Per-unit transformer impedance same on both sides

The statement is true. The per-unit impedance of a transformer is the same on both sides, provided the base voltages on the two sides are in the ratio of the transformer turns (with the same VA base).

Proof

Consider a transformer of rating $S$ (VA), with voltages $V_1$ (LV) and $V_2$ (HV) and turns ratio $a = V_1/V_2 = N_1/N_2$. Choose:

• the same base VA, $S_b$, on both sides
• base voltages $V_{b1} = V_1$ and $V_{b2} = V_2$, so that $V_{b1}/V_{b2} = a$

Base impedances:

$$Z_{b1} = \frac{V_{b1}^2}{S_b},\qquad Z_{b2} = \frac{V_{b2}^2}{S_b} \;\Rightarrow\; \frac{Z_{b1}}{Z_{b2}} = a^2$$

Let $Z_1$ be the equivalent impedance referred to the LV side. Referred to the HV side, it becomes:

$$Z_2 = Z_1\left(\frac{N_2}{N_1}\right)^2 = \frac{Z_1}{a^2}$$

Per-unit values:

$$Z_{1,pu} = \frac{Z_1}{Z_{b1}},\qquad Z_{2,pu} = \frac{Z_2}{Z_{b2}} = \frac{Z_1/a^2}{Z_{b1}/a^2} = \frac{Z_1}{Z_{b1}} = Z_{1,pu}$$

So $Z_{pu}$ is the same whichever side it is referred to.

Numerical check

Take a 10 kVA, 200/400 V transformer with $Z = 0.4\ \Omega$ referred to the LV side.

• LV side: $Z_b = 200^2/10000 = 4\ \Omega$, so $Z_{pu} = 0.4/4 = 0.1$.
• HV side: $Z = 0.4\times(400/200)^2 = 1.6\ \Omega$ and $Z_b = 400^2/10000 = 16\ \Omega$, so $Z_{pu} = 1.6/16 = 0.1$.

Significance

The ideal transformer disappears from the per-unit equivalent circuit, leaving only a series impedance. Manufacturers can therefore quote a single % or p.u. impedance without naming a side. The result holds only when the bases follow the turns ratio. If they do not, an off-nominal tap must be included.