InterviewSolution
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InterviewSolution
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An `LC` circuit contains a `20 mH` inductor asn a `50 mu F` capacitor with initial change of `10 mC`. The resistance of the circuit is negligible. Let the instant the circuit is closed be `t = 0`. A. Energy stored in the circuit in completely electrical at `t = (n pi)/(2000)`B. Energy stored in the circuitin completely magnetic at `t = ((2n + 1)pi)/(2000)`C. Energy stored in the circuit in shared equally between the inductor and capacitor at `t = ((2n + 1)pi)/(4000)`D. Energy stored in the circuit is shared euqally between the inductor and capacitor at `t = (n pi)/(2000)` |
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Answer» Correct Answer - 1,2,3 Instantaneous electrical energy `U_(E) = (q_(0)^(2) cos^(2) omega t)/(2C)` At `omega t = 0, pi, 2 pi, 3 pi`….. The energy is completely electrical. `t = (2 pi)/(2 pi f) = (n)/(2 f) = (n pi)/(1000)` sec , `n = 1, 2, 3, 4` or `t = 0, T//2, 3T//2`.... Instantaneous magnetic energy `U_(B) = (1)/(2) Lq_(0)^(2) omega^(2) sin^(2) omega t` or `U_(B) = (q_(0)^(2))/(2C) sin^(2) omega t` So, at `omega t = pi//2, 3 pi//2, 5pi //2` The energy is completely magnetic `t = ((2n + 1) pi)/(2(2pi f)) = ((2n + 1))/(4f) = ((2n + 1))/(2000) sec` Where `n = 0,1,2,3,4`...... or `t = T//4, 3T//4, 5T//4`....... Timings for energy shared equally between inductor and capacitor. `U_(B) = U_(E)` `(q_(0)^(2))/(2C ) sin^(2) omega t = (q_(0)^(2))/(2C) cos^(2) omega t` `tan^(2) omega t = 1` or `tan omega t = tan pi//4` `t = (pi)/(4 omega), (3 pi)/(4 omega),(5 pi)/(4 omega)`,............or `t = (T)/(8), (3T)/(8), (5T)/(8)`............ |
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