Split-phase motor winding currents at starting

Both windings are connected directly across the 230 V supply at starting, so each current is $V/Z$ and the line current is their phasor sum.

Given: $Z_m = 4 + j7.5\ \Omega$, $Z_s = 7.5 + j4\ \Omega$, $V = 230\angle 0^\circ$ V.

$$\begin{aligned} Z_m &= \sqrt{4^2 + 7.5^2}\angle\tan^{-1}\tfrac{7.5}{4} = 8.5\angle 61.93^\circ\ \Omega \\ Z_s &= \sqrt{7.5^2 + 4^2}\angle\tan^{-1}\tfrac{4}{7.5} = 8.5\angle 28.07^\circ\ \Omega \end{aligned}$$

a) Main winding current

$$I_m = \frac{230\angle 0^\circ}{8.5\angle 61.93^\circ} = 27.06\angle -61.93^\circ\ \text{A} = 12.73 - j23.88\ \text{A}$$

b) Starting winding current

$$I_s = \frac{230\angle 0^\circ}{8.5\angle 28.07^\circ} = 27.06\angle -28.07^\circ\ \text{A} = 23.88 - j12.73\ \text{A}$$

c) Phase angle between $I_s$ and $I_m$

$$\alpha = 61.93^\circ - 28.07^\circ = 33.86^\circ \quad (I_s \text{ leads } I_m)$$

d) Line current

$$\begin{aligned} I_L &= I_m + I_s = (12.73 + 23.88) - j(23.88 + 12.73) \\ &= 36.61 - j36.61 = 51.77\angle -45^\circ\ \text{A} \end{aligned}$$

e) Power factor

$$\text{pf} = \cos 45^\circ = 0.707\ \text{lagging}$$

Answer: $I_m = 27.06$ A, $I_s = 27.06$ A, $\alpha = 33.86^\circ$, $I_L = 51.77$ A, pf $= 0.707$ lagging.