Transformer at resistive, inductive, capacitive loads

The transformer is treated as a practical one with winding resistances $R_1,R_2$ and leakage reactances $X_1,X_2$. For any load, with the turns ratio $a = N_1/N_2$:

$$V_2 = E_2 - I_2(R_2 + jX_2),\qquad I_1 = I_0 + \frac{I_2}{a},\qquad V_1 = -E_1 + I_1(R_1 + jX_1)$$

Circuit diagram

 I1   R1   X1        R2   X2   I2
 o--/\/\--UUU--+  +--/\/\--UUU--o
               |  |             |
   V1     E1 ( )  ( ) E2        Z (load)
               |  |             |
 o-------------+--+-------------o
        ideal transformer N1:N2

Phasor diagrams (drawn with $V_2$ as the reference)

For each load, add to $V_2$ the drop $I_2R_2$ (parallel to $I_2$) and then $I_2X_2$ (90° ahead of $I_2$) to obtain $E_2$.

 Inductive          Resistive         Capacitive
 (I2 lags V2)       (I2 = V2)         (I2 leads V2)

   E2 (big)           E2                I2 /
   /|                 /|                  /
  / | I2X2           / | I2X2        E2 /__ V2 (> E2)
 /__|                /__|
 V2 I2R2             V2 I2R2
 (I2 below V2)      (I2 along V2)

Primary side: in each case, the load component $I_1' = I_2/a$ has the same phase angle $\phi_2$ relative to $-E_1$ as $I_2$ has relative to $E_2$. It is added to the no-load current $I_0$ (which lags $-E_1$ by about 78°) to give $I_1$. Then $V_1$ is the phasor sum of $-E_1$, $I_1R_1$ and $jI_1X_1$.

Result: the inductive load gives the poorest regulation, the resistive load an intermediate value, and a capacitive load can give a voltage rise. The primary power factor follows the load: lagging for an inductive load, nearly unity for a resistive load, and leading for a strongly capacitive load.