A circuit containing capacitors `C_(1)` and `C_(2)` shown in the figure is in the steady state with key `K_(1)` cloesed. At the instant `t=0` , `K_(1)` is opened and `K_(2)` is closed.
(a) Find the angular frequency of oscillation of the `LC` circuit.
(b) Determine the first instant `t` , when enregy in the indctor becomes one third of that in the capacitor.
(c) Calculate the change on the plates of the capacitor at that instant.

Explanation

When `K_(1)` is close, `C_(eq)=(C_(1)C_(2))/(C_(1)+c_(2))=(3xx6)/(3+6)=2muF`
Change on capacitor `q_(0)=C_(eq)V=2xx20=40muC`
(a) `omega=(1)/(sqrt(LC_(1)))=(1)/(sqrt(1.2xx10^(-3)xx3xx10^(-6)))=(10^(5))/(6)rad//sec`
(b) Change on capacitor
`q=q_(0)cosomegat`
`U_(L)+U_(C)=(q_(0)^(2))/(2C_(1))`
`U_(L)=(1)/(3)U_(C)`
`(4)/(3).(q^(2))/(2C_(1))=(q_(0)^(2))/(2C_(1))` implies `q=(sqrt(3))/(2)q_(0)`
`q=q_(0)cosomegat`
`(sqrt(3))/(2)q_(0)=q_(0)cosomegat` implies `cosomegat=(sqrt(3))/(2)=cos(pi)/(6)`
`omegat=(pi)/(6)` implies `t=(pi)/(6omega)=(pixx6)/(6xx10^(%))=10^(-5)pisec`
`=10pimusec=31.4musec`
(c) `q=(sqrt(3))/(2)q_(0)=(sqrt(3))/(2)xx40=20sqrt(3)muC`