The switch S is closed aty t = 0. the capacitor C is uncharged but `C_0` has a charge `q_0` at t =0. Calculate the current i(t) in the circuit

Explanation

Let `q_(0)` and q be the instantaneous charges on `C_(0)` and C, respectively. Applying KVL to the circuit, we have
`(q_(0))/(C_(0)) +(q)/(C ) +iR = epsilon`
Differentiating this equation, we get
`(1)/(C_(0))(dq_(0))/(dt) +(1)/(C ) (dq)/(dt) +R (di)/(dt) = 0`
or `i[(1)/(C_(0) +(1)/(C ) ] = R(di)/(dt) or i[(C +C_(0))/(C C_(0))] = -R (di)/(dt)`
`("where" i = (dq_(0))/(dt)) = (dq)/(dt)`
or Intergrating this expression, we have
`int_(i_0)^(i(t)) (dt)/(i) = int_0^(t) (dt)/(RC_(eq)) or [log_ei]_(i_0)^(i(t)) = -(t)/(RC_(eq))`
or `i(t)=i_(0)e^(-(t//R)C_(eq))`
where `i_(0)` is the inital current.
Further, `i_(0)R +(q_(0))/(C_(0)) = epsilon or i_(0) = (E-(eq_(0))/(C_(0)))//R` (iii)
Substituting `i_(0)` form Eq. (iii) into Eq. (ii)we get
`i(t) = [(E-(q_(0)//C_(0)))/(R )]e^(-((t)/(RC_(eq)))),` where `C_(eq)=(C C_(0))/((C+C_(0)))`