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InterviewSolution
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An L-C circuit contains 20 mH inductor and a `50 muF` capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t = 0. what is the total energy stored initially ? At what times is the total energy shared equally between the inductor and the capacitor ? |
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Answer» Given Inductance `L = 20 mH = 20xx10^(-3)H` Capacitance of capacitor `C=50 mu f=50xx10^(-6)F` Initial charge on the capacitor, `Q_(i)=10mc =10xx10^(3)C` Equal sharing of energy between inductor and capacitor means the energy stored in capacitor `= (1)/(2)xx` Maximum energy `(Q^(2))/(2C)=(1)/(2)xx(Q_(0)^(2))/(2C) " " Q=(Q_(0))/(sqrt(2))` .....(ii) From `Q=Q_(0)cos omega t=Q_(0)cos.(2pi)/(T)t` `(Q_(0))/(sqrt(2))=Q_(0).cos.(2pi)/(T)t` `(1)/(sqrt(2))=cos.(2pi t)/(T)` or `cos(2n+1)pi//4 = cos.(2pi t)/(T)` `((2n+1)pi)/(4)=(2pi t)/(T)` `t=T//8(2n+1) (n=0,1,2,3,....)` Hence the energy will be shared half on capacitor and half on inductor. `t=(T)/(8), (3T)/(8), (5T)/(8)`, ....... |
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