An inductor 200 mH, capactior `500 mu F`, resistor 10 ohm are connected in series with a 100 V, varialbe frequency a.c. source. Calculate (i) frequency at which power factor of the circuit is unity (ii) current amplitude at this frequency (iii) Q factor.

Explanation

Here, `L = 200 mH = 0.2 H, C = 500 xx 10^(-6) F`,
`R = 10 Omega`,
`E_(v) = 100 V`
`X_(L) = X_(C )` or `omega L (1)/(omega C)` or `omega = (1)/(sqrt(LC))`
`v = (1)/(2 pi sqrt(LC)) = (1)/(2 pi sqrt(0.2 xx 500 xx 10^(-6))) = (50)/(pi) Hz`
`I_(0) = (E_(0))/(Z) = (sqrt2 E_(v))/(R ) = (1.414 xx 100)/(10) = 14.14 A`
Q factor ` = (1)/(R ) sqrt((L)/(C )) = (1)/(10) sqrt((0.2)/(5 xx 10^(-4))) = 2`