ignored. Wattmeter reading gives full-load copper loss.
Q13. During short-circuit test applied voltage is:
A. Full rated voltage
B. 50% rated voltage
C. 5 to 10% of rated voltage
D. 200% rated voltage
Correct Option: C
Explanation:
Only 5–10% of rated voltage is required to circulate full-load current due to small leakage impedance.
Q14. In S.C test core losses are:
A. Maximum
B. Negligible
C. Equal to copper losses
D. Zero only at unity pf
Correct Option: B
Explanation:
Applied voltage is very small, so flux is very low and iron losses are negligible.
Q15. A transformer has %R = 2%, %X = 6%, pf = 0.8 lag. Voltage regulation ≈:
A. 2.8%
B. 5.2%
C. 6.4%
D. 7.6%
Correct Option: B
Explanation:
Voltage regulation = vr cosφ + vx sinφ
= 2×0.8 + 6×0.6 = 1.6 + 3.6 = 5.2%.
Q16. A 230/460 V transformer supplies 10 A at 0.8 pf lag. R₀₂ = 1.55 Ω, X₀₂ = 3.8 Ω. Secondary terminal voltage approximately is:
A. 380 V
B. 400 V
C. 425 V
D. 460 V
Correct Option: C
Explanation:
Voltage drop ΔV = I(Rcosφ + Xsinφ)
= 10(1.55×0.8 + 3.8×0.6) ≈ 35.2 V.
V₂ ≈ 460 − 35.2 ≈ 425 V.
Q17. In a no-load test: V = 220 V, I = 0.5 A, W = 30 W. Core loss =:
A. 15 W
B. 25 W
C. 30 W
D. 110 W
Correct Option: C
Explanation:
At no-load, copper loss is negligible. Wattmeter directly reads core loss, hence core loss = 30 W.
Q18. From the above test the working component of current is:
A. 0.068 A
B. 0.12 A
C. 0.136 A
D. 0.5 A
Correct Option: C
Explanation:
cosφ = W/(VI) = 30/(220×0.5) = 0.2727
Working component Iw = I cosφ = 0.5×0.2727 = 0.136 A.
Q19. In separation of losses method, total core loss is expressed as:
A. Af² + Bf
B. Af + Bf²
C. Af³ + Bf
D. Af − Bf²
Correct Option: B
Explanation:
Total core loss = hysteresis loss (∝ f) + eddy current loss (∝ f²).
Therefore Pc = Af + Bf².
Q20. In short circuit test, wattmeter reading equals:
A. Iron loss only
B. Copper loss at full load
C. Hysteresis loss only
D. Stray losses only
Correct Option: B
Explanation:
In SC test, flux is very small and iron losses are negligible. Wattmeter reading represents I²R loss at full-load, i.e., copper loss.