91Ó°ÊÓ

We need to find that What value resistor will discharge.

The capacitor is discharged when the switch is off and the current flows through the resistor. To discharge the capacitor, the charges flow through the circuit in time $r$. The initial charge$Q_o$is related to the final charge $Q$by equation in the form

localid="1648920367106" $Q=Q_o{e}^{-t/RC}\left(1\right)$

$R$is the resistance and $C$is the capacitance. The initial charge $Q_o$is reduced to its of its $10\%$value, so the final charge is

$Q=0.10Q_o$

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

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

Putting the value for $C,tandQ=0.1Q_o$into equation $\left(2\right)$to get $R$

$$\begin{gathered} R=\frac{1}{C}\left(\frac{-t}{In(Q/Q_o)}\right) \\ =\frac{-1}{1×{10}^{-6}F}\left(\frac{-(2×{10}^{-3}s)}{In(0.1Q_o/Q_o)}\right) \\ =0.87k\mathrm{Î\copyright } \end{gathered}$$