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Figure shows a battery with emf 15 V in a circuit with $R_{1} = 30 Ω, R_{2} = 10 Ω, R_{3} = 20 Ω$ and capacitance $C = 10 μ F .$ The switch S is initially in the open position and is then closed at time t=0. What will be the final steady-state charge on capacitor?

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$75 μ C$

$10 μ C$

none of these

$50 μ C$

answer is A.

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Detailed Solution

$i = \frac{15}{60} = \frac{1}{4} A$

$V_{A B} = \frac{1}{4} \times 30 = 7.5 V$

$q = C V = 10 \times 7.5 = 75 μ C$

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