Question

A capacitor of 4 μF is connected as shown in the circuit Figure. The internal resistance of the battery is 0.5 Ω. The amount of charge on the capacitor plates will be :
Question-Image

(a) 0 μC
(b) 4 μC
(c) 16 μC
(d) 8 μC
(d) Explanation : As capacitor offer infinite resistance for DC circuit. So current from cell will not flow across branch of 4 μF and 10 Ω. So current will flow across 2 ohm branch.
So current flows through across 2 Ω resistance from left to right is,
Answer-Image

So Potential Difference (PD) across 2 Ω resistance V = RI = 2×1 = 2 Volt
As battery, capacitor and 2 branches are in parallel. So PD will remain same across all three branches. As current does not flow through capacitor branch so no potential drop will be across 10 Ω.
So PD across 4 F capacitor = 2 Volt
Q = CV = 2 μF× 2 V = 8 μC

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