An LCR series circuit with `R = 100 Omega` is connected to a 300V, 50Hz a.c. source. If the capacitance is removed from the circuit then the current lags behind the voltage by `30^(@)`. But if the inductance is removed form the circuit the current leads the voltage by `30^(@)`. What is the current in the circuit?

Options

A. 2A
B. 3A
C. 1.5A
D. 4.5A

Answer

B

Explanation

If C is removed, then the remaining circuit becomes an L-R circuit and the phase angle is given by
`tan phi = (X_L)/(R) = (omega L)/(R)`
Similarly if the capacitative reactance `X_(C) = 1/(omega C)`
Since both L and C are producing the same change in the phase angle, `:. omega L = 1/(omega C)`.
`:. I_(rms) = (E_(rms))/(R) = 300/100 = 3A`.