Find the amount by which the total energy stored in the capacitors will increase $(i n μ J)$ in the circuit shown in the figure after switch $K$ is closed? $[C = 3 μ F, V = 10 V o l t]$

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Detailed Solution

With key open circuit is

$u_{i} = \frac{1}{2} (C) V^{2}$

With key closed

$u_{f} = \frac{1}{2} (\frac{5}{3} C) V^{2} u_{f} - u_{i} = \frac{1}{2} \frac{2}{3} C V^{2} = \frac{1}{2} \times \frac{2}{3} \times 3 \times 10^{- 6} \times 10^{2} = 100 μ J$

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