In the R–C circuit shown in the figure the total energy of 3.6 ×10^–3 J is dissipated in the 10 ohms resistor when the switch S is closed. The initial charge on the capacitor is

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$\frac{60}{\sqrt{2}} μC$

60 µC

120 µC

$60 \sqrt{2} μC$

answer is B.

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Detailed Solution

Maximum energy dissipated by the capacitor
$\begin{array}{l} U = \frac{q_{0}^{2}}{2 C} = 3 .6 \times 10^{- 3} \\ \Rightarrow q_{0}^{2} = 2 \times 2 \times 10^{- 6} \times 36 \times 10^{- 4} \Rightarrow q_{0} = 120 μC \end{array}$

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