Increase in starting torque with 1000 μF capacitor

Starting torque of a two-winding single-phase motor is proportional to the product of the winding currents and the sine of the angle between them:

$$T_{st} = K I_m I_a \sin\alpha, \qquad I = \frac{V}{|Z|}$$

Given: $V = 120$ V, $f = 50$ Hz, $Z_m = 1.5 + j4\ \Omega$, $Z_a = 3 + j6\ \Omega$, $C = 1000\ \mu\text{F}$.

Main winding (unchanged)

$$Z_m = 1.5 + j4 = 4.272\angle 69.44^\circ\ \Omega,\quad I_m = 28.09\angle -69.44^\circ\ \text{A}$$

Case 1: without capacitor

$$\begin{aligned} Z_a &= 3 + j6 = 6.708\angle 63.43^\circ\ \Omega \\ I_a &= \frac{120}{6.708} = 17.89\angle -63.43^\circ\ \text{A} \\ \alpha_1 &= 69.44^\circ - 63.43^\circ = 6.01^\circ \\ T_1 &= K(28.09)(17.89)\sin 6.01^\circ = 52.62\,K \end{aligned}$$

Case 2: with 1000 μF capacitor

$$\begin{aligned} X_C &= \frac{1}{2\pi \times 50 \times 1000\times10^{-6}} = 3.183\ \Omega \\ Z_a' &= 3 + j(6 - 3.183) = 3 + j2.817 = 4.115\angle 43.20^\circ\ \Omega \\ I_a' &= \frac{120}{4.115} = 29.16\angle -43.20^\circ\ \text{A} \\ \alpha_2 &= 69.44^\circ - 43.20^\circ = 26.25^\circ \\ T_2 &= K(28.09)(29.16)\sin 26.25^\circ = 362.3\,K \end{aligned}$$

Ratio and percentage increase

$$\begin{aligned} \frac{T_2}{T_1} &= \frac{29.16 \sin 26.25^\circ}{17.89 \sin 6.01^\circ} = 6.886 \\ \%\ \text{increase} &= (6.886 - 1)\times 100 = 588.6\% \end{aligned}$$

Answer: starting torque increases about 6.89 times, i.e. an increase of about 589%.