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Electrical Engineering QUESTION #2168
Question 1
In the two-wattmeter method for a balanced three-phase load, $W_1 = 1500\,\text{W}$ and $W_2 = 500\,\text{W}$. The load power factor is:
  • $0.866$✔️
  • $0.707$
  • $0.5$
  • $0.75$
Correct Answer Explanation
$\tan\phi = \frac{\sqrt{3}(W_1-W_2)}{W_1+W_2} = \frac{\sqrt{3}(1000)}{2000} = \frac{\sqrt{3}}{2}$, so $\phi = 30°$ and $\text{pf} = \cos 30° \approx 0.866$.