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$\frac{2 .5}{π} μF$ capacitor and 3000-ohm resistance are joined in series to an ac source of 200 volt and $50 \sec^{- 1}$ frequency. The power factor of the circuit and the power dissipated in it will respectively

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0.6, 0.06 W

0.06, 0.6 W

0.6, 4.8 W

4.8, 0.6 W

answer is C.

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Detailed Solution

$Z = \sqrt{R^{2} + {(\frac{1}{2 πνC})}^{2}} = \sqrt{{(3000)}^{2} + \frac{1}{{(2 π \times 50 \times \frac{2 .5}{π} \times 10^{- 6})}^{2}}}$

$\Rightarrow Z = \sqrt{{(3000)}^{2} + {(4000)}^{2}} = 5 \times 10^{3} Ω$

So power factor $cosφ = \frac{R}{Z} = \frac{3000}{5 \times 10^{3}} = 0 .6$ and power $P = V_{rms} i_{rms} cosφ = \frac{V_{rms}^{2} cosφ}{Z} \Rightarrow P = \frac{{(200)}^{2} \times 0 .6}{5 \times 10^{3}} = 4 .8 W$

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