In the circuit shown in figure $C_1=2C_2$. Switch $S$ is closed at time $t=0$. Let $i_1$ and $i_2$ be the currents flowing through $C_1$ and $C_2$ at any time $t$, then the ratio $i_1/ i_2$
Diffcult
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Here $I_{1}=\frac{V}{R} e^{\frac{-t}{R C_{1}}}$
and $\quad \mathrm{I}_{2}=\frac{\mathrm{V}}{\mathrm{R}} \mathrm{e}^{\frac{-\mathrm{t}}{\mathrm{RC}_{2}}}$
$\therefore \quad \frac{I_{1}}{I_{2}}=\frac{e^{\frac{-t}{R C_{1}}}}{e^{\frac{-t}{RC_{2}}}}=e^{\frac{-t}{R}\left(\frac{1}{C_{1}}+\frac{1}{C_{2}}\right)}$
$\therefore \quad \frac{I_{1}}{I_{2}}=e^{\frac{-t\left(C_{1} + C_{2}\right)}{R C_{1} C_{2}}}$
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