In one line
A practical transformer has winding resistance, leakage flux, core losses and finite permeability, so even at no load it draws a small current \( I_0 \) made of a magnetising part and a core-loss part.
Explained simply
A real cycle has friction in its chain and air resistance, so you need some effort just to keep it moving, even with no passenger. A real transformer similarly takes a small current from the supply even when nothing is connected to the secondary, just to magnetise the core and cover iron losses.
Technically, the no-load current \( I_0 \) (typically only about 2–5% of rated current) has two components: \( I_\mu \) (magnetising component) lagging \( V_1 \) by 90°, which produces flux; and \( I_w \) (active or core-loss component) in phase with \( V_1 \), which supplies hysteresis and eddy-current losses.
How it works
- Apply \( V_1 \) with secondary open; \( I_0 \) flows in the primary.
- \( I_0 \) lags \( V_1 \) by angle \( \phi_0 \), slightly less than 90°, so no-load power factor \( \cos\phi_0 \) is very low (around 0.1–0.2 is typical).
- \( I_w = I_0\cos\phi_0 \) and \( I_\mu = I_0\sin\phi_0 \).
- No-load input \( W_0 = V_1I_0\cos\phi_0 \) is practically equal to the iron (core) loss.
- Other non-ideal effects: winding resistances \( R_1, R_2 \) cause copper loss and voltage drop; leakage flux is modelled as leakage reactances \( X_1, X_2 \). So on load, \( V_2 \) is slightly less than \( E_2 \).
| Ideal | Practical |
|---|---|
| no winding resistance | \( R_1, R_2 \) present: copper loss |
| no leakage flux | leakage reactances \( X_1, X_2 \) |
| no core loss | hysteresis + eddy current loss |
| efficiency 100% | typically 95–99% |
Key formula
\( I_0 = \sqrt{I_w^2 + I_\mu^2}, \quad I_w = I_0\cos\phi_0, \quad I_\mu = I_0\sin\phi_0, \quad W_0 = V_1I_0\cos\phi_0 \)
\( I_0 \) = no-load current (A), \( I_w \) = core-loss component (A), \( I_\mu \) = magnetising component (A), \( \phi_0 \) = no-load phase angle, \( W_0 \) = no-load input power (W) ≈ iron loss.
Quick example
A distribution transformer left energised overnight with almost no load still hums and stays slightly warm: it is drawing \( I_0 \) and dissipating its iron loss continuously, which is why utilities care about low core loss in such transformers.
Common mistakes
- Saying no-load current is zero in a practical transformer.
- Drawing \( I_0 \) in phase with \( V_1 \); it lags by almost 90°.
- Saying the no-load input is mainly copper loss; copper loss at no load is negligible.
How MU asks it
MU states numericals are not expected in this module, so questions are theory. Typical: "Explain the practical transformer on no load with phasor diagram" or "Differentiate between ideal and practical transformer" (5 marks). For full marks draw the phasor diagram with \( V_1, I_0, I_w, I_\mu, \phi, E_1, E_2 \), explain both components of \( I_0 \), and list the non-ideal effects.