Q1. The r.m.s. value of induced emf per turn in a transformer is:
A. 2.22 f Φm
B. 4 f Φm
C. 4.44 f Φm
D. 1.11 f Φm
Correct Option: C
Explanation:
RMS emf per turn = 4.44 f Φm.
Average emf per turn = 4 f Φm and form factor of sine wave = 1.11.
Therefore, 4 × 1.11 = 4.44.
Q2. In a transformer, emf equation is:
A. E = 2π f N Φm
B. E = 4.44 f N Φm
C. E = 2 f N Φm
D. E = N Φm / f
Correct Option: B
Explanation:
For sinusoidal flux, the standard emf equation of a transformer is
E = 4.44 f N Φm, where f is frequency, N is number of turns and Φm is maximum flux.
Q3. If flux waveform is purely sinusoidal, form factor value assumed in emf equation is:
A. 1.0
B. 1.11
C. 1.732
D. 2.22
Correct Option: B
Explanation:
Form factor = RMS value / average value.
For a sinusoidal waveform, form factor = 1.11, which leads to the constant 4.44 in emf equation.
Q4. In an ideal transformer on no-load:
A. V1 = E1 and V2 = E2
B. V1 > E1
C. V1 < E1
D. V2 = 0
Correct Option: A
Explanation:
In an ideal transformer, there are no losses and no voltage drops. Hence applied voltage equals induced emf on both primary and secondary sides.
Q5. Voltage transformation ratio K is defined as:
A. N1 / N2
B. I1 / I2
C. E2/E1 = N2/N1
D. V1 / V2
Correct Option: C
Explanation:
Voltage ratio equals turns ratio. Therefore transformation ratio
K = E2/E1 = N2/N1.
Q6. If K > 1, the transformer is:
A. Step-down
B. Constant voltage
C. Step-up
D. Auto-transformer
Correct Option: C
Explanation:
K > 1 means secondary voltage is greater than primary voltage, hence the transformer is step-up.
Q7. In an ideal transformer:
A. Input VA > Output VA
B. Input VA < Output VA
C. Input VA = Output VA
D. Losses are large
Correct Option: C
Explanation:
An ideal transformer is lossless, therefore input apparent power equals output apparent power.
Q8. In an ideal transformer currents are related as:
A. I1 / I2 = N1 / N2
B. I1 / I2 = V1 / V2
C. I1 / I2 = N2 / N1
D. I1 = I2
Correct Option: C
Explanation:
From power balance V1I1 = V2I2 and V ∝ N,
therefore I1/I2 = N2/N1.
Q9. No-load primary current of transformer consists of:
A. Only reactive component
B. Only active component
C. Active and magnetising components
D. Only magnetising component
Correct Option: C
Explanation:
No-load current has magnetising component to establish flux and working (active) component to supply iron loss.
Q10. On no-load, primary current lags supply voltage by angle:
A. Exactly 90°
B. Less than 90°
C. More than 90°
D. Zero
Correct Option: B
Explanation:
Because no-load current has both reactive and active components, it lags voltage by slightly less than 90°.
Q11. No-load current magnitude is approximately:
A. 10–20% of full load current
B. 0.1% of full load current
C. 1–5% of full load current
D. 50% of full load current
Correct Option: C
Explanation:
Transformer exciting current is very small, typically about 1–5% of full-load current.
Q12. No-load input power mostly represents:
A. Copper loss
B. Mechanical loss
C. Iron loss
D. Leakage loss
Correct Option: C
Explanation:
At no-load, secondary current is zero so copper loss is negligible. Wattmeter reading mainly represents core (iron) loss.
Q13. Iron loss in transformer consists of:
A. Hysteresis loss only
B. Eddy current loss only
C. Copper loss and hysteresis loss
D. Hysteresis and eddy current loss
Correct Option: D
Explanation:
Iron or core loss is the sum of hysteresis loss and eddy current loss.
Q14. Copper loss varies as:
A. Load current
B. Square of load current
C. Flux
D. Frequency
Correct Option: B
Explanation:
Copper loss Pc = I²R, therefore it varies as the square of load current.
Q15. Transformer with losses but without leakage assumes:
A. No copper loss
B. Only mechanical loss
C. Iron and copper loss but no leakage flux
D. Leakage present
Correct Option: C
Explanation:
For simplified analysis, transformer is assumed to have iron and copper losses while leakage flux is neglected.
Q16. A single-phase transformer has 1000 turns, Φm = 0.01 Wb, f = 50 Hz. Find E1.
A. 22 V
B. 111 V
C. 222 V
D. 444 V
Correct Option: C
Explanation:
E = 4.44 f N Φm
= 4.44 × 50 × 1000 × 0.01
= 222 V.
Q17. A transformer has emf per turn 8 V and primary voltage 400 V. Primary turns are:
A. 25
B. 40
C. 50
D. 100
Correct Option: C
Explanation:
E = N × (emf per turn)
⇒ 400 = N × 8
⇒ N = 50 turns.
Q18. A 50 Hz transformer gives 200 V on secondary with 500 turns. Emf per turn is:
A. 0.4 V
B. 1 V
C. 2 V
D. 4 V
Correct Option: A
Explanation:
Emf per turn = 200 / 500 = 0.4 V.
Q19. Voltage transformation ratio is 5, primary voltage = 400 V. Secondary voltage is:
A. 40 V
B. 80 V
C. 2000 V
D. 100 V
Correct Option: C
Explanation:
K = V2/V1 = 5
⇒ V2 = 5 × 400 = 2000 V.
Q20. Full-load primary copper loss = 400 W. At half load copper loss is:
A. 50 W
B. 100 W
C. 200 W
D. 400 W
Correct Option: B
Explanation:
Copper loss ∝ I².
At half load, current = 0.5I,
Loss = (0.5)² × 400 = 100 W.