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Answer :

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

For this circuit the parameters present in Ohm.s law `triangle Psi =I R` <br> take the form <br> `triangle Psi=q/C, i=underset(triangle t to 0)lim (-(triangle q)/(triangle t))=-(dq)/(dt)` <br> 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. <br> `q/(RC)=-(dq)/(dt), or (dt)/(RC)=- (dq)/q` <br> Integrating, we obtain <br> `t/(RC)=-ln q+ ln A` <br> 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 <br> `t/(RC)=-ln q+ln q_(0)` <br> Denoting the time constant (the relaxation time) `tau=RC,` we obtain <br> `ln q/q_(0)=- t/tau," giving "q=q_(0) e^(-t//tau)` <br> 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`