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Loss assumptions in OC and SC tests

Short answer

In the OC test current is only 2-5% of rated value, so I0^2R is negligible and the wattmeter reads full iron loss. In the SC test applied voltage is only 5-10% of rated value, so flux and iron loss are negligible while rated current gives full copper loss. Hence the assumptions are valid.

While performing transformer test, copper loss is assumed to be negligible during no-load test and iron loss is assumed to be negligible during short circuit test. Justify that these assumptions are correct with detail explanation.

In the OC test the measured power is taken as iron loss, and in the SC test as copper loss. Both assumptions are justified because in each test the neglected loss is a tiny fraction of the measured one.

No-load (OC) test: copper loss is negligible

 rated V  -->[W]--(A)--+ LV winding || HV winding  (open)
  • The test is done on the LV side at rated voltage, with the HV side open.
  • Rated voltage gives rated flux (ϕm=V4.44fN\phi_m = \frac{V}{4.44fN}), so the full normal iron loss occurs (hysteresis loss ∝Bm1.6\propto B_m^{1.6}, eddy-current loss ∝Bm2\propto B_m^2, so PiP_i varies roughly as Bm2B_m^2).
  • The only current is the no-load current I0I_0, which is only 2-5% of rated current. The copper loss in the energised winding is I02R1I_0^2R_1.
  • Since copper loss varies as the square of current, it is only about (0.02(0.02 to 0.05)2=0.04%0.05)^2 = 0.04\% to 0.25%0.25\% of the full-load copper loss, which is negligible. The HV winding carries no current at all.
  • Hence P0=Pi+I02R1≈PiP_0 = P_i + I_0^2R_1 \approx P_i.

Example: in a 10 kVA, 250 V transformer with I0=0.8I_0 = 0.8 A and R1=0.04 ΩR_1 = 0.04\ \Omega, the copper loss is 0.82×0.04=0.0260.8^2 \times 0.04 = 0.026 W against a wattmeter reading of 80 W, i.e. 0.03%.

Short-circuit (SC) test: iron loss is negligible

 low V  -->[W]--(A)--+ HV winding || LV winding (shorted)
  • The LV side is shorted and a low voltage is applied to the HV side, raised until rated current flows. This voltage is only 5-10% of rated voltage, because only the small leakage impedance Z01Z_{01} limits the current.
  • Rated current in both windings gives the full-load copper loss I12R01I_1^2R_{01}.
  • The flux is proportional to the applied voltage, so the core works at only 5-10% of its normal flux density. Iron loss varies roughly as Bm2B_m^2, so it is only (0.05(0.05 to 0.1)2=0.25%0.1)^2 = 0.25\% to 1%1\% of the normal iron loss, which is negligible compared with the full-load copper loss.
  • Hence Psc=Pcu,FL+Pi,sc≈Pcu,FLP_{sc} = P_{cu,FL} + P_{i,sc} \approx P_{cu,FL}.

Summary

ItemOC testSC test
Applied voltageRated (100%)5-10% of rated
Current2-5% of ratedRated (100%)
Flux levelNormal5-10% of normal
Iron lossFull value~0.25-1% of normal
Copper loss~0.1% of FL valueFull-load value
Wattmeter readsIron lossCopper loss

Because each neglected loss is far below 1% of the measured one, the error is within instrument accuracy. This lets the two losses be found separately while using little energy, so the efficiency and regulation at any load can be predicted.

✓ Answer checked by Claude Opus 5.5 (high effort)

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