InterviewSolution
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InterviewSolution
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An `L-C` circuit consists of an inductor coil withh `L=50.0mH` and `20.0muF` capacitor. There is negligible resistance in the circuit. The circuit is driven by a voltage source with `V=V_0cosomegat`. If `V_0=5.00mV` and the frequency is twice the resonance frequency, determine. a. the maximum charge on the capacitor. b. the maximum current in the circuit. c. the phase relationship between the voltages across the inductor, the capacitor and the sourse. |
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Answer» Correct Answer - A::B::C::D `omega=2/(sqrtLC)=2/(sqrt(5xx10^-3xx20xx10^-6))` `632.4.5rad//s` `X_L=omegaL=(6324.5)(5xx10^-3)=31.62Omega` `X_C=1/(omegaL)=1/(6324.5xx20xx10^-6)=7.9Omega` `:. Z=X_L-X_C=23.72Omega` (a) Maximum voltage across capacitor `=i_0X_C=(0.211)(7.9)=1.67mV` `:.` Maximum charge `q_0=(20xx10^-6)(1.67xx10^-3)=33.4nC` (b) `i_0=V_0/Z=5/23.72mA=0.211mA` (c) Since `X_LgtX_C`, current in the circuit will lag behind the applied voltage by `pi//2`. Further voltage across the inductor will lead this current by `pi//2`. Therefore, applied voltage and voltage across inductor are in phase. Voltage across the capacitor will lag the circuit current by `pi//2`. Therefore, phase difference between `V_L` and `V_C` will be `180^@`. |
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