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A circuit containing capacitors `C_1` and `C_2`, shown in the figure is in the steady state with key `K_1` and `K_2` opened. At the instant `t = 0`, `K_1` is opened and `K_2` is closed. (a) Find the angular frequency of oscillations of `L- C` circuit. (b) Determine the first instant `t`, when energy in the inductor becomes one third of that in the capacitor. (c) Calculate the charge on the plates of the capacitor at that instant. |
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Answer» Correct Answer - A::C With key `K_1` closed `C_1` and `C_2` are in series with the battery in steady state. `:. C_("net")=1muF` or `q_0=C_("net")V=20muC` a. With `K_1` opened and `K_2` closed, charge on `C_2` will remain as it is while charge on `C_1` will oscillate in `L-C_1` circuit. `omega=1/(sqrt(LC_1))` `=1/sqrt(0.2xx10^-3xx2xx10^-6)` `=5xx10^4rad//s` b. Since at `t=0`, charge is maximum `(=q_0)` Therefore, current will be zero. `1/2Li^2=1/3(1/2q^2/C)` or `i=q/sqrt(3CL)=(qomega)/sqrt3` From the expression `i=omegasqrt(q_0^2-q^2)` We have `(qomega)/sqrt3=omegasqrt(q_0^2-q^2)` or `q=sqrt3/2q_0` Since at `t=0` charge is maximum or `q_0,` so we can write `q=q_0cosomegat` or `(sqrt3q_0)/2=q_0cosomegat` or `omegat=pi/6` or `t=pi/(6omega)=pi/(6xx5xx10^4)` `=1.05xx10^-5s` c. `q=sqrt3/2q_0=sqrt3/2=xx20=10sqrt3muC` |
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