---
title: "Per-unit transformer impedance same on both sides"
subject: "Power System Analysis I"
chapter: 3
key: "semester_4/power_system_analysis_i:pu-transformer-impedance-same-sides"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/pu-transformer-impedance-same-sides
site: BCT Engineering
---

# Per-unit transformer impedance same on both sides

Power System Analysis I · Chapter 3: Computational Technique

- Asked in: 2080 Chaitra, 4 marks; 2079 Chaitra, 4 marks; 2075 Bhadra, 6 marks; 2074 Bhadra, 2+4 marks (asked 4 times)
- Video: [Why Transformer Per Unit Impedance is Same on Both Sides | Proof + Example](https://www.youtube.com/watch?v=TOqxFs9W9bg) — Circuit Masterclass
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/pu-transformer-impedance-same-sides
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3

## Question

Verify mathematically the following statement: "The per unit impedance of a transformer remains the same when referred to the high voltage side or low voltage side of the transformer."

## Short answer

With base voltages chosen in the transformer turns ratio and a common base MVA, the per-unit impedance is identical on both sides because the impedance and base impedance scale by the square of the turns ratio, cancelling out. Thus Z_pu,LV = Z_pu,HV.

## Answer

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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The p.u. impedance is the same on both sides provided
   A. Base voltages are in the ratio of transformer turns, with the same VA base
   B. Base voltages are equal on both sides
   C. Base VA differs on each side
   D. The load is resistive
   Answer: A. Bases must follow the turns ratio with a common VA base. (Key point: Condition)

2. With V_b1/V_b2 = a, the base impedance ratio Z_b1/Z_b2 is
   A. a
   B. a^2
   C. 1/a
   D. 1
   Answer: B. Z_b = V_b^2/S_b, so the ratio is a^2. (Key point: Base impedance ratio)

3. Z referred from the LV side to the HV side is
   A. Z1 x a^2
   B. Z1/a^2
   C. Z1/a
   D. Z1 x a
   Answer: B. Z2 = Z1 (N2/N1)^2 = Z1/a^2. (Key point: Referred impedance)

4. For 10 kVA, 200/400 V and Z = 0.4 ohm on the LV side, Z_pu is
   A. 0.4
   B. 0.2
   C. 0.1
   D. 1.6
   Answer: C. 0.4/4 = 0.1, and on HV side 1.6/16 = 0.1. (Key point: Numerical check)

## Related in Power System Analysis I

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- [Per unit values: 100 MVA generator, two motors](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aper-unit-100mva-6kv-two-motors) — Power System Analysis I, chapter 3
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- [Per unit advantages and choice of bases](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aper-unit-advantages-base-selection) — Power System Analysis I, chapter 3, asked 3×
- [Per unit circuit of three-generator system, 30 MVA base](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Apu-circuit-30mva-g1-g2-g3) — Power System Analysis I, chapter 3
- [Major advantages of per unit system](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aper-unit-major-advantages) — Power System Analysis I, chapter 3

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