Open-Circuit and Short-Circuit Tests

The turns ratio is the open-circuit voltage ratio. With no current flowing there is no impedance drop, so the terminal voltages are the induced EMFs:

This is the one measurement in the whole test set that is exact rather than inferred — which is why the open-circuit test is also the standard way of checking a nameplate ratio.

Extract the no-load power factor from the three meter readings:

Resolve the no-load current into its two components:

Separate the iron loss from the wattmeter reading. Strictly, the reading also contains the primary copper loss caused by the exciting current — normally ignored, and here is why:

The correction is 0.5% of the reading — smaller than the accuracy of the wattmeter itself. This is precisely why the open-circuit wattmeter is taken as reading iron loss directly.

The shunt-branch parameters follow, referred to the primary because that is the excited winding:

Open-circuit test — the shunt branch. The instruments are on the 200 V winding, so everything obtained here is referred to the primary:

Short-circuit test — the series branch. Here the instruments have been placed on the secondary, that is on the high-voltage winding, while the low-voltage primary was short-circuited. Every parameter therefore comes out referred to the secondary:

Refer them to the primary as well, so that both tests describe the same circuit. This transformer steps up, so \(a < 1\) and the primary-side values are the smaller ones:

Find the secondary current at the stated load. The output is given in kilowatts, so the power factor must be divided out before the current can be found:

The 400 V used here is the no-load secondary voltage, since 200 V is applied to the primary and the ratio is 1:2. Using the as-yet-unknown loaded voltage would make the calculation circular; the small error this introduces is corrected below.

The total drop referred to the secondary, with the parameters obtained from the short-circuit test on that same winding:

Check the size of the neglected term. Solving the exact phasor relation \(|\bar{V}_2 + \bar{I}_2Z_{02}| = 400\) with \(\bar{I}_2 = 12.5 - j9.375\) gives \(V_2 = 377.7\ \text{V}\) — one tenth of a volt below the approximate answer, or 0.03%.

5.55% is a large regulation, and it comes almost entirely from \(X_{02}\): the reactive term contributes 11.6 V of the 22.2 V total at this power factor.

Confirm that the tests were run at the right operating points. This check takes ten seconds and catches most bad data:

The open-circuit test was at rated LV voltage (200 V) and the short-circuit test at rated HV current (10 A). Both are as they should be, so the iron loss is the true rated-voltage value and the copper loss is the true full-load value.

Open-circuit test. The exciting current is 4 A against a rated 100 A on that winding — 4%, which is normal:

Move the shunt branch to the HV side, where the short-circuit data already live:

A check that costs nothing: the iron loss computed on either side must agree. \(200^2/333.3 = 120\ \text{W}\) and \(2000^2/33\,330 = 120\ \text{W}\).

Short-circuit test — series parameters referred to the HV winding, since that is where the meters were:

The same parameters on the LV side, obtained by division rather than by repeating the work:

Efficiency at full load and 0.8 power factor. Both losses are now known at that operating point — the iron loss from the first test, the copper loss from the second:

Collect the equivalent circuit in one place. Six numbers describe the machine completely, and every one of them came from two meter readings and a stopwatch's worth of arithmetic:

Series parameters from the short-circuit test. The rated HV current is \(10\,000/2000 = 5\ \text{A}\), which is exactly the test current, so 150 W is the full-load copper loss:

Express them as percentages of rated volts, which is the form regulation wants:

A 2.5% impedance also says that a dead short on the secondary would draw \(100/2.5 = 40\) times rated current — the number a protection engineer takes from this test.

Regulation at the three power factors. Only the sign and size of the reactive term change:

At 0.8 leading the two drops cancel exactly, because \(\%R/\%X = 1.5/2.0 = 0.75 = \tan(36.87^\circ)\). The terminal voltage at full load is then identical to its no-load value.

The load fraction for maximum efficiency is where the variable loss has grown to equal the fixed loss:

The maximum efficiency itself, at 0.8 power factor. At this load the copper loss has fallen to \(0.8^2 \times 150 = 96\ \text{W}\), equal to the iron loss as required:

Compare with full load at the same power factor:

Only 0.07 percentage points separate them. The efficiency curve is extremely flat near its peak, which is why designers place the maximum well below full load — around 75–85% for a distribution transformer that spends most of its life part-loaded.

Note where every reading was taken before touching it. Both tests are on the HV winding, so both sets of parameters emerge referred to the HV side and both must later be divided by \(a^2\):

The short-circuit test was run at 10 A, which is rated HV current — so 360 W is the full-load copper loss and no scaling of the wattmeter reading is needed.

Open-circuit test, HV side. Exciting the HV winding is unusual and demands care: the current is only 0.5 A and the power factor 0.2, so a low-power-factor wattmeter is essential:

Refer the shunt branch to the LV side by dividing by \(a^2 = 100\):

The iron loss must be the same number whichever side it is computed on. If it is not, the referral went the wrong way — the commonest error in this type of problem, and one that this single line catches every time.

Short-circuit test, HV side:

Terminal voltage and regulation at full load, 0.8 lagging. All quantities are now on the LV side, so the calculation is a single line:

Alternative method — stay on the HV side throughout. The percentages are side-independent, so the same answer follows without ever referring anything:

Working in percentages sidesteps the referral entirely, which is why per-unit notation is the professional's default for any problem where the side of a measurement is in doubt.

Understand what each wattmeter is measuring. With the secondaries in opposition their induced EMFs cancel round the loop, so the 500 V supply drives no load current — it sees only two excited cores in parallel. The first wattmeter therefore reads the combined iron loss:

The 2 A drawn is the sum of two exciting currents, 1 A each against a rated primary current of \(10\,000/500 = 20\ \text{A}\) — 5%, exactly what an open-circuit test on a single unit would have shown.

The auxiliary source sees no EMF at all. It is injected into a loop whose net induced voltage is zero, so it needs only to overcome the impedance of the two secondary circuits. The current it drives is a genuine load current, circulating between the transformers, and the wattmeter reads the combined copper loss:

The circulating current is 40 A, which is exactly rated secondary current — so 240 W is the full-load copper loss and needs no scaling.

Halve the auxiliary voltage before using it. The 20 V is shared between two identical secondary circuits in series, so each transformer sees 10 V — this is the step the test is designed to catch:

The percentage values, which show this to be an entirely ordinary machine:

Regulation at 0.8 lagging would be \(2.4(0.8) + 3.2(0.6) = 3.84\%\).

Efficiency at full load, 0.8 power factor:

The load for maximum efficiency:

Why the test is worth the trouble. Both transformers are carrying rated current and rated flux, so both are heating exactly as they would on full load — yet the total input drawn from the two supplies is only:

Three percent of the apparent power that a real load bank would have demanded, because the useful power circulates between the two machines and only the losses are drawn from the mains. That is the whole idea of a back-to-back test, and it is the only practical way to run a temperature-rise test on a large transformer.