SSE JE Electrical Engineering 24 March 2021 Shift 1 Paper

Question : 148 of 200

Marks: +1, -0

The main and auxiliary winding impedance of a 50-Hz, capacitor-start single-phase induction motor is

\Z_\m = (3 + \j3) \Ω

and

\Z_\a= (7 + \j3) \Ω

. Determine the value of the capacitor to be connected in series with the auxiliary winding to achieve a phase difference of 90° between the currents of the two windings at the start.

225 μF
22.5 μF
318 μF
31.8 μF

Validate

Solution:

Given that

\∅_{ \a }+\∅_{ \m }=90°

\tan ^{-1}({\X_{\a}-\X_{\c}}/{\R_{\a}})-\tan ^{-}({\X_{\m}}/{\R_{\m}})=90°

⇒\tan ^{-1}({3-\X_{\c}}/{7})-\tan ^{-}({3}/{3})=90°

⇒\tan ^{-1}({3-\X_{\c}}/{7})=135

⇒{3-\X_{\c}}/{7}=-1

⇒\X_{\c}=10

⇒{1}/{2 \π\fC }=10 ⇒\C =318 \µ\F

Go to Question:

Prev Question

Next Question