Ideal transformer and coupled circuit model

Characteristics of an ideal transformer

1. Winding resistances are zero, so there is no copper loss.
2. There is no leakage flux; all the flux links both windings (coupling coefficient k = 1).
3. The core has infinite permeability, so no magnetising current is needed.
4. There is no core loss (no hysteresis or eddy current loss).
5. Hence efficiency is 100%, voltage regulation is zero, and

$$\frac{V_1}{V_2} = \frac{N_1}{N_2} = \frac{I_2}{I_1}, \qquad Z_1 = \left(\frac{N_1}{N_2}\right)^2 Z_2$$

Coupled circuit model of a transformer

A real (linear) transformer is treated as two magnetically coupled coils with self inductances $L_1$, $L_2$, mutual inductance $M$, and winding resistances $R_1$, $R_2$.

   i1    R1            M             R2    i2
  o-->--/\/\--+    .        .    +--/\/\--<--o
              |   (L1)    (L2)   |
  v1          )    )      (      (         v2
              |                  |
  o-----------+                  +-----------o

The flux linking each coil has a self and a mutual part:

$$\lambda_1 = L_1 i_1 + M i_2, \qquad \lambda_2 = M i_1 + L_2 i_2$$

Voltage equations (with both currents entering the dotted terminals):

$$\begin{aligned} v_1 &= R_1 i_1 + L_1\frac{di_1}{dt} + M\frac{di_2}{dt} \\ v_2 &= R_2 i_2 + L_2\frac{di_2}{dt} + M\frac{di_1}{dt} \end{aligned}$$

In sinusoidal steady state: $V_1 = (R_1 + j\omega L_1)I_1 + j\omega M I_2$ and $V_2 = j\omega M I_1 + (R_2 + j\omega L_2) I_2$.

Link with leakage and magnetising inductances

With $a = N_1/N_2$, the self inductances are split into leakage and magnetising parts:

$$\begin{aligned} L_1 &= L_{l1} + aM, \qquad L_2 = L_{l2} + \frac{M}{a}, \qquad L_m = aM \\ M &= k\sqrt{L_1 L_2}, \quad k \le 1 \end{aligned}$$

Substituting gives the familiar T-equivalent circuit: $R_1$ and $X_{l1} = \omega L_{l1}$ in series, the magnetising reactance $X_m = \omega aM$ in shunt, and $R_2' = a^2R_2$, $X_{l2}' = a^2 X_{l2}$ followed by an ideal transformer of ratio $a$.

  R1   Xl1           a^2 R2  a^2 Xl2      ideal a:1
 o-/\/-mmm--+--------/\/\----mmm-----+--||--o
            |                        |  ||
           Xm=w aM                   |  ||  V2
            |                        |  ||
 o----------+------------------------+--||--o

When $k \to 1$, $R \to 0$ and $L_1, L_2 \to \infty$ with $L_1/L_2 = a^2$, the coupled-circuit model reduces to the ideal transformer. Core loss is added later as a shunt resistance $R_c$ across $X_m$.