Equivalent Circuit Parameters

Establish the turns ratio and the scaling rule. Voltages on the secondary side are multiplied by \(a\) and currents divided by \(a\), so any ratio of the two — an impedance — is multiplied by \(a^2\):

The angle is untouched: \(36.87^\circ\) on both sides. Referring rescales an impedance but never rotates it, so the load power factor is the same whichever side you stand on.

Currents, worked from the primary side. With the ideal transformer replaced by the referred load, the primary sees a single 1000 Ω impedance:

The same currents from the secondary side, as a check on the referral:

Power is invariant, which is the whole point of the \(a^2\). Compute it on each side:

Identical, and necessarily so: \(I'^2R' = (I/a)^2(a^2R) = I^2R\). The referred network is not an approximation — it is an exact replacement that no external measurement can distinguish from the original.

Referring the other way divides by \(a^2\). A 400 Ω resistor sitting in the primary circuit, viewed from the secondary, becomes

The direction of the scaling is fixed by one question: which side has more turns? Impedances always look larger when viewed from the high-voltage side.

Refer the secondary winding to the primary and add, element by element:

The two windings contribute almost equally, which is not a coincidence — a well-designed transformer splits the copper and the leakage flux roughly evenly between them, and the referral makes that symmetry visible.

Now refer the primary winding to the secondary instead, dividing by the same \(a^2\):

There is no need to repeat the work in practice: once \(Z_{01}\) is known, dividing by \(a^2\) gives \(Z_{02}\) directly. The long route is done here only to show that the two agree.

Full-load currents, from the rating and the two rated voltages:

The copper loss must come out the same from either side, and this is the standard check on a referral:

468.75 W on a 25 kVA rating is 1.9% — a normal figure for a distribution transformer of this size.

Express the parameters per unit of rated voltage, which makes them independent of the side chosen:

Computing \(\%R\) from the secondary side gives \((125 \times 0.030)/200 = 1.875\%\) as well — the percentage values are the one description of the transformer that does not need the phrase "referred to".

Split the exciting current and form the shunt branch. Both elements sit across the rated primary voltage during the no-load test:

The core loss follows immediately: \(P_i = V_1I_w = 2400 \times 0.1 = 240\ \text{W}\), or 1.0% of the rating.

Set up the loading condition with the secondary terminal voltage as the phasor reference, referred to the primary:

Exact circuit — work outwards from the load. With the split \(R_2' = 1.5\ \Omega\), \(X_2' = 4.0\ \Omega\), the voltage across the shunt branch is

The exciting current is driven by \(\bar{E}_1\), not by \(\bar{V}_1\) — this is the only structural difference between the two circuits:

Add the primary winding drop to reach the applied voltage:

Approximate circuit — the shunt branch is moved to the terminals. The two series impedances then merge into \(R_{01} + jX_{01}\), and the exciting current keeps its no-load value:

Compare, and notice how small the difference is:

Even the further simplification \(V_1 \simeq V_2' + I_2'(R_{01}\cos\phi + X_{01}\sin\phi) = 2400 + 10(2.4+4.8) = 2472\ \text{V}\), which drops the quadrature term entirely, is within 0.1%.

Alternative reading of why the error is so small. Moving the shunt branch changes only the current that flows through \(Z_1\) — by \(\bar{I}_0\), which is 0.5 A against 10.4 A — and the extra drop it causes is \(I_0Z_1 \approx 0.5 \times 4.27 = 2.1\ \text{V}\) out of 2475 V. The exact circuit is worth the trouble only when the exciting current is a large fraction of the load current, which means light load or a very small machine.

Refer the load impedance to the primary. Nothing can be combined until every element is on the same side:

Form the series branch carrying the load current — the referred secondary winding in series with the referred load:

Convert the shunt branch to an impedance. \(R_0\) and \(X_0\) are in parallel, so add them as admittances and invert once:

Reduce the network. \(Z_m\) is in parallel with the series branch, and \(Z_1\) is in series with that combination:

Primary current and input power factor follow at once:

Split \(\bar{I}_1\) between the two branches by current division:

Check: \(\bar{I}_0 + \bar{I}_2' = (9.04 - j1.91) + (36.42 - j28.38) = 45.46 - j30.29 = 54.63\angle-33.67^\circ\ \checkmark\)

Powers. Input from the terminal quantities, output from the resistive part of the referred load:

Losses, and the balance that confirms them:

The core loss can be checked independently: the shunt branch voltage is \(I_0|Z_m| = 9.24 \times 245.1 = 2265\ \text{V}\), and \(2265^2/250 = 20\,520\ \text{W}\).

Efficiency and regulation. The referred secondary terminal voltage is the drop across the load itself:

On the actual secondary this is \(2233/10 = 223.3\ \text{V}\) against a no-load 230 V. Quoting the regulation on the no-load base instead gives \(67/2300 = 2.9\%\); always state which base is used.

Fix the no-load secondary voltage and the full-load current. With the primary held at rated voltage, the no-load secondary voltage is the rated 400 V:

Convert to percentages, which makes all three parts one line each:

0.8 power factor lagging. Both terms add, because a lagging current makes the reactive drop subtract from the terminal voltage:

Unity power factor. The reactive term vanishes and only the resistive drop remains:

Note that \(X_{02}\) is more than twice \(R_{02}\) yet contributes nothing here — the reactance only matters when the current has a quadrature component.

0.8 power factor leading. The sine term changes sign, and here it more than cancels the resistive drop:

A negative regulation is a real effect, not an arithmetic slip: a leading load makes the secondary terminal voltage rise above its no-load value.

The power factor for zero regulation is where the two terms cancel exactly:

Confirm one case exactly, without the approximation, by requiring the phasor magnitude of \(\bar{E}_2\) to be 400 V. At 0.8 lagging, \(\bar{I}_2 = 200 - j150\):

The approximate formula gave 388.8 V and 2.80% — a discrepancy of 0.05 V. The quadrature term it neglects enters only through \(6.6^2/(2\times400)\), about eight hundredths of a volt.