Question

In an `LCR` circuit `R=100 ohm`. When capacitance `C` is removed, the current lags behind the voltage by `pi//3`. When inductance `L` is removed, the current leads the voltage by `pi//3`. The impedence of the circuit is
A. `50` ohm
B. `100` ohm
C. `200 ohm`
D. `400` ohm

Answer 1

Correct Answer - B
When `C` is removed circuit becomes `RL` circuit hence
`tan ((pi)/3)=(X_(L))/R.......(i)`
When `L` is removed circuit becomes `RC` circuit hence
`tan(pi/3) =(X_(C))/R.....(ii)`
from equation `(i)` and `(ii)` we obtain `X_(L)=X_(C)`. This is the condition of resonance and in resonance `Z=R=100 Omega`