First slide

Inductance

Question

In the circuit, 1&2 are joined for long time. After disconnecting 1&2, 2&3 are connected. The charge on the capacitor when the current in the inductor (ideal) decreases by 3A is [nearly]

Easy

A

$11 m C$
B

$12 m C$
C

$13 m C$
D

$14 m C$

Solution

When 1&2 are closed for long time $i_{m a x} = \frac{5}{1} = 5 a m p$ through inductor .
When 2&3 are joined, by law of conservation of energy $\frac{1}{2} L i_{\max}^{2} = \frac{1}{2} L i^{2} + \frac{q^{2}}{2 C}$
$\frac{1}{2} (2 \times 10^{- 3}) \times 25 = \frac{1}{2} (2 \times 10^{- 3}) \times 4 + \frac{q^{2}}{2 \times 4 \times 10^{- 3}}$ $q = \sqrt{168} \times 10^{- 3} \approx 13 \times 10^{- 3} C$

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