Without using the formula of equivalent. Find out charge on capacitor and current in all the branches as a function of time.

Explanation

Applying KVL in ABDEA
` in - iR =(q)/(2C) " "rArr " "I =(in )/® -(q)/( 2CR) =( 2Cin-q)/( 2CR) `
` (dq)/( 2in C-q)= ( dt)/( 2CR) rArr " " int _0^( q) ( dq)/( ( 2in C- q)=( t)/( 2CR) `
` ( 2in C-q)/( 2in C) = e ^(-t//RC) rArr " " q=2in C( 1-e^(-t//2RC)) `
`q_1 =(q)/(2) =in C(1-e^(-t//RC) ) " "rArr " " i_ 1 = ( in )/(2R) e^(-t//2RC)`
` q_2 =(q)/(2)= in C( 1- e^(-t//2RC) ) " "rArr " " i_2 =(in )/(2R) e^(-t//RC)`
Alternate solution
By equivalent
Time constant of circuit `=2C xx R = 2RC `
maximum charge on capacitor `= 2C xx in 2Cin `
Hence equations of charges and current are as given below
` q= 2in C( 1-e^(-t//2RC) `
`q_1 =(q)/(2) = in C( 1- e^(-t//2RC) `
` rArr " " i_1 = (in )/( 2R) e^(-t//2RC) `
` q_2 -(q)/(2) =in C ( 1-e^(-t//2RC) ) `
` rArr " "i_ 2 =(in )/(2R) e^(-t//RC) `
` (##MOT_CON_JEE_PHY_C25_SLV_039_S01.png" width="80%">