Starting capacitor and change in starting torque

The capacitor must make the starting-winding current lead the main-winding current by $90^\circ$. Starting torque is $T_{st} = K I_m I_s \sin\alpha$, where $\alpha$ is the angle between the two currents. The change in torque is found by comparing with the motor without the capacitor (plain resistance split-phase start), with the same supply voltage $V$.

Capacitor value

$$\theta_m = \tan^{-1}\frac{3}{3} = 45^\circ \ (I_m \text{ lags } V)$$

For $\alpha = 90^\circ$, $I_s$ must lead $V$ by $90^\circ - 45^\circ = 45^\circ$, so the starting branch angle is $-45^\circ$:

$$\begin{aligned} \frac{3 - X_C}{7.5} &= \tan(-45^\circ) = -1 \\ X_C &= 3 + 7.5 = 10.5\ \Omega \\ C &= \frac{1}{2\pi \times 50 \times 10.5} = 303.2\ \mu\text{F} \end{aligned}$$

Starting torque without capacitor

$$\begin{aligned} \theta_s &= \tan^{-1}\frac{3}{7.5} = 21.80^\circ,\quad |Z_s| = 8.0777\ \Omega \\ \alpha_1 &= 45^\circ - 21.80^\circ = 23.20^\circ \\ I_{s1} &= \frac{V}{8.0777} = 0.1238V \\ T_1 &\propto I_m \times 0.1238V \times \sin23.20^\circ = 0.04877\,V I_m \end{aligned}$$

Starting torque with capacitor

$$\begin{aligned} |Z_s'| &= |7.5 - j7.5| = 10.6066\ \Omega,\quad I_{s2} = 0.0943V \\ T_2 &\propto I_m \times 0.0943V \times \sin90^\circ = 0.09428\,V I_m \end{aligned}$$

$I_m$ is the same in both cases, so

$$\frac{T_2}{T_1} = \frac{0.09428}{0.04877} = 1.9333$$

Percentage change $= (1.9333 - 1) \times 100 = 93.33\ \%$ (increase).

        Is (with C, leads V by 45 deg)
       /
      / 90 deg
 ----+-----------------> V
     |\  Is (no C, lags 21.8 deg)
     | \
     |  Im (lags 45 deg)

Answer: $C = 303.2$ µF (series with starting winding); starting torque increases by about 93.3 % compared with the same motor started without the capacitor.