Real transformer no-load operation and amp-turn balance

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°.

        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%.)