First slide

Capacitance

Question

A 2 $μ$ F capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch S is turned to position 2 is :

Moderate

A

0 %
B

20 %
C

75 %
D

80 %

Solution

$U_{i} = \frac{1}{2} {CV}^{2} = \frac{1}{2} \times 2 \times V^{2} = V^{2} q_{i} = 2 V$
When switch S is turned to position 2
$\frac{2 V - q}{2} = \frac{q}{8} \Rightarrow q = \frac{8 V}{5} ∴ Energy dissipated = V^{2} - (\frac{64 V^{2}}{2 \times 25 \times 8} + \frac{4 V^{2}}{2 \times 25 \times 2}) = \frac{4 V^{2}}{5}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App