Transformer no-load and load operation, constant core flux

An ideal transformer has no winding resistance, no leakage flux, no core loss and a core of infinite permeability (so it needs negligible magnetising current). Its primary and secondary are linked by the same mutual flux φ.

No-load operation

          phi (mutual flux)
     +---------------------+
  I0 |                     |
 o-->|]  N1           N2  [|---o
 V1  |]  E1           E2  [|   open
 o---|]               [   |---o
     +---------------------+

• The primary is connected to $V_1$; the secondary is open, so $I_2 = 0$.
• Since the winding has no resistance, the applied voltage is balanced only by the induced emf: $V_1 = E_1 = 4.44 f N_1\phi_m$. This fixes the flux.
• A very small current $I_0$ (the magnetising current $I_m$) flows. It is in phase with φ and lags $V_1$ by 90°, so the input power is zero.
• The same flux induces $E_2 = 4.44 f N_2\phi_m$ in the secondary. $E_1$ and $E_2$ are in phase, both lagging φ by 90°, and

$$\frac{E_1}{E_2} = \frac{N_1}{N_2}$$

Phasor diagram (no load): $\phi$ along the reference, $I_0$ in phase with φ, $V_1$ leading φ by 90°, $E_1$ and $E_2$ opposite to $V_1$.

Loaded operation

1. A load is connected; current $I_2$ flows in the secondary. It produces mmf $N_2 I_2$ which, by Lenz's law, opposes the mutual flux.
2. If φ fell, $E_1$ would fall below $V_1$. The primary therefore draws an extra load component of current $I_2'$ whose mmf exactly cancels the secondary mmf:

$$N_1 I_2' = N_2 I_2 \Rightarrow I_2' = \frac{N_2}{N_1}I_2$$

3. Total primary current $I_1 = I_0 + I_2'$. Power drawn from the supply rises to match the load: $V_1 I_1 \cos\phi_1 = V_2 I_2\cos\phi_2$.

Proof that the core flux remains constant at any load

Net mmf acting on the core:

$$\begin{aligned} F_{net} &= N_1 I_1 - N_2 I_2 \\ &= N_1 (I_0 + I_2') - N_2 I_2 \\ &= N_1 I_0 + (N_1 I_2' - N_2 I_2) \\ &= N_1 I_0 \end{aligned}$$

So the net mmf, and hence the flux $\phi = N_1 I_0 / S$, is the same as on no load. The same result follows from the voltage equation: since $V_1 = E_1 = 4.44 f N_1\phi_m$ at every load and $V_1$, $f$ and $N_1$ are fixed,

$$\phi_m = \frac{V_1}{4.44 f N_1} = \text{constant}$$

Hence the mutual flux of an ideal transformer is independent of load, from no load to full load. In a practical transformer it falls only very slightly because of the small leakage impedance drop.