GATE 2020 EE Q.21: solution and answer

GATE 2020 · EEQ.21 · 1 mark · NAT

A single-phase, 4 kVA, 200 V/100 V, 50 Hz transformer with laminated CRGO steel core has rated no-load loss of 450 W. When the high-voltage winding is excited with 160 V, 40 Hz sinusoidal ac supply, the no-load losses are found to be 320 W. When the high-voltage winding of the same transformer is supplied from a 100 V, 25 Hz sinusoidal ac source, the no-load losses will be _______ W (rounded off to 2 decimal places).

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Answer and solution

ANSWER

The no-load loss at 100 V, 25 Hz is 162.50 W.

Official answer: 162.41 to 162.59Source: Official answer key, GATE 2020 EE (IIT Delhi)

Since $E = 4.44 f N B_{m} A$ for a fixed winding, the peak flux density obeys $B_{m} \propto V / f$ . Check the $V / f$ ratio in all three given conditions: rated $200/50 = 4$ ; test $160/40 = 4$ ; target $100/25 = 4$ . All three give the same ratio, so $B_{m}$ is the SAME constant value in each case.
With $B_{m}$ held constant, hysteresis loss $P_{h} \propto f$ and eddy current loss $P_{e} \propto f^{2}$ — the $B_{m}$ -dependence in each loss law contributes only an unchanging multiplicative factor, so only the frequency dependence survives when comparing conditions.
Let the rated-condition ( $200 V, 50 Hz$ ) losses be $P_{h}$ and $P_{e}$ , so $P_{h} + P_{e} = 450 W$ .
At $160 V, 40 Hz$ , the frequency ratio relative to rated is $40/50 = 0.8$ , so $0.8 P_{h} + (0.8)^{2} P_{e} = 320 \Rightarrow 0.8 P_{h} + 0.64 P_{e} = 320.$
Substituting $P_{e} = 450 - P_{h}$ : $0.8 P_{h} + 0.64 (450 - P_{h}) = 320 \Rightarrow 0.16 P_{h} = 32 \Rightarrow P_{h} = 200 W, P_{e} = 250 W .$
At $100 V, 25 Hz$ (same $B_{m}$ ), the frequency ratio relative to rated is $25/50 = 0.5$ : $P_{h}^{'} = 200 (0.5) = 100 W, P_{e}^{'} = 250 (0.5)^{2} = 62.5 W .$
Total no-load loss $= 100 + 62.5 = 162.50 W$ .

KEY IDEA

No-load (core) loss splits into hysteresis loss (∝ f · B_m^n) and eddy current loss (∝ f² · B_m²). Since E = 4.44 f N B_m A, the flux density B_m depends only on the ratio V/f. Whenever V/f is held at the same value across different test conditions, B_m stays constant, so the two loss components reduce to simple functions of frequency alone: P_h ∝ f and P_e ∝ f². This lets two known (V, f, loss) data points fix P_h and P_e at a reference condition, from which the loss at any other same-(V/f) condition follows by scaling.

WATCH OUT FOR

Forgetting to check that V/f is actually the same in all three conditions before dropping B_m from the proportionality — if V/f changed, B_m would change too and P_h ∝ f alone (or P_e ∝ f² alone) would no longer hold. A second common slip is mixing up which loss scales as f versus f² (swapping hysteresis and eddy current), or using absolute frequencies in the ratio instead of the ratio relative to the condition where P_h, P_e were computed.

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