In the circuit shown in fin. Calculate the capacitance C of the capacitor, if the power factor of the circuit is unity. Also, calculate Q factor of the circuit.

Explanation

Given P.F `cos varphi= 1 rArr varphi = 0^(@)`
`rArr ` Circuit has only and inductor and capacitor cancels effect of each other
Using relation `omega_(L)=1/(omegaC)`
we have `100 xx 200 xx 100^(-3) = 1/(100xxc) rArr1/c = 2 xx 10^(6) xx 10^(-3)`
`rArr c= 1/(2 xx 10^(6)xx 10^(-3))=1/(2 xx10^(3))=5 xx 10^(-4) F = 500 xx 10^(-6)F = 500 mu F`
Q factor=` (omega0L)/R=( cancel(100) xx 200 xx 10^(-3))/(cancel (10))=2`
Q factor basically gives selectivity at the resonance circuit.
It signifies that current level increases to maximum. It is also the ratio of apparent power of ac circuit to that of power loss in ac circuit.
`P_(v)=E_(v) xx I_(v) = 50 xx (E_(v))/R = cancel (50) xx 50/cancel (10)= 250 W`