Transformer on load and no load, phasor diagrams

A practical transformer has winding resistances $R_1, R_2$, leakage reactances $X_1, X_2$ and a core that draws an exciting current $I_0$.

No-load operation

The secondary is open ($I_2 = 0$). The primary draws only $I_0$, which has two components:

$$I_0 = I_c + I_\mu,\qquad I_c = I_0\cos\phi_0\ (\text{core loss}),\qquad I_\mu = I_0\sin\phi_0\ (\text{magnetising})$$

              V1 (= -E1 approx)
               ^
               |
          Ic   |      . I0
     (along V1)|    .
               |  .     phi0 = angle between V1 and I0
               |.          (about 75-80 deg)
               +-------------------> Im, flux phi
               |
               v
             E1, E2 (lag phi by 90 deg)

• $I_c$ is in phase with $V_1$ and supplies the core loss.
• $I_\mu$ lags $V_1$ by 90°, and the flux $\phi$ is in phase with $I_\mu$.
• $E_1$ and $E_2$ lag $\phi$ by 90°, so $E_1$ is opposite to $V_1$.
• $I_0$ is small (2 to 5% of full-load current) and lags $V_1$ by $\phi_0$, so the no-load power factor is low (about 0.1 to 0.25).

On load

The secondary current $I_2$ flows, and the primary takes the additional load component $I_1' = I_2/a$, so $I_1 = I_0 + I_1'$ (phasor sum). The voltage equations are

$$V_1 = -E_1 + I_1(R_1 + jX_1),\qquad V_2 = E_2 - I_2(R_2 + jX_2)$$

Resistive load. $I_2$ is in phase with $V_2$. $E_2$ is found by adding to $V_2$ the drop $I_2R_2$ (in phase with $I_2$) and $I_2X_2$ (90° ahead of $I_2$). So $E_2 > V_2$ and the terminal voltage falls on load. $I_1'$ is in phase with $-I_2$ (it is drawn in phase with $-E_1$), and $I_1$ is close to $I_1'$. The primary power factor is nearly unity.

  Resistive load (phasors referred to secondary)

          E2
         /|
        / | I2X2
       /  |
      /___|______
      V2  I2R2    I2 along V2

Capacitive load. $I_2$ leads $V_2$ by $\phi_2$. The drops $I_2R_2$ and $I_2X_2$ are drawn along $I_2$ and 90° ahead of it. Because $I_2$ is now ahead of $V_2$, the drop $I_2X_2$ points partly opposite to $V_2$, and $E_2$ becomes smaller than $V_2$: approximately $E_2 \approx V_2 + I_2R_2\cos\phi_2 - I_2X_2\sin\phi_2$. So the secondary terminal voltage rises on load (negative regulation) when the leading power factor is low enough.

  Capacitive load (V2 as reference, I2 leads V2 by phi2)

               E2 tip <-----------+
                 .'      I2X2     /
              .'                 / I2R2 (along I2)
   O ----------------------------+ V2
      E2 runs from O to E2 tip and is shorter than V2

In each case, $I_1$ is obtained as $I_0 + I_1'$, and $V_1 = -E_1 + I_1(R_1 + jX_1)$ completes the primary diagram. For the capacitive load $I_1'$ leads $-E_1$, so the primary power factor is leading when the load current is large compared with $I_0$.