---
title: "No-load current components, iron loss and core flux"
subject: "Electrical Machines I"
chapter: 2
key: "semester_4/electrical_machines_i:no-load-current-10a-0p3pf-500turns"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/no-load-current-10a-0p3pf-500turns
site: BCT Engineering
---

# No-load current components, iron loss and core flux

Electrical Machines I · Chapter 2: Transformer

- Asked in: 2070 Magh, 8 marks
- Video: [No load losses of transformer calculation | magnetising current | How to calculate iron loss](https://www.youtube.com/watch?v=Vq3B_JyN7oI) — Bright Future Tutorials
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/no-load-current-10a-0p3pf-500turns
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2

## Question

The no-load current of a transformer is 10 A at a p.f. of 0.3 lagging when connected to a 400 V, 50 Hz power supply. If the primary winding has 500 turns, calculate: (a) the magnetizing and working component of no-load current (b) iron loss (c) maximum and rms value of flux in the core.

## Short answer

Working component Iw = 10×0.3 = 3 A, magnetizing component Iμ = 10×sin(arccos0.3) = 9.54 A. Iron loss = 400×10×0.3 = 1200 W. Maximum flux = 400/(4.44×50×500) = 3.604 mWb, rms flux = 2.548 mWb.

## Answer

Given: $I_0 = 10$ A, $\cos\phi_0 = 0.3$ lagging, $V_1 = 400$ V, $f = 50$ Hz, $N_1 = 500$.

$$
\sin\phi_0 = \sqrt{1 - 0.3^2} = 0.9539
$$

#### (a) Components of no-load current
Working (core-loss) component, in phase with $V_1$:
$$
I_w = I_0\cos\phi_0 = 10 \times 0.3 = 3\ \text{A}
$$
Magnetizing component, lagging $V_1$ by 90°:
$$
I_\mu = I_0\sin\phi_0 = 10 \times 0.9539 = 9.54\ \text{A}
$$
Check: $\sqrt{3^2 + 9.54^2} = 10$ A.

```text
   V1 ^
      |
 Iw=3 +- - - - - - - - - - - - -+  I0 = 10 A
      |               ___...--'
      |      ___...--'
      | _..-'   phi0 = 72.5 deg (between V1 and I0)
      O-------------------------+----> flux
                          Im = 9.54 A (along flux)
```

#### (b) Iron loss
At no load the input power is almost entirely iron loss:
$$
P_i = V_1I_0\cos\phi_0 = 400 \times 10 \times 0.3 = 1200\ \text{W}
$$

#### (c) Maximum and rms flux in the core
From the EMF equation (taking $E_1 \approx V_1$ at no load):
$$
\begin{aligned}
E_1 &= 4.44\,f\,\phi_m N_1 \\
\phi_m &= \frac{400}{4.44 \times 50 \times 500} = 3.604 \times 10^{-3}\ \text{Wb} = 3.604\ \text{mWb}
\end{aligned}
$$
For sinusoidal flux:
$$
\phi_{rms} = \frac{\phi_m}{\sqrt2} = \frac{3.604}{1.4142} = 2.548\ \text{mWb}
$$

| Quantity | Value |
|---|---|
| Working component $I_w$ | 3 A |
| Magnetizing component $I_\mu$ | 9.54 A |
| Iron loss | 1200 W |
| Maximum flux | 3.604 mWb |
| RMS flux | 2.548 mWb |

**Answer:** (a) $I_w = 3$ A, $I_\mu = 9.54$ A; (b) iron loss = **1200 W**; (c) $\phi_m = 3.604$ mWb, $\phi_{rms} = 2.548$ mWb.

## Check yourself

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

1. With I0 = 10 A and pf 0.3, the magnetising component is:
   A. 3 A
   B. 9.54 A
   C. 10 A
   D. 7 A
   Answer: B. Imu = I0 sin(phi0) = 10 x 0.9539 = 9.54 A. (Key point: Magnetising component)

2. The working component Iw = I0 cos(phi0) is responsible for:
   A. Producing the flux
   B. Leakage flux
   C. Iron loss
   D. Copper loss only
   Answer: C. Iw is in phase with V1 and supplies the core loss; Pi = V1 I0 cos(phi0) = 1200 W. (Key point: Working component)

3. The maximum flux is found from E1 = 4.44 f phi_m N1 as:
   A. 2.548 mWb
   B. 3.604 mWb
   C. 7.2 mWb
   D. 0.36 mWb
   Answer: B. phi_m = 400/(4.44 x 50 x 500) = 3.604 mWb. (Key point: Max flux)

4. The rms flux is:
   A. phi_m x sqrt(2)
   B. phi_m / 2
   C. phi_m / sqrt(2) = 2.548 mWb
   D. phi_m
   Answer: C. For sinusoidal flux rms = peak/sqrt2. (Key point: RMS flux)

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