GATE 2026 EE Q.31: solution and answer

GATE 2026 · EEQ.31 · 1 mark · MSQ

A 220V/12V single-phase transformer is designed for use in India and rated 100 VA at 50 Hz. Later, this unit is shipped to the USA where it is used as a 110V/6V transformer at 60 Hz.

Which of the following statements is/are correct?

(A)
No-load current drawn would be smaller for operation in the USA compared to that in India
(B)
For the same load current, the losses would be higher in the USA compared to that in India
(C)
The peak magnetic flux density in the core would be higher while operating in the USA compared to that in India
(D)
The eddy current losses in the core would be approximately 44% higher while operating in the USA compared to that in India

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Answer and solution

ANSWER

The correct option is (A) only.

Official answer: ASource: Official answer key, GATE 2026 EE (IIT Guwahati)

Since it is the same physical transformer, only re-rated for the USA supply, its turns ratio is unchanged: $220/12 = 18.33 = 110/6$ . Only the applied voltage $V$ and frequency $f$ change; the core, the windings, the magnetizing inductance $L_{m}$ , and the core-loss constants all stay the same.
From the EMF equation $E = 4.44 f N ϕ_{m}$ , for a fixed winding the peak core flux (and hence peak flux density $B_{m}$ , since the core area is fixed) satisfies $ϕ_{m} \propto \frac{V}{f}$ .
Compute the $V / f$ ratio: $(\frac{V}{f})_{India} = \frac{220}{50} = 4.4$ , $(\frac{V}{f})_{USA} = \frac{110}{60} = 1.833$ . So $\frac{B _{m, USA}}{B _{m, India}} = \frac{1.833}{4.4} \approx 0.417$ — the USA flux density is only about 42% of the India value, i.e. LOWER, not higher. Hence (C) is false.
For an unsaturated (linear) core, the magnetizing current is $I_{m} = \frac{V}{2 π f L _{m}}$ with $L_{m}$ fixed, so $I_{m} \propto \frac{V}{f}$ , the same ratio $\approx 0.417$ . Since core loss also falls (next steps), the in-phase (core-loss) component of the no-load current falls too, so the total no-load current is smaller in the USA. Hence (A) is true.
Core loss components scale as: hysteresis loss $P_{h} \propto f B_{m}^{1.6}$ , eddy-current loss $P_{e} \propto f^{2} B_{m}^{2}$ .
Eddy-current loss ratio $= (\frac{60}{50})^{2} \times (0.417)^{2} \approx 1.44 \times 0.174 \approx 0.25$ — eddy loss in the USA is about 25% of the India value, a decrease, not a 44% increase. Hence (D) is false.
Hysteresis loss ratio $\approx (\frac{60}{50}) \times (0.417)^{1.6} \approx 1.2 \times 0.246 \approx 0.30$ — also a decrease.
For the same load current, copper loss ( $I^{2} R$ ) is unchanged, while total core loss decreases in the USA. So the total loss for the same load current is lower in the USA, not higher. Hence (B) is false.
Only (A) survives: the no-load current is smaller in the USA because the reduced $V / f$ drives both the magnetizing current and the core-loss current down to about 42% and 25–30% of their India values respectively.

KEY IDEA

Same physical transformer means same turns ratio, same core, same magnetizing inductance and same core-loss constants. Only the applied V and f change, so every core-dependent quantity (peak flux density, magnetizing current, hysteresis loss, eddy-current loss) scales with the ratio V/f, while copper loss for a given load current is unaffected.

WATCH OUT FOR

Assuming the transformer is "under-utilized" or "over-fluxed" in the USA just because the voltage is lower, without actually computing V/f; or forgetting that eddy current loss scales as $f^{2} B_{m}^{2}$ (not $f B_{m}^{2}$ ) and hysteresis loss as $f B_{m}^{1.6}$ , which is needed to see that both core losses actually fall in the USA rather than rise.

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