A circuit consists of a source of a constant `emf xi` and a resistance `R` amd a capacitor with capacitance `C` connected in series. The internal resistance of the source is negligible. At a moment `t = 0` the capacitance of the capacitor is abruply decreased `eta`-fold. FInd the current flowing through the circuit as a function of time `t`.

Explanation

Suppose `q` is the charge at the time `t`, Initially `q = C xi`, at `t = 0`.
Then at time `t`.
`(eta q)/(C) - i R - xi = 0`
But, `i = - (dq)/(dt)` (-sign because charge decreases)
So `(eta q)/(C) + R (dq)/(dt) = xi`
`(dq)/(dt) + (eta)/(R C) q = (xi)/(R)`
or, `(d)/(dt) q e^(t eta//R C) = (xi)/(R) e^(t eta//R C)`
`q = (C xi)/(eta) + A e^(-t eta//R C)`
`A = C xi (1 - (1)/(eta))`, from `q = C xi` at `t = 0`
Hence, `q = C xi ((1)/(eta) + (1 - (1)/(eta)) e^(-t eta//R C))`
Finally, `i = - (dq)/(dt) = (xi (eta - 1))/(R) e^(-t eta//R C)`