OC/SC tests and autotransformers

Why two separate tests, and not one. A transformer's losses split into core (iron) loss — nearly constant, set by flux and hence by applied voltage — and copper loss — proportional to current squared. No single test at one operating point can separate the two cleanly, because normal full-load operation has both rated voltage and rated current at once. The trick is to run two tests that each isolate one loss by starving the other of its driving quantity.

Open-circuit test. Rated (or near-rated) voltage $V_{OC}$ is applied to one winding — usually the low-voltage side, since it needs less test-bench insulation — while the other winding is left open. With no secondary current, the primary draws only the small no-load current $I_0$, so primary copper loss $I_0^2R_1$ is negligible; almost the entire measured input power $W_{OC}$ is core loss:

where: $P_i$ is core (iron) loss, $\phi_0$ is the no-load power-factor angle, and $V_{OC}, I_{OC}, W_{OC}$ are the OC-test meter readings. Resolving $I_{OC}$ along and perpendicular to $V_{OC}$ splits it into the working component $I_w = I_{OC}\cos\phi_0$ (feeds core loss) and the magnetising component $I_\mu = I_{OC}\sin\phi_0$ (builds flux), from which $R_0 = V_{OC}/I_w$ and $X_0 = V_{OC}/I_\mu$ give the shunt magnetising-branch resistance and reactance.

Short-circuit test. The secondary is shorted, and only a small, reduced voltage $V_{SC}$ is applied to the primary — just enough to drive rated current $I_{SC}$ through the short. Because $V_{SC}$ is a small fraction of rated voltage, the core flux (and hence core loss) at this test point is negligible; almost all the measured input power $W_{SC}$ is copper loss at the test current:

where: $Z_{01}, R_{01}, X_{01}$ are the equivalent series impedance, resistance and reactance referred to whichever side was energised in the SC test. If the test current equals full-load rated current, $W_{SC}$ is the full-load copper loss $P_{cu,FL}$ directly — no scaling needed. Because the OC and SC tests are usually run on different sides for convenience, their results must be referred to a common side (via $a^2$ scaling of impedance) before being combined into one equivalent circuit.

Autotransformers. A two-winding transformer keeps primary and secondary as physically separate coils, isolated but each carrying its own full share of copper. An autotransformer instead uses one continuous, tapped winding: the section between the tap and the common terminal carries the load current shared by both sides, while only the remaining section carries the difference between primary and secondary current. Wherever current is shared rather than duplicated, less copper is needed for the same VA rating — the source of the autotransformer's economy, developed formally below.

A 20 kVA, 2000 V/200 V, 50 Hz transformer was tested: OC test (LV side energised) gave $V_{OC}=200\ \text{V}$, $I_{OC}=4\ \text{A}$, $W_{OC}=180\ \text{W}$; SC test (HV side energised, rated current) gave $V_{SC}=120\ \text{V}$, $I_{SC}=10\ \text{A}$, $W_{SC}=600\ \text{W}$. Find the four equivalent-circuit parameters, and then estimate the copper this same transformer would save if it were rebuilt as an autotransformer of the same 2000 V/200 V ratio.

Step 1 — core loss and no-load branch (from the OC test). $P_i \approx W_{OC} = 180\ \text{W}$. The no-load power factor is $\cos\phi_0 = 180/(200\times4) = 0.225$, giving $\sin\phi_0 \approx 0.9743$. The branch currents are $I_w = 4\times0.225 = 0.9\ \text{A}$ and $I_\mu = 4\times0.9743\approx3.897\ \text{A}$, so $R_0 = 200/0.9 \approx 222.2\ \Omega$ and $X_0 = 200/3.897\approx51.3\ \Omega$ (both referred to the LV side, where the OC test ran).

Step 2 — series impedance (from the SC test). $Z_{01} = 120/10 = 12\ \Omega$, $R_{01} = 600/100 = 6\ \Omega$, and $X_{01} = \sqrt{12^2-6^2} = \sqrt{108}\approx10.39\ \Omega$, all referred to the HV side, where the SC test ran. Since $I_{SC}$ was the rated 10 A, $W_{SC} = 600\ \text{W}$ is directly the full-load copper loss.

Step 3 — autotransformer copper saving at the same 2000 V/200 V ratio. Here $a = 2000/200 = 10$, so from the boxed formula the saving is $\left(1-\tfrac1{10}\right)\times100\% = 90\%$: rebuilding this particular transformer as an autotransformer of the same ratio would need only about a tenth of its present winding copper. That large a saving is unusual precisely because $a=10$ is far from 1 — it is why this specific transformer, with its 10:1 ratio, is a case where the autotransformer is copper-attractive but the loss of isolation between a 2000 V and a 200 V terminal would rule it out for most practical, safety-conscious uses.

Answer. $P_i = 180\ \text{W}$; $R_0\approx222.2\ \Omega$, $X_0\approx51.3\ \Omega$ (LV side); $R_{01}=6\ \Omega$, $X_{01}\approx10.39\ \Omega$, $Z_{01}=12\ \Omega$ (HV side); autotransformer copper saving at $a=10$ is 90 percent.

• Mixing up which side each test is run on and forgetting to refer OC-side and SC-side parameters to a common side before combining them into one equivalent circuit.
• Assuming SC-test power directly equals full-load copper loss no matter what test current was used — it only holds when the SC test was run at rated current; otherwise scale by (current ratio)$^2$.
• Treating the autotransformer copper-saving fraction $1-1/a$ as a saving in cost or loss rather than strictly a saving in winding copper quantity — core loss and other design factors are separate.
• Forgetting the tradeoff: the copper saving is real, but an autotransformer has no galvanic isolation between primary and secondary, which rules it out wherever isolation is a safety requirement.

Ask the tutor: if the referred-side scaling between OC- and SC-test parameters, or the step-by-step autotransformer copper-count in the derivation above, still feels shaky, ask the AI tutor to redo it with a fresh numeric example.