Loaded transformer and phasor diagram for inductive load

Loaded operation of a transformer

When a load is connected to the secondary, a current $I_2$ flows, set by the load: $I_2 = V_2/Z_L$.

1. $I_2$ in $N_2$ turns produces an MMF $N_2I_2$ that, by Lenz's law, opposes the main flux $\phi$.
2. The flux tends to fall, so the back EMF $E_1$ falls slightly and the primary draws an extra current $I_2'$ from the supply.
3. $I_2'$ is just large enough for its MMF to cancel the secondary MMF: $N_1I_2' = N_2I_2$, so $I_2' = KI_2$ (with $K = N_2/N_1$).
4. The total primary current is the phasor sum $\bar I_1 = \bar I_0 + \bar I_2'$.
5. Because the net MMF is still $N_1I_0$, the core flux stays almost constant from no load to full load, so the iron loss is constant.

In a real transformer the windings have resistance and leakage reactance, so:

$$\begin{aligned} \bar V_1 &= -\bar E_1 + \bar I_1(R_1 + jX_1) \\ \bar V_2 &= \bar E_2 - \bar I_2(R_2 + jX_2) \end{aligned}$$

As the load increases, these drops increase, so $V_2$ falls (voltage regulation).

Phasor diagram for inductive (lagging) load

Steps: draw $\phi$ as the reference; $E_1$ and $E_2$ lag $\phi$ by 90°; $I_0$ leads $\phi$ by a small angle (because of core loss); $I_2$ lags $V_2$ by $\phi_2$; $V_2 = E_2 - I_2R_2 - jI_2X_2$; $I_2'$ is drawn equal and opposite to $I_2$; $I_1 = I_0 + I_2'$; $V_1 = -E_1 + I_1R_1 + jI_1X_1$.

                   V1
                   ^
                  /|  I1.X1 (perpendicular to I1)
                 / |
       -E1 ^    /--+  I1.R1 (parallel to I1)
           |   /
           |  /         ^ I2' (equal and opposite to I2)
           | /        /
           |/       /   ^ I1 = I0 + I2'
           |      /   /
           |    /   /
           |  /   /  ___...---> I0 (leads phi slightly)
           |/  /..--'
  ---------O-------------------------> phi (main flux, reference)
          /|
         / |
        /  |
       v   |
     I2    v V2 = E2 - I2(R2 + jX2)
 (lags V2  |
  by phi2) v E2   (E1 also lies along this line)

  phi1 = angle between V1 and I1; phi2 = angle between V2 and I2

(For clarity many textbooks take $K = 1$, so $E_1 = E_2$ and $I_2' = I_2$ in magnitude.)

Key features for an inductive load:

• $I_2$ lags $V_2$ by $\phi_2$, so it points down and to the left of $V_2$; $I_2'$ is exactly opposite to it, lagging $-E_1$.
• $I_1$ lags $V_1$ by $\phi_1$, with $\phi_1$ slightly greater than $\phi_2$ because of $I_0$ and the drops.
• $V_2 < E_2$ and $V_1 > E_1$; the drop is largest for lagging loads.
• On full load $I_0$ is small compared with $I_2'$, so $I_1 \approx I_2'$ and $\frac{I_1}{I_2} \approx K$.