91Ó°ÊÓ

We need to find the value of capacitor.

When the switch is off, the capacitor-discharge, and the current flows through the resistor. To discharge the capacitor, the charges flow through the circuit in time $r$. The initial current $I_o$is related to the final charge $I$by equation in the form

localid="1648921959204" $I=I_o{e}^{-t/RC}\left(1\right)$

Where $R$ is the resistance and $C$is the capacitance. The initial current $I_o$is reduced to its $25\%$value, so the final current is

$I=0.25I_o$

After solving the equation $\left(1\right)$for $C$to be in the form

$$\begin{gathered} Q=Q_o{e}^{-t/RC} \\ In\left(\frac{Q}{Q_o}\right)=\frac{-t}{RC} \left(2\right) \\ C=\frac{1}{R}\left(\frac{-t}{In(Q/Q_o)}\right) \end{gathered}$$

Putting the values localid="1648922115145" $C,tandI=0.25I_{}$into equation $\left(2\right)$to get $R$

localid="1648922333327" $$\begin{gathered} C=\frac{1}{R}\left(\frac{-t}{In(I/I_o)}\right) \\ =\frac{1}{1000\mathrm{Î\copyright }}\left(\frac{-(2.5×{10}^{-3}s)}{In(0.25I_o/I_o)}\right) \\ =18×{10}^{-6}F \\ =18\mathrm{μ}F \end{gathered}$$