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

Explanation

`epsilon-iR=(q)/(2C)`
`i=(epsilon)/(R)-(q)/(2RC)=(2Cepsilon-q)/(2CR)`
`(dq)/(2 epsilonC-q)=(dt)/(2CR)`
` int_(0)^(q) (dq)/((2 epsilonC-q))=(t)/(2CR)`
`(2 epsilon C-q)/(2 epsilon C)=e^(-t//2RC)`
`q=2 epsilon C (1-e^(-t//2RC))`
`q_(1)=(q)/(2)=epsilon C (1-e^(-t//2RC)) rArr i_(1)=(epsilon)/(2R) e^(-t//2RC)`
`q_(2)=(q)/(2)=epsilon C (1-e^(-t//2RC)) rArr i_(2)=(epsilon)/(2R) e^(-t//2RC)`
Alternate solution
by equivalent
Time constant of circuit `=2CxxR=2RC`
maximum charge on capacitor `=2Cxx epsilon = 2C epsilon`
Hence equations of charge and current are as given below
`q=2 epsilon C (1-e^(-t//2RC))`
`q_(1)=(q)/(2)=epsilon C (1-e^(-t//2CR)) rArr i_(1)=(epsilon)/(2 R) e^(-t//2CR)`
`q_(2)=(q)/(2)=epsilon C (1-e^(-t//2RC)) rArr i_(2)=(epsilon)/(2R)e^(-e^(-t//2RC)`