InterviewSolution
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InterviewSolution
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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 isA. `50` ohmB. `100` ohmC. `200 ohm`D. `400` ohm |
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Answer» 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` |
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