A `2.5//pi` `muF` capacitor and a 3000 ohm resistance are joined in series to an a.c. source of 200 volt and `50 "sec"^(-1)` frequency. The power factor of the circuit and the power dissipated in it will respectively be-

Options

A. 0.6, 0.06W
B. 0.06, 0.6W
C. 0.6, 4.8W
D. 4.8, 0.6W

Answer

C

Explanation

The capacitive reactance is
`X_(C) = (1)/(omegaC) = (1)/(2pifC)`
`=(1)/(2pi xx 50 xx ((25)/(pi) xx 10^(-6)))=4000 ohm`.
The impedance of the circuit is
`Z = sqrt((R^(2) + X_(C)^(2)))=sqrt([(3000)^(2) + (4000)^(2)])=500 ohm`
Power factor, `cos phi = (R)/(Z) = (3000)/(5000) = 0.6`
Power dissipation, `overset(-)p = V_(rms) xx i_(rms) xx cos phi`.
`= V_(rms) xx (V_(rms))/(Z) xx cos phi`.
`= 200 xx (2000)/(5000) xx 0.6`
`= 4.8 "watt"`.