Two capacitors of capacitances `2C` and `C` are connected in series with an inductor of inductance `L`. Initially, capacitors have charge such that `V_B - V_A = 4V_0` and `V_C- V_D = V_0`. Initial current in the circuit is zero. Find

(a) maximum current that will flow in the circuit,
(b) potential difference across each capacitor at that instant,
(c) equation of current flowing towards left in the inductor.

Answer

A, B, C

Explanation

`q_1=8CV_0:q_2=CV_0`
`q_1+q_2=9CV_0`
In the absence of inductor this `9C_0V` will distributed `6CV_0` in `2c` and `3CV_0` in `C`. Thus, mean position of `q_1` is `6CV_0` and mean position of `q_2` is `CV_0`

At `t=0`, `q_1` is `2CV_0` more than its mean position and `q_2` is `2CV_0` less
Thus `q_0=2CV_0`
`C_("net")=(2C)/3`
`:. omega=(1)/(sqrt(LC_("net")) =sqrt(3/(2LC))`
a. `I_(max)=q_0omega`
`V_1=(6CV_0)/(2C)=3V_0`
and `V_2=(3CV_0)/C`
`=3V_0`
`c. i=q_0sinomegat`