GATE 2017 EE – Question 55
A 375 W, 230 V, 50 Hz, capacitor start single-phase induction motor has the following constants for the main and auxiliary windings (at starting): $Z_m = (12.50 + j15.75)\ \Omega$ (main winding), $Z_a = (24.50 + j12.75)\ \Omega$ (auxiliary winding). Neglecting the magnetizing branch, the value of the capacitance (in $\mu$F) to be added in series with the auxiliary winding to obtain maximum torque at starting is ________.
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Correct answer: 145 to 155
Explanation
The starting torque is proportional to $I_mI_a\sin\alpha$, where $\alpha$ is the angle between the two winding currents. Write $\theta_m = \tan^{-1}\frac{15.75}{12.5} = 51.57°$ for the main winding and $\theta_a$ for the auxiliary branch, whose impedance is $24.5 + j(12.75 - X_c)$. The torque is proportional to $\frac{\sin(\theta_m - \theta_a)}{|Z_a|} \propto \cos\theta_a\sin(\theta_m - \theta_a)$. This is greatest when $\theta_m - 2\theta_a = 90°$, so $\theta_a = -19.2°$. Then $12.75 - X_c = 24.5\tan(-19.2°) = -8.53$, so $X_c = 21.28\ \Omega$ and $C = \frac{1}{2\pi \times 50 \times 21.28} = 149.6\ \mu$F.