GATE 2023 IN Q.58: solution and answer

GATE 2023 · INQ.58 · 2 marks · NAT

A short-circuit test is conducted on a single-phase transformer by shorting its secondary. The frequency of input voltage is 1 kHz. The corresponding wattmeter reading, primary current and primary voltage are 8 W, 2 A and 6 V respectively. Assume that the no-load losses and the no-load currents are negligible, and the core has linear magnetic characteristics. Keeping the secondary shorted, the primary is connected to a 2 V (RMS), 1 kHz sinusoidal source in series with a $\frac{1}{2 π 5}$ mF capacitor. The primary current (RMS) will be _____ A (rounded off to two decimal places).

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Answer and solution

ANSWER

The primary current (RMS) is 1.00 A.

Official answer: 0.95 to 1.05Source: Official answer key, GATE 2023 IN (IIT Kanpur)

Short-circuit test data (secondary shorted): $V_{sc} = 6 V$ , $I_{sc} = 2 A$ , $P_{sc} = 8 W$ .
$Z_{sc} = \frac{V _{sc}}{I _{sc}} = \frac{6}{2} = 3 Ω$ .
$R_{sc} = \frac{P _{sc}}{I _{sc}^{2}} = \frac{8}{2 ^{2}} = 2 Ω$ .
$X_{sc} = Z_{sc}^{2} - R_{sc}^{2} = 9 - 4 = 5 Ω$ . With the secondary shorted, the transformer's impedance as seen from the primary is $R_{sc} + j X_{sc} = 2 + j 5 Ω$ .
New circuit: a $2 V$ (RMS), $1 kHz$ source in series with a capacitor $C = \frac{1}{2 π 5} mF$ , feeding the primary while the secondary stays shorted.
$ω = 2 π \times 1000 = 2000 π rad/s$ .
$X_{C} = \frac{1}{ω C} = \frac{1}{2000 π \times \frac{1}{2 π 5} \times 1 0 ^{- 3}} = 5 Ω$ .
Total series impedance = capacitor + shorted transformer impedance: $Z = - j 5 + (2 + j 5) = 2 Ω$ — the capacitive and leakage reactances cancel exactly (series resonance).
Primary current: $I = \frac{V}{Z} = \frac{2}{2} = 1.00 A$ .

KEY IDEA

A short-circuit test gives the transformer's series (leakage) impedance as seen from the primary; when a capacitor is placed in series with that impedance and its reactance is chosen to exactly cancel the leakage reactance, the circuit becomes purely resistive (series resonance), so the current is simply the source voltage divided by the short-circuit resistance.

WATCH OUT FOR

Forgetting that with the secondary shorted, the transformer itself still contributes its short-circuit impedance $R_{sc} + j X_{sc}$ to the new circuit (treating the capacitor as connected to an ideal short instead of to the transformer's leakage impedance), or making a sign/algebra slip when combining $- j X_{C}$ with $+ j X_{sc}$ and not noticing they are equal in magnitude (resonance), leading to a wrong non-real or mismatched impedance.

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GATE 2023 IN Q.58: solution and answer

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