---
title: "Equivalent impedance and copper loss of 50 kVA transformer"
subject: "Electrical Machines I"
chapter: 2
key: "semester_4/electrical_machines_i:50kva-4400-220-equivalent-copper-loss"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/50kva-4400-220-equivalent-copper-loss
site: BCT Engineering
---

# Equivalent impedance and copper loss of 50 kVA transformer

Electrical Machines I · Chapter 2: Transformer

- Asked in: 2075 Baisakh, 4+4 marks
- Video: [(E)EM Ex 2.2 || Equivalent Circuit of a Transformer || Refer to Pri & Secy sides ||Open-Short Test](https://www.youtube.com/watch?v=uJP_rFndPtU) — Electrical Engineering Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/50kva-4400-220-equivalent-copper-loss
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2

## Question

A 50 kVA, 4400/220 V transformer has R1 = 3.45 Ω, R2 = 0.009 Ω, X1 = 5.2 Ω and X2 = 0.015 Ω. Calculate (i) equivalent resistance, reactance and impedance as referred to both primary and secondary sides (ii) total copper loss using individual resistance of the two windings and using equivalent resistances as referred to each side.

## Short answer

With K = 0.05, equivalent values referred to primary: R01 = 7.05 Ω, X01 = 11.2 Ω, Z01 = 13.23 Ω; referred to secondary: R02 = 0.01763 Ω, X02 = 0.028 Ω, Z02 = 0.0331 Ω. Full-load copper loss is 910.4 W.

## Answer

Transformation ratio $K = \frac{220}{4400} = 0.05$, so $K^2 = 0.0025$.

#### (i) Equivalent resistance, reactance and impedance
**Referred to primary (4400 V side):**
$$
\begin{aligned}
R_{01} &= R_1 + \frac{R_2}{K^2} = 3.45 + \frac{0.009}{0.0025} = 3.45 + 3.6 = 7.05\ \Omega \\
X_{01} &= X_1 + \frac{X_2}{K^2} = 5.2 + \frac{0.015}{0.0025} = 5.2 + 6.0 = 11.2\ \Omega \\
Z_{01} &= \sqrt{7.05^2 + 11.2^2} = 13.23\ \Omega
\end{aligned}
$$
**Referred to secondary (220 V side):**
$$
\begin{aligned}
R_{02} &= R_2 + K^2R_1 = 0.009 + 0.0025 \times 3.45 = 0.01763\ \Omega \\
X_{02} &= X_2 + K^2X_1 = 0.015 + 0.0025 \times 5.2 = 0.028\ \Omega \\
Z_{02} &= \sqrt{0.01763^2 + 0.028^2} = 0.03309\ \Omega
\end{aligned}
$$
Check: $Z_{02} = K^2 Z_{01} = 0.0025 \times 13.23 = 0.0331\ \Omega$.

#### (ii) Full-load copper loss
Full-load currents:
$$
I_1 = \frac{50000}{4400} = 11.364\ \text{A}, \qquad I_2 = \frac{50000}{220} = 227.27\ \text{A}
$$
**Using individual winding resistances:**
$$
\begin{aligned}
I_1^2R_1 &= 11.364^2 \times 3.45 = 445.5\ \text{W} \\
I_2^2R_2 &= 227.27^2 \times 0.009 = 464.9\ \text{W} \\
P_{cu} &= 445.5 + 464.9 = 910.4\ \text{W}
\end{aligned}
$$
**Using $R_{01}$:**
$$
P_{cu} = I_1^2R_{01} = 11.364^2 \times 7.05 = 910.4\ \text{W}
$$
**Using $R_{02}$:**
$$
P_{cu} = I_2^2R_{02} = 227.27^2 \times 0.017625 = 910.4\ \text{W}
$$

All three methods give the same value, which confirms that referring resistances through $K^2$ keeps the power loss unchanged.

| Quantity | Primary side | Secondary side |
|---|---|---|
| Equivalent R | 7.05 Ω | 0.01763 Ω |
| Equivalent X | 11.2 Ω | 0.028 Ω |
| Equivalent Z | 13.23 Ω | 0.0331 Ω |
| Full-load Cu loss | 910.4 W | 910.4 W |

**Answer:** $R_{01}=7.05\ \Omega$, $X_{01}=11.2\ \Omega$, $Z_{01}=13.23\ \Omega$; $R_{02}=0.01763\ \Omega$, $X_{02}=0.028\ \Omega$, $Z_{02}=0.0331\ \Omega$; full-load copper loss = **910.4 W** by every method.

## Check yourself

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

1. How is R01 found?
   A. R1 + R2 = 3.459 ohm
   B. R1 + K^2 R2
   C. R1 x R2/K
   D. R1 + R2/K^2 = 7.05 ohm
   Answer: D. Secondary resistance referred to primary is divided by K^2 = 0.0025. (Key point: Referring to primary)

2. How is X02 found?
   A. X2 + K X1
   B. X2 + X1/K^2
   C. X1 + X2
   D. X2 + K^2 X1 = 0.028 ohm
   Answer: D. Primary values are multiplied by K^2 when referred to the secondary. (Key point: Referring to secondary)

3. How is Z01 found?
   A. sqrt(R01^2 + X01^2) = 13.23 ohm
   B. R01 + X01
   C. R01 - X01
   D. R01 x X01
   Answer: A. Resistance and reactance combine as a phasor sum. (Key point: Impedance)

4. Why does copper loss come to 910.4 W by all three methods?
   A. Referring through K^2 preserves power
   B. The windings have equal loss by design
   C. Because copper loss is neglected
   D. Because K = 1
   Answer: A. I^2R is invariant when R is referred through K^2 and current is scaled correspondingly. (Key point: Copper loss consistency)

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