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Question

A capacitor of capacitance 5μF is connected as shown in the figure. The internal resistance of the cell is 0.5Ω. The amount of charge on the capacitor plates is?
741878_05668142d3484362beeac6ef0a5819ea.png

A
80μC
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B
40μC
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C
20μC
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D
10μC
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Solution

The correct option is C 10μC
In steady state, there will be no current in the capacitor branch. Net resistance of the circuit
R=1+1+0.5=2.5Ω
Current drawn from the cell
i=VR=2.52.5=1A
Potential drop across two parallel branches
V=Eir=2.51×0.5
=2.50.5=2.0V
So, charge on the capacitor plates
q=CV=5×2=10μC.

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