Main winding current and pf at 5% slip

Using double field revolving theory, the rotor and magnetising branches are split into forward and backward halves. Core loss is neglected (no core-loss resistance is given).

Given: $V = 230$ V, $r_1 = 2.51\ \Omega$, $x_1 = 4.62\ \Omega$, $r_2' = 7.81\ \Omega$, $x_2' = 4.62\ \Omega$, $X_m = 150.88\ \Omega$, $s = 0.05$.

  I ->  r1     jx1
 o--/\/\/--mmm--+--------+---------+
                |        |         |
 V           0.5jXm  0.5r2'/s     |  Forward
                |     +j0.5x2'    |  Z_F
                +--------+---------+
                |        |         |
             0.5jXm  0.5r2'/(2-s) |  Backward
                |     +j0.5x2'    |  Z_B
 o--------------+--------+---------+

Forward impedance

$$\begin{aligned} 0.5 X_m &= 75.44\ \Omega,\quad \frac{0.5r_2'}{s} = \frac{3.905}{0.05} = 78.1\ \Omega,\quad 0.5x_2' = 2.31\ \Omega \\ Z_F &= \frac{(j75.44)(78.1 + j2.31)}{78.1 + j77.75} \\ &= 36.60 + j39.01\ \Omega \end{aligned}$$

Backward impedance

$$\begin{aligned} \frac{0.5 r_2'}{2-s} &= \frac{3.905}{1.95} = 2.003\ \Omega \\ Z_B &= \frac{(j75.44)(2.003 + j2.31)}{2.003 + j77.75} \\ &= 1.884 + j2.290\ \Omega \end{aligned}$$

Total input impedance

$$\begin{aligned} Z &= (r_1 + jx_1) + Z_F + Z_B \\ &= (2.51 + 36.60 + 1.884) + j(4.62 + 39.01 + 2.290) \\ &= 40.99 + j45.92 \\ &= 61.55\angle 48.24^\circ\ \Omega \end{aligned}$$

Main winding current and power factor

$$\begin{aligned} I_m &= \frac{V}{Z} = \frac{230\angle 0^\circ}{61.55\angle 48.24^\circ} = 3.737\angle -48.24^\circ\ \text{A} \\ \text{pf} &= \cos 48.24^\circ = 0.666\ \text{lagging} \end{aligned}$$

(For reference: air-gap powers are $P_{gF} = I^2 R_F = 511.0$ W and $P_{gB} = I^2 R_B = 26.3$ W.)

Answer: main winding current $I_m \approx 3.74$ A, power factor $\approx 0.666$ lagging.