Capacitor for quadrature currents (4.5+j3.7 Ω)

For the two winding currents to be in quadrature, the starting-winding current must lead the main-winding current by $90^\circ$. The capacitor makes the starting-winding branch capacitive.

Given: $Z_m = 4.5 + j3.7\ \Omega$, $Z_s = 9.5 + j3.5\ \Omega$, $f = 50$ Hz. (The rating, 250 W or "250 kW", does not affect the result.)

Angle of main winding current

$$\theta_m = \tan^{-1}\frac{3.7}{4.5} = 39.43^\circ \quad (I_m \text{ lags } V \text{ by } 39.43^\circ)$$

Required angle of starting branch

$I_s$ must lead $I_m$ by $90^\circ$, so it must lead $V$ by $90^\circ - 39.43^\circ = 50.57^\circ$. The impedance angle of the starting branch must be $-50.57^\circ$:

$$\begin{aligned} \tan(-50.57^\circ) &= \frac{X_s - X_C}{R_s} = \frac{3.5 - X_C}{9.5} \\ 3.5 - X_C &= 9.5 \times (-1.2162) = -11.554 \\ X_C &= 3.5 + 11.554 = 15.054\ \Omega \end{aligned}$$

Capacitance

$$C = \frac{1}{2\pi fX_C} = \frac{1}{2\pi \times 50 \times 15.054} = 211.4\ \mu\text{F}$$

           Is (leads V by 50.6 deg)
          /
         /  90 deg between Is and Im
  ------+--------------------> V
         \
          \ Im (lags V by 39.4 deg)

Check: angle between currents $= 50.57^\circ + 39.43^\circ = 90^\circ$.

Answer: $X_C = 15.05\ \Omega$, so a capacitor of about 211 µF must be connected in series with the starting winding.