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

`I_(0) = V_(0) sqrt((6C)/(L))`; b. `V_(1) = 3V_(0), V_(2) = 3V_(0)`; c. `V_(0) sqrt((6C)/(L)) sin sqrt((3)/(2CL))t`

Explanation

a. Applying loop equation in Fig (b)
`(q'_(1))/(2C) - (q'_(2))/(C ) - L(dI)/(dt) = 0`
or `L (dI)/(dt) = ((q_(1) - 2q_(2))/(2C)) - (3q)/(2C)` (i)

Differeentiating (i), we get
`L(d^(2) I)/(dt^(2)) = - (3)/(2C) I` (ii)
Solution of (ii) is `I = I_(0) sin (omega t +phi)`
where `omega^(2) = (3)/(2CL) rArr omega = sqrt((3)/(2CL))`
but initial current is zero, i.e., at `t =0, i = 0`, putting this above we get `I = I_(0) sin omega t` (iii)
`rArr (dI)/(dt) = I_(0) omega cos omega t`
putting `dI//dt` in Eq.(i), `LI_(0) omega cos omegat = ((q_(1) - 2q_(2))/(2C)) - (3q)/(2C)` (iv)
At `t = 0, q= 0`, putting this in the above equation, we get
`LI_(0) omega = ((q_(1) - 2q_(2))/(2C)) rArr LI_(0) sqrt((3)/(2CL)) = ((8 CV_(0) - 2CV_(0))/(2C))`
`rArr I_(0) = V_(0) sqrt((6C)/(L)) rarr` the maximum value of current
b. From (iii) maximum current occurs at `omega t = pi//2, 3pi//2`, ...etc,
and at these instants from (i) we get `q = ((q_(1) - 2q_(2))/(3))`
Potential difference across each capacitor at this instant,
`V_(1) = (q'_(1))/(2C) = (1)/(2C) [q_(1) -((q_(1) - 2q_(2))/(3))]`
`V_(1) = ((q_(1) + q_(2)))/(3C) = 3V_(0)`
`V_(2) = (q'_(2))/(C) = (1)/(C) [q_(2) -((q_(1) - 2q_(2))/(3))]`
`V_(2) = ((q_(1) + q_(2)))/(3C) = 3V_(0)`
c. Equation of current:
`I = I_(0) sin omegat = V_(0) sqrt((6C)/(L)) sin sqrt((3)/(2CL)) t`