Capacitor for quadrature currents (7+j5 Ω)

For quadrature, $I_a$ must lead $I_m$ by $90^\circ$, so the auxiliary branch angle must be $90^\circ$ less than that of the main winding.

Given: $Z_m = 7 + j5\ \Omega$, $Z_a = 11.5 + j5\ \Omega$, $V = 230$ V, $f = 50$ Hz.

$$\begin{aligned} \theta_m &= \tan^{-1}\frac{5}{7} = 35.54^\circ \\ \theta_a' &= 35.54^\circ - 90^\circ = -54.46^\circ \\ 5 - X_C &= 11.5\tan(-54.46^\circ) = -11.5 \times \frac{7}{5} = -16.1 \\ X_C &= 21.1\ \Omega \\ C &= \frac{1}{2\pi \times 50 \times 21.1} = 150.9\ \mu\text{F} \end{aligned}$$

Currents: $I_m = \dfrac{230}{8.602\angle 35.54^\circ} = 26.74\angle -35.54^\circ$ A, $I_a = \dfrac{230}{19.78\angle -54.46^\circ} = 11.62\angle 54.46^\circ$ A.

 Circuit:                 Phasors:
 o---+--------+              I_a (54.5 deg)
     |        |               ^
   Z_m      Z_a               |  90 deg
     |        |        -------+-------> V
 230V|       [C]              \
     |        |                v I_m (-35.5 deg)
     |       CS
 o---+--------+

Answer: $C \approx 150.9\ \mu\text{F}$ ($X_C = 21.1\ \Omega$) in series with the auxiliary winding.