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

Options

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

C

Explanation

`Z=sqrt(R^(2)+(1/(2pivC))^(2))`
`=sqrt((3000)^(2)+1/((2pixx50xx2.5/(pi)xx10^(-6))^(2))`
`Z=sqrt((3000)^(2)+(4000)^(2))=5xx10^(3)Omega`
So power factor `cos varphi=R/Z=3000/(5xx10^(3))=0.6` and power
`P=V'_(rms)i_(rms)cos varphi=(V_(rms)^(2) cos varphi)/Z`
`implies P=((200)^(2)xx0.6)/(5xx10^(3))=4.8W`