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In the circuit shown in figure the capacitor is charged with a cell of $5 V$ . If the switch is closed at t = 0, then at $t = 12 s$ , charge on the capacitor is

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$(0.37) 10 μ C$

${(0.37)}^{2} 10 μ C$

$(0.63) 10 μ C$

${(0.63)}^{2} 10 μ C$

answer is B.

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Detailed Solution

$\begin{array}{l} q = q_{0} e^{- \frac{t}{C R}} \\ q = (10 μ C) e^{\frac{- 12}{6}} \\ q = (10 μ C) \cdot \frac{1}{e^{2}} = (0.37)^{2} (10 μ C) \end{array}$

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