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The power factor of the circuit shown in the figure is(a) 0.2(b) 0.8(c) 0.4(d) 0.6 |
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Answer» Answer is : (d) 0.6 R = 40+20 = 60 Ω Power factor = \(\frac{R}{\sqrt{R^2+(X_L-X_C)^2}}\) \(=\frac{60}{\sqrt{60^2+(100-20)^2}}\) = 0.6 |
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