The circuit shown in the figure consists of a battery of $emf$ $\varepsilon = 10 \,V$ ; a capacitor of capacitance $C = 1.0$ $ \mu F$ and three resistor of values $R_1 = 2$ $\Omega$ , $R_2 = 2$ $\Omega$ and $R_3 = 1$ $\Omega$ . Initially the capacitor is completely uncharged and the switch $S$ is open. The switch $S$ is closed at $t = 0.$
Diffcult
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At $t=,$ the equivalent circuit is
$I=\frac{\varepsilon}{R_{3}+\frac{R_{1} R_{2}}{R_{1}+R_{2}}}=\frac{10}{1+\frac{(2)(2)}{2+2}}=\frac{10}{1+1}=5 A$
Also, $I_{R_{1}} R_{1}=I_{R_{2}} R_{2} \Rightarrow \frac{I_{R_{1}}}{I_{R_{2}}}=\frac{R_{2}}{R_{1}}$
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