A capacitor of 4 microF is connected as shown in the circuit shown below .The internal resistance of the battery is 0.5 ohms . The amount of chargeon the capacitor plates will be
(a) 0 C (b) 4 microC (c) 16 microC (d) 8 microC
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Dear Student ,
please find below the solution to the asked query:
once the capacitor is fully charged , the curent flow in capacitor branch stops and therefore there is no potential drop across the 10 ohm resistor, and the potential across the capacitor is sam as the potential across the 2 ohm resistor. 
when the current flow stops in the capacitor branch, the internal resistance of the battery and the 2 ohm resistance are in series with the battery emf. 
therefore,
$$\begin{gathered} If the current through \mathbf{2}\mathbf{ }\mathit{\Omega } resis\tan ce be I, \\ \mathbf{2}\mathbf{.}\mathbf{5}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{)}\mathit{I} \\ \mathit{O}\mathit{r}\mathbf{,}\mathbf{ }\mathit{I}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0} \\ so, potential across \mathbf{2}\mathit{\Omega } resis\tan ce \mathbf{=}\mathbf{ }\mathbf{2}\mathit{I}\mathbf{=}\mathbf{2}\mathit{V}\mathbf{ }\mathbf{=}potential across capacitor. \\ So, charge on capacitor =\mathbf{ }\mathbf{2}\mathbf{\times }\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{ }\mathit{C}\mathbf{=}\mathbf{8}\mathit{\mu }\mathit{C} \end{gathered}$$
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