When you have learned to integrate, try to derive a formula for the instantaneous current from a capacitor discharged through a resistor.

Answer

`i=i_(0) e^(-t//x)" where "tau=RC`

Explanation

For this circuit the parameters present in Ohm.s law `triangle Psi =I R`
take the form
`triangle Psi=q/C, i=underset(triangle t to 0)lim (-(triangle q)/(triangle t))=-(dq)/(dt)`
The minus sign is due to the decrease in the capacitor.s charge in the process of its discharge. Substituting into the expression for Ohm.s law, we obtain.
`q/(RC)=-(dq)/(dt), or (dt)/(RC)=- (dq)/q`
Integrating, we obtain
`t/(RC)=-ln q+ ln A`
where A is a constant. Nothing that for t=0m `q=q_(0)`, we obtain `0=-ln q_(0)` + ln A, from which `A=q_0`. Hence
`t/(RC)=-ln q+ln q_(0)`
Denoting the time constant (the relaxation time) `tau=RC,` we obtain
`ln q/q_(0)=- t/tau," giving "q=q_(0) e^(-t//tau)`
For the current we have `l=- (dq)/(dt)=(q_(0))/(tau) e^(-t//pi)=l_(0) e^(-t//tau)" where "i_(0)=(q_(0))/(tau)=(q_(0))/(RC)=U_(0)/R`