First slide

Inductance

Question

A circuit containing capacitors $C_{1}$ and $C_{2}$ as shown in the figure are in steady state with key $K_{1}$ closed and $K_{2}$ is opened. At the instant t = 0, if $K_{1}$ is opened and $K_{2}$ is closed then the maximum current in the circuit will be :

Moderate

A

1 A
B

$\frac{1}{2} A$
C

2A
D

$\frac{3}{2} A$

Solution

When switch $K_{1}$ is opened and $K_{2}$ is closed it becomes
L–C circuit so applying energy
conservation :
$V o l t a g e a c r o s s 2 μ f = 10 V$
$∴ \frac{1}{2} C V^{2} = \frac{1}{2} L i^{2} [i = \max c u r r e n t a c r o s s i n d u c t o r]$
$= 2 \times 10^{- 6} \times 10^{2} = 0.2 \times 10^{- 3} \times i^{2}$
$\frac{2}{10} = 0.2 i^{2}$
$i = 1 A$

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