---
title: "Real transformer no-load operation and amp-turn balance"
subject: "Electrical Machines I"
chapter: 2
key: "semester_4/electrical_machines_i:real-transformer-noload-amp-turn-balance"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/real-transformer-noload-amp-turn-balance
site: BCT Engineering
---

# Real transformer no-load operation and amp-turn balance

Electrical Machines I · Chapter 2: Transformer

- Asked in: 2068 Bhadra, 8 marks
- Video: [Transformer Lecture 3 (Ampere-Turn Relation)](https://www.youtube.com/watch?v=qiniAfjSA6g) — sourabh sharma
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/real-transformer-noload-amp-turn-balance
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2

## Question

Explain no-load operation of a real transformer. What do you mean by Amp-turn balance in loaded transformer? Prove that the magnetic flux in the core remains constant irrespective of load on transformer.

## Short answer

At no load, primary draws I0 consisting of magnetizing Iμ and core-loss Iw components, with low pf. When loaded, secondary ampere-turns N2I2 are balanced by primary load component I2' (N1I2'= N2I2), so net MMF equals N1I0. Thus core flux φm = V1/(4.44fN1) remains essentially constant from no load to full load.

## Answer

#### No-load operation of a real transformer
At no load the secondary is open, so $I_2 = 0$. The primary draws a small **no-load current** $I_0$ (2-5% of rated current). In a real transformer $I_0$ does two jobs, so it has two components:
- **Magnetizing component** $I_\mu = I_0\sin\phi_0$: in phase with flux, lags $V_1$ by 90°, sets up the core flux. It is the larger part.
- **Working (core-loss) component** $I_w = I_0\cos\phi_0$: in phase with $V_1$, supplies hysteresis and eddy-current losses (plus a tiny $I_0^2R_1$).
$$
I_0 = \sqrt{I_\mu^2 + I_w^2}, \qquad P_0 = V_1I_0\cos\phi_0 \approx P_i
$$
The no-load pf $\cos\phi_0$ is low (0.1-0.3), so $\phi_0$ is close to 90°.

```text
        V1 = -E1
          ^
          |   . I0
       Iw |  /
          | /  phi0 (near 90 deg)
          |/
  --------O----------> phi
          |   Im
          v E1, E2
```

#### Amp-turn balance in a loaded transformer
When load is connected, secondary current $I_2$ produces MMF $N_2I_2$ which opposes the flux (Lenz's law). The primary immediately draws an extra **load component** $I_2'$ whose MMF exactly cancels it:
$$
N_1I_2' = N_2I_2 \quad\Rightarrow\quad I_2' = \frac{N_2}{N_1}I_2 = KI_2
$$
This equality of opposing ampere-turns is called **ampere-turn (MMF) balance**. The total primary current is $\bar I_1 = \bar I_0 + \bar I_2'$; since $I_0$ is small, $N_1I_1 \approx N_2I_2$.

#### Proof that core flux is (nearly) constant at all loads
1. The applied voltage $V_1$ is fixed. Neglecting the small primary drop, $V_1 \approx E_1 = 4.44fN_1\phi_m$. Hence
$$
\phi_m = \frac{V_1}{4.44fN_1} = \text{constant}
$$
2. MMF view: net core MMF with load is
$$
N_1\bar I_1 - N_2\bar I_2 = N_1(\bar I_0 + \bar I_2') - N_2\bar I_2 = N_1\bar I_0 + (N_1\bar I_2' - N_2\bar I_2) = N_1\bar I_0
$$
because the bracket is zero by amp-turn balance. The net MMF equals the no-load MMF $N_1I_0$, so the core flux $\phi = \frac{N_1I_0}{\mathcal R}$ is the same as at no load.
3. Self-regulation: if load rises, $I_2$ rises, flux tends to fall, $E_1$ falls slightly, so $(V_1 - E_1)$ rises and the primary draws more current until the flux is restored.

Hence the core flux, and therefore **iron loss, remain practically constant from no load to full load**. (Strictly, $\phi$ falls a little at full load because of the $I_1Z_1$ drop, about 1-2%.)

## Check yourself

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

1. The magnetizing component Imu of the no-load current is:
   A. In phase with V1 and supplies core loss
   B. In phase with flux and lagging V1 by 90 degrees
   C. Equal to the load current
   D. Leading V1 by 90 degrees
   Answer: B. Imu establishes the core flux, so it is in phase with flux and lags V1 by 90 degrees; Iw is in phase with V1. (Key point: No-load current components)

2. Why is the no-load power factor of a transformer low (0.1-0.3)?
   A. Imu is much larger than Iw, so phi0 is near 90 degrees
   B. The core loss is large
   C. Copper loss is high
   D. Secondary is shorted
   Answer: A. Imu >> Iw, so I0 lags V1 by an angle close to 90 degrees. (Key point: No-load power factor)

3. Amp-turn balance in a loaded transformer means:
   A. N1 I2' = N2 I2, the primary load component cancels secondary MMF
   B. N1 I0 = N2 I2
   C. N1 I1 = 0
   D. V1 = V2
   Answer: A. The extra primary current I2' produces MMF that exactly opposes N2 I2, so net core MMF stays N1 I0. (Key point: Ampere-turn balance)

4. The main reason core flux is nearly constant from no load to full load is:
   A. V1 is fixed and V1 ~ E1 = 4.44 f N1 phi_m
   B. Load current is constant
   C. Secondary MMF is zero
   D. Core reluctance rises with load
   Answer: A. With fixed V1, f and N1, phi_m = V1/(4.44 f N1) is constant; equivalently, net MMF equals N1 I0. (Key point: Constant flux proof)

## Related in Electrical Machines I

- [Transformer no-load and load operation, constant core flux](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aideal-transformer-noload-load-constant-flux) — Electrical Machines I, chapter 2, asked 5×
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