First slide

Capacitance

Question

In the network shown initially $S_{1}$ is closed and $S_{2}$ is open. Now $S_{1}$ is opened and at the same time $S_{2}$ is closed. Find the loss of energy in the process. [Given $C = 2 μ F, V = 6 volt$ ]

Moderate

A

24 $μ J$
B

18 $μ J$
C

12 $μ J$
D

36 $μ J$

Solution

Initially energy stored in the system
$U_{i} = \frac{1}{2} C V^{2}$
Finally energy stored in the system
$U_{f} = \frac{{(C V)}^{2}}{2 (C + C)} = \frac{1}{4} C V^{2}$
$∴$ Lost energy $= U_{i} - U_{f} = \frac{1}{2} C V^{2} - \frac{1}{4} C V^{2} = \frac{1}{4} C V^{2}$
$= \frac{1}{4} \times 2 \times {(6)}^{2} μ J = 18 μ J$

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