Switch `S` is cloesd in the circuit at time `t = 0`. Find the current through the capacitor and the inductor at any time `t`

Answer

`(varepsilon)/(R_(1))e^(-((R_(1) + R_(2))/(R_(1)R_(2)))(t)/(C)`

Explanation

Circuit can be analyzed separately


i. Inductor
`varepsilon = I_(L)R_(3) + L(dI_(L))/(dt)`
On integrating, we get
`I_(L) = (E)/(R_(3)) (1 - e^((R_(2)t)/(L)))`
ii. In loop `ABCDEFA`, `varepsilon = IR_(1) + (I - I_(1)) R_(2)`,

`(varepsilon + I_(1) R_(2))/(R_(1) + R_(2)) = 1`(1)
In loop `ABGEFA`, `varepsilon = IR_(1) + (q)/(C)`
`varepsilon = ((varepsilon + I_(1) R_(2)))/(R_(1) + R_(2)) R_(1) + (q)/(C)`
Diffferentiating w.r.t. time, we get
`0 = 0 + (dI_(1))/(dt) xx ((R_(1)R_(2))/(R_(1)+ R_(2))) + (dq)/(dt) xx (1)/(C)`
`(dI_(1))/(dt) = - ((R_(1) + R_(2))/(R_(1) R_(2))) xx (1)/(C) xx I_(1)`
`(dI_(1))/(I_(1)) = - ((R_(1) + R_(2))/(R_(1) R_(2))) xx (1)/(C) xx dt`
`[In I_(1)]_(epsilon//RI)^(I1) = ((-R_(1) + R_(2))/(R_(1)R_(2))) xx ([t]_(0)^(t))/(C)`
`(:.` at `t = 0, C` acts as a conducting wire)
In`(I_(1)/(varepsilon//R_(1))) = - ((R_(1) + R_(2))/(R_(1)R_(2))) xx (t)/(C)`
`I_("through capactior") = (varepsilon)/(R_(1)) e ^(-((R_(1) + R_(2))/(R_(1) R_(2)))(t)/(C))`