---
title: "25 kVA ideal transformer currents, flux"
subject: "Basic Electrical and Electronics Engineering"
chapter: 2
key: "semester_1/basic_electrical_and_electronics_engineering:ideal-25kva-transformer"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/ideal-25kva-transformer
site: BCT Engineering
---

# 25 kVA ideal transformer currents, flux

Basic Electrical and Electronics Engineering · Chapter 2: Electrical Machines

- Asked in: 2081 Chaitra (new course), 6 marks
- Video: [ECE202_Lec25_Part 2: Voltage and Current Ratios of Ideal Transformer](https://www.youtube.com/watch?v=cvxeoQqJvME) — Bingsen Wang
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/ideal-25kva-transformer
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_and_electronics_engineering/chapter-2

## Question

An ideal 25 kVA transformer has 500 turns on the primary winding and 40 turns on the secondary winding. The primary is connected to 3,000 V, 50 Hz supply. Calculate i) Primary and secondary currents on full load. ii) Secondary emf iii) The maximum core flux.

## Short answer

Full-load currents: primary8.33 A, secondary104.17 A. Secondary emf is240 V. Maximum core flux is27.03 mWb.

## Answer

**Given:** $S=25$ kVA, $N_1=500$, $N_2=40$, $V_1=3000$ V, $f=50$ Hz. The transformer is ideal.

**Turns ratio:** $K=\frac{N_2}{N_1}=\frac{40}{500}=0.08$.

**(i) Full-load currents.**

$$
I_1=\frac{S}{V_1}=\frac{25000}{3000}=8.33\ \text{A}
$$

$$
I_2=\frac{I_1}{K}=\frac{8.33}{0.08}=104.17\ \text{A}
$$

**(ii) Secondary emf.**

$$
E_2=V_1\times\frac{N_2}{N_1}=3000\times0.08=240\ \text{V}
$$

(Check: $25000/240=104.17$ A.)

**(iii) Maximum core flux.** From the emf equation $E_1=4.44fN_1\Phi_m$ (for an ideal transformer, $E_1=V_1$):

$$
\Phi_m=\frac{3000}{4.44\times50\times500}=0.02703\ \text{Wb}=27.03\ \text{mWb}
$$

**Answer: (i) $I_1=8.33$ A, $I_2=104.17$ A; (ii) $E_2=240$ V; (iii) $\Phi_m=27.03$ mWb.**

## Check yourself

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

1. What is the full-load primary current?
   A. I1 = 3000/25000 = 0.12 A
   B. I1 = 25000 x 3000
   C. I1 = S/V1 = 25000/3000 = 8.33 A
   D. I1 = 104.17 A
   Answer: C. Rated current = VA rating / voltage. (Key point: Primary current)

2. How is the secondary full-load current found?
   A. I2 = I1 x K = 0.67 A
   B. I2 = I1 = 8.33 A
   C. I2 = I1 x 500 = 4165 A
   D. I2 = I1/K = 8.33/0.08 = 104.17 A
   Answer: D. I2/I1 = N1/N2, so the current on the low-voltage side is larger. (Key point: Secondary current)

3. What is the secondary emf?
   A. 240 V
   B. 37.5 V
   C. 600 V
   D. 3000 V
   Answer: A. E2 = V1 x N2/N1 = 3000 x 0.08 = 240 V. (Key point: Secondary emf)

4. Which gives the maximum core flux?
   A. Phi_m = E1/(f N1) = 0.12 Wb
   B. Phi_m = E1/(4.44 f N1) = 3000/(4.44 x 50 x 500) = 27.03 mWb
   C. Phi_m = E2/(4.44 f N1) = 2.2 mWb
   D. Phi_m = E1 x 4.44 f N1
   Answer: B. Rearranging E1 = 4.44 f N1 Phi_m with E1 = V1 for an ideal transformer. (Key point: Maximum flux)

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