Practical vs ideal transformer, phasor diagram

Practical vs ideal transformer

Operation of a practical transformer on load

When load current $I_2$ flows, the secondary terminal voltage and primary current are given by

$$V_2 = E_2 - I_2(R_2 + jX_2),\qquad I_1 = I_0 + I_1',\ \ I_1' = \frac{N_2}{N_1}I_2,\qquad V_1 = -E_1 + I_1(R_1 + jX_1)$$

Phasor diagram (inductive load, lagging $\phi_2$):

          V1 (applied)
         /|
   I1X1/ |             Steps:
      /  | I1R1        1. Take flux phi as reference.
 -E1 /___|              2. E1, E2 lag phi by 90 deg.
     |                  3. I2 lags V2 by phi2.
     |  E2              4. E2 = V2 + I2R2 + jI2X2.
     |   \ I2X2         5. I1' = I2/a opposes the I2 mmf.
 phi |    \ I2R2        6. I1 = I0 + I1'.
     v     \V2          7. V1 = -E1 + I1R1 + jI1X1.

Steps in words:

1. Draw the flux $\phi$ as reference. The emfs $E_1$ and $E_2$ lag it by 90°, and $-E_1$ is drawn opposite to $E_1$.
2. Draw $V_2$, then $I_2$ lagging it by $\phi_2$. Add $I_2R_2$ (parallel to $I_2$) and $I_2X_2$ (90° ahead) to reach $E_2$.
3. The load component $I_1'$ is drawn opposite to $I_2$ (scaled by $N_2/N_1$). The no-load current $I_0$ is drawn leading the flux by the hysteresis angle, with $I_\mu$ along $\phi$ and $I_c$ along $-E_1$.
4. $I_1 = I_0 + I_1'$. Add $I_1R_1$ and $I_1X_1$ to $-E_1$ to get $V_1$.

The angle between $V_1$ and $I_1$ is the primary power factor angle $\phi_1$, slightly different from $\phi_2$ because of $I_0$ and the drops.