In the circuit shown here, at the steady state, the charge on the capacitor is $C (\frac{x E}{R_{1} + R_{3}}) R_{3}$ . Find $x$

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Solution

As the capacitor C is fully charged in the steady state so no current will pass through the resistor $R_{2}$ .
Thus the current in the circuit is $I = \frac{E}{R_{1} + R_{2}}$
The potential across $R_{3}$ is equal to the potential across C. i.e
$V = I R_{3} = \frac{E}{R_{1} + R_{2}} R_{3}$
Thus charge on the capacitor $Q = C V = C (\frac{E}{R_{1} + R_{2}}) R_{3}$
thus $x = 1$

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