Starting line current of capacitor start motor

At starting both windings are across the 230 V supply. The capacitor is in series with the auxiliary winding. Line current is the phasor sum of the two winding currents.

Given: $Z_m = 15 + j22.5\ \Omega$, $Z_a = 50 + j120\ \Omega$, $C = 12\ \mu\text{F}$, $V = 230\angle 0^\circ$ V, $f = 50$ Hz.

Capacitive reactance

$$X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times 50 \times 12\times10^{-6}} = 265.26\ \Omega$$

Branch impedances

$$\begin{aligned} Z_m &= 15 + j22.5 = 27.04\angle 56.31^\circ\ \Omega \\ Z_a' &= 50 + j(120 - 265.26) = 50 - j145.26 \\ &= 153.62\angle -71.01^\circ\ \Omega \end{aligned}$$

Winding currents

$$\begin{aligned} I_m &= \frac{230\angle 0^\circ}{27.04\angle 56.31^\circ} = 8.505\angle -56.31^\circ\ \text{A} \\ &= 4.718 - j7.077\ \text{A} \\ I_a &= \frac{230\angle 0^\circ}{153.62\angle -71.01^\circ} = 1.497\angle 71.01^\circ\ \text{A} \\ &= 0.487 + j1.416\ \text{A} \end{aligned}$$

Line current

$$\begin{aligned} I_L &= I_m + I_a \\ &= (4.718 + 0.487) + j(-7.077 + 1.416) \\ &= 5.205 - j5.661 \\ &= 7.69\angle -47.40^\circ\ \text{A} \end{aligned}$$

Phase angle between $I_a$ and $I_m$ $= 71.01^\circ + 56.31^\circ = 127.3^\circ$; starting power factor $= \cos 47.40^\circ = 0.677$ lagging.

        I_a (1.50 A, +71.0 deg)
         ^
         |
  -------+--------------> V
         | \
         |  \  I_L (7.69 A, -47.4 deg)
         |   \
         v I_m (8.51 A, -56.3 deg)

Answer: starting line current $I_L \approx 7.69\angle -47.4^\circ$ A, i.e. 7.69 A at 0.677 lagging power factor.