Discussion
A 50 kW single-phase motor is supplied by an AC source of 50 Hz. If the terminal voltage of the motor is 230 V angle(\(0^\circ\)), the power factor of the motor is 0.5 lag, find the source current.
Answer: C
For a single phase load the real power is \(P = VI\cos\phi\), so the line current is
\[ \begin{aligned} I = \frac{P}{V\cos\phi} \\ &= \frac{50 \times 10^3}{230 \times 0.5} \\ &= \frac{50000}{115} = 434.8\ \text{A} \end{aligned} \]
This is the value of 434.7 A in the options, option C. A current of 43.47 A or 4.34 A would be a slip of one or two powers of ten, and 0.4347 A would come from taking the power as 50 W. The current lags the voltage by \(\cos^{-1}0.5 = 60^\circ\).
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