---
title: "50 kVA transformer: equivalent parameters, copper loss"
subject: "Electrical Machines"
chapter: 2
key: "semester_4/electrical_machines:tf-50kva-4400-220-equivalent-resistance"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-50kva-4400-220-equivalent-resistance
site: BCT Engineering
---

# 50 kVA transformer: equivalent parameters, copper loss

Electrical Machines · Chapter 2: Transformer

- Asked in: 2075 Baisakh (old course), 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/tf-50kva-4400-220-equivalent-resistance
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2

## Question

A 50 kVA, 4400/220 V transformer has $R_1 = 3.45\ \Omega$, $R_2 = 0.009\ \Omega$, $X_1 = 5.2\ \Omega$ and $X_2 = 0.015\ \Omega$. 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

Referred to primary: Req = 7.050 Ω, Xeq = 11.200 Ω, Zeq = 13.234 Ω. Referred to secondary: Req = 0.01762 Ω, Xeq = 0.02800 Ω, Zeq = 0.03309 Ω. Total copper loss = 910.4 W.

## Answer

The equivalent values are found by moving one winding's impedance to the other side using the square of the turns ratio.

Turns ratio: $a = \dfrac{V_1}{V_2} = \dfrac{4400}{220} = 20$

Rated currents:

$$
I_1 = \frac{50000}{4400} = 11.364\ \text{A}, \qquad I_2 = \frac{50000}{220} = 227.27\ \text{A}
$$

#### (i) Equivalent resistance, reactance and impedance

**Referred to primary:**

$$
\begin{aligned}
R_{01} &= R_1 + a^2R_2 = 3.45 + 400(0.009) = 7.050\ \Omega \\
X_{01} &= X_1 + a^2X_2 = 5.2 + 400(0.015) = 11.200\ \Omega \\
Z_{01} &= \sqrt{R_{01}^2 + X_{01}^2} = \sqrt{7.050^2 + 11.200^2} = 13.234\ \Omega
\end{aligned}
$$

**Referred to secondary:**

$$
\begin{aligned}
R_{02} &= R_2 + \frac{R_1}{a^2} = 0.009 + \frac{3.45}{400} = 0.01762\ \Omega \\
X_{02} &= X_2 + \frac{X_1}{a^2} = 0.015 + \frac{5.2}{400} = 0.02800\ \Omega \\
Z_{02} &= \sqrt{R_{02}^2 + X_{02}^2} = 0.03309\ \Omega
\end{aligned}
$$

#### (ii) Total copper loss

**Using individual resistances:**

$$
P_{cu} = I_1^2R_1 + I_2^2R_2 = (11.364)^2(3.45) + (227.27)^2(0.009) = 445.5 + 464.9 = 910.4\ \text{W}
$$

**Using equivalent resistance referred to primary:**

$$
P_{cu} = I_1^2R_{01} = (11.364)^2(7.050) = 910.4\ \text{W}
$$

**Using equivalent resistance referred to secondary:**

$$
P_{cu} = I_2^2R_{02} = (227.27)^2(0.01762) = 910.4\ \text{W}
$$

All three methods give the same loss.

**Answer:** $R_{01} = 7.050\ \Omega$, $X_{01} = 11.200\ \Omega$, $Z_{01} = 13.234\ \Omega$; $R_{02} = 0.01762\ \Omega$, $X_{02} = 0.02800\ \Omega$, $Z_{02} = 0.03309\ \Omega$; total copper loss $\approx 910\ \text{W}$.

## Check yourself

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

1. With a = 20, the equivalent resistance referred to the primary is
   A. R1 + R2/a^2 = 3.45 Ω
   B. R1 + a^2 R2 = 7.05 Ω
   C. R1 + a R2 = 3.63 Ω
   D. R1 a^2 + R2 = 1380 Ω
   Answer: B. Secondary quantities are multiplied by a^2 when referred to the primary: 3.45 + 400 x 0.009 = 7.05 Ω. (Key point: Referring to primary)

2. Equivalent reactance referred to the secondary, X02, is
   A. X2 + a^2 X1 = 2080 Ω
   B. X2 + X1/a = 0.275 Ω
   C. X2 + X1/a^2 = 0.028 Ω
   D. X1 + a^2 X2 = 11.2 Ω
   Answer: C. Primary values are divided by a^2: 0.015 + 5.2/400 = 0.028 Ω. (Key point: Referring to secondary)

3. Copper loss using individual resistances, I1^2 R1 + I2^2 R2, compared with I1^2 R01 and I2^2 R02 is
   A. Different, since referred values change the power
   B. Same, about 910 W in all three methods
   C. Larger in the individual method only
   D. Same only if the load is at unity power factor
   Answer: B. Referring resistance with a^2 keeps the I^2R power unchanged, so all three give about 910.4 W. (Key point: Copper loss consistency)

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