A series L-C-R circuit with a resistance of `500 Omega` is connected to an a.c. source of 250 V. When only the capacitance is removed, the current lags behind the voltage by `60^(@)`. When only the inductance is removed, the current leads the voltage by `60^(@)`. What is the impedance of the circuit?

Options

A. `250 Omega`
B. `500 Omega`
C. `500sqrt(3) Omega`
D. `500/(sqrt(3)) Omega`

Answer

B

Explanation

Given `R = 500 Omega, V = 250 V`
(i) When only the capacitor is removed, the circuit consists of L and R in series and the phase difference between I and V is given by `tan phi_(1) = (X_L)/(R)`
`:. Tan 60^(@) = (X_L)/R :. X_(L) = R tan 60^(@) = sqrt(3) R`
Thus, `X_(L) = X_(C)`. Hence the given circuit is an L-C-R series resonance circuit. For this circuit,
Impedance = `Z =R = 500 Omega`.