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

The charges on both capacitors are shown at `t=0 sec`.and any time `t` later in figure `a` and `b`.Applying `KVL` to the loop at any time when current `i` flows through it.
`L(di)/(dt)+(9CV_(0)-Q)/(2C)-Q/(2C)=0`
for cuurent to be maximum `(di)/(dt)=0` or `Q=6 CV_(0)`
Total energy in the system at `t=0` is `U=1/2L(0)^(2)+1/2CV_(0)^(2)=(-Q)/(2C)+(9CV_(0)-Q)/C=0` or `Q=6 CV_(0)`
Total energy in the system at `t=0` is `U_(i)=1/2L(0)^(2)+1/2(2C)(4V_(0))^(2)+1/2CV_(0)^(2)=33/2CV_(0)^(2)`
Total energy when current `i` is maximum `U_(1)=1/2L i_(max)2+1/2(6CV_(0))^(2)/(2C)+1/2(3CV_(0))^(2)/C=1/2L i_(max)2+27/2(6CV_(0))^(2)`
from conservation of energy `33/2CV_(0)^(2)=1/2Li_(max)2+27/2 CV_(0)^(2), i_(max)=sqrt((6CV_(0)^(2))/L)`