A `50 mu F` capacitor, 0.05 H inductor and `48 Omega` resistor are connected in series with an a.c. source of e.m.f. `E = 310 sin 314 t`. Calculate reactance of the circuit. What is its nature ?
What is phase angle between current and applied e.m.f. ?

Explanation

Here, `C = 50 mu F = 50 xx 10^(-6) F, L = 0.5 H`
`R = 48 Omega, E = 310 sin 314 t , E_(0) = 310 V`,
`omega = 314 rad//s`
`X_(L) = omega = 314 xx 0.05 = 15.7 Omega`
`X_(C ) = (1)/(omega C) = (1)/(314 xx 50 xx 10^(-6)) = 63.7 Omega`
Reactance of the circuit `= X_(C ) - X_(L)`
`= 63.7 - 15.7 = 48 Omega`
It is capactive in nature.
If `phi` angle by which current in RLC circuit
leads the applied e.m.f., then
`tan phi = (X_(C ) = X_(L))/(R ) = (48)/(48) = 1 , phi = 45^(@)`