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2.5π μFcapacitor and 3000 ohm resistor 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 be

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a

0.6, 0.06 W

b

0.06, 0.6 W

c

0.6, 4.8 W

d

4.8, 0.6 W

answer is C.

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Detailed Solution

Z=R2+12πvC2=(3000)2+12π×50×2.5π×10−62Z=(3000)2+(4000)2=5×103 ΩSo, power factor, cosϕ=RZ=30005×103=0.6Now, P=Vrmsirmscosϕ=Vrms2cosϕZ⇒P=(200)2×0.65×103=4.8 W

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