In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance $C$ will be :

C E \frac{r_{1}}{(r_{1} + r)}

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C E

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C E \frac{r_{1}}{(r_{2} + r)}

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C E \frac{r_{2}}{(r + r_{2})}

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Solution

The correct option is D $C E \frac{r_{2}}{(r + r_{2})}$
At steady state, no current will flow through capacitor

Current through the battery.

$i = \frac{E}{r + r_{2}}$

Potential diff. across $P_{2}$

$V = i r_{2}$

$V = \frac{E r_{2}}{r + r_{2}}$

Thus charge on capacitor $Q = C V$

$Q = C E \frac{r_{2}}{r + r_{2}}$

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C

E = 5 V, r = 1 W, R_{2} = 4 W, R_{1} = R_{3} = 1 W

C = 3 m F

R_{1} = 1 Ω, R_{2} = 2 Ω,, C_{1} = 1 μ F, C_{2} = 2 μ F

E = 6 V

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