GATE 2018 EE – Question 64
The equivalent circuit of a single phase induction motor is shown in the figure, where the parameters are $R_1=R_2'=X_{l1}=X_{l2}'=12\ \Omega$, $X_M=240\ \Omega$ and $s$ is the slip. At no-load, the motor speed can be approximated to be the synchronous speed. The no-load lagging power factor of the motor is ________ (up to 3 decimal places).
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Correct answer: 0.10 to 0.12
Explanation
At no load $s\approx0$, so the forward rotor branch $\frac{R_2'}{2s}\to\infty$ and the forward branch is just $j\frac{X_M}{2}=j120\ \Omega$. The backward branch has $\frac{R_2'}{2(2-s)}=3\ \Omega$ with $j6\ \Omega$, in parallel with $j120$: $\frac{j120(3+j6)}{3+j126}=2.72+j5.78\ \Omega$. The total is $12+j12+j120+2.72+j5.78=14.72+j137.8\ \Omega$, so the power factor is $\frac{14.72}{138.6}=0.106$.