A series LCR circuit is made by taking
`R = 100 Omega, L = (2)/(pi)` and `C = (100)/(pi) mu F`. The
series combination is connected across an a.c. source of 220 V, 50 Hz. Calculate impedance of the circuit and peak value of current flowing n the circuit. What is power factor of the circuit ?
Compare it with one at resonance frequency.

Explanation

Here, `R = 100 Omega,L = (2)/(5) H`
and `C = (100)/(pi) xx 10^(-6) F`
`E_(v) = 220 V, v = 50 Hz Z = ?, I_(0) = ?, cos phi ?`
`X_(L) = omega L = 2 pi v L = 2 pi xx 50 xx (2)/(pi) H = 200 Omega`
`X_(C ) = (1)/(omega C) = (1)/(2 pi v C)`
`= (1)/(2 pi xx 50 xx (100//pi) xx 10^(-6)) ohm = 10^(2) ohm`
`Z = sqrt(R^(2) + (X_(L) - X_(C ))^(2))`
`= sqrt(100^(2) + (200 - 100)^(2)) = 141.4 Omega`
`I_(0) = (E_(0))/(Z) = (sqrt2 E_(v))/(Z) = (1.414 xx 220)/(141.4) = 2.2 A`
Power factor, `cos phi = (R )/(Z) = (100)/(141.4) = (1)/(sqrt2)`
At resonant frequency, `cos phi = (R )/(Z) = (R )/(R ) = 1`