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.

Answer

A, C

Explanation

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`