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. 2A
B. 1A
C. `(sqrt(3))/(2) A`
D. `(2)/(sqrt(3))A`

Answer

A

Explanation

As `tan phi = (X_(L) - X_(C))/(R)`
When capacitance is removed,
`tan 60^(@) = (X_(L))/(R) rArr X_(L) = sqrt(3) R `
When inductance is removed
`tan 60^(@) = (X_(C))/(R) rArr X_(C) = sqrt(2) R `
Since ` X_(L) = X_(C) ` (Resonance) `rArr phi = 0^(@)`
`Z_("min") = R rArr I_(rms)= (V_(rms))/(Z) = (200)/(100) = 2A `