In the figure shown below, find out the current as function of time and charge on capacitor `C_1 `and also plot the graph of charge on plate A and B of capacitor `C_2` as a function of time.

Explanation

Let at any time charge on ` C_1 =q ` Now apply , `K.V.L`
` -iR -(q)/( c) -((q-2in c))/( C) +in =0 `
` 3in =( 2q)/( c)+ iR " "rArr " " 3in C = 2q + iRC `
` (##MOT_CON_JEE_PHY_C25_SLV_038_S01.png" width="80%">
` 3in C = 2q + iRC rArr " "int _( 0)^(q) ( dq)/( 3in C - 2q) =int _(0 )^(t) ( dt)/( RC) `
` ( - l n ( 3in C- 2q))/( 2) |_0^(q) =(t)/(RC) `
` rArr " " ln ((3in C -2q)/( 3in C) ) = ( -2t)/( RC) `
` 3in C -2q= 3in C e^(-2t//RC) `
`rArr " " q= (3in C ) /( 2 ) (1-e^(2t//RC)) `
` i= (dq)/( dt) =(3in )/(R) e ^(-2t//RC) ` At plate A ,
Charge ` q_A=q - 2in C = (2)/(3) in C (1-e^(2t//RC) ) -2in C `
` (##MOT_CON_JEE_PHY_C25_SLV_038_S02.png" width="80%">
` =- ( inc)/( 2) - (2in c) /(2) e^(-2t//RC) `
`q_A=-( in c)/( 2) ( 1+3e^(-2t//RC) ) `
` q_B =(in C)/(2) (1+3e^(-2t//RC)) `