Transformer operation; flux constant at any load

A transformer works on mutual induction. Its operation is explained at no-load and on load, and the same net flux is present in both.

No-load operation

 I0 <-- Primary           Secondary (open)
 o--+  +-------------+  +--o
 V1 |N1|   core  Φ   |N2|  E2 = V2 (no current)
 o--+  +-------------+  +--o

• The primary is connected to the supply V1 and the secondary is open.
• A small no-load current I0 flows. It has a magnetising component Iμ (in phase with Φ, produces the flux) and a core-loss component Iw (in phase with V1, supplies the iron loss). So I0 = √(Iμ² + Iw²) = 2 to 5 % of full-load current.
• The alternating flux Φ links both windings: E1 = 4.44 f N1 Φm, E2 = 4.44 f N2 Φm. Since E1 ≈ V1, the flux is fixed by V1.
• Primary current lags V1 by nearly 90°, so the no-load power factor is very low.

Loaded operation

• A load is connected to the secondary. A secondary current I2 flows and produces an mmf N2 I2 which, by Lenz's law, opposes the flux Φ.
• The flux tends to fall, so E1 falls slightly and the primary draws a larger current. The extra current I2′ gives an mmf N1 I2′ that cancels N2 I2: N1 I2′ = N2 I2 or I2′ = (N2/N1) I2.
• Total primary current: I1 = I0 + I2′ (phasor sum).

Proof that the net flux is constant

Net mmf acting on the core = N1 I1 − N2 I2
 = N1 (I0 + I2′) − N2 I2
 = N1 I0 + N1 I2′ − N2 I2
 = N1 I0 + 0                (because N1 I2′ = N2 I2)

So the net mmf is N1 I0, the same as at no-load. Since the flux produced by this mmf is Φ, the core flux is the same at all loads. This is also clear from V1 ≈ E1 = 4.44 f N1 Φm. V1 and f are constant, so Φm is constant.