Question

A circuit containing capacitors $C_1$ and $C_2$ as shown in figure are in steady state with key $k_1$ closed at the instant t=0, if $k_1$ is opened and $K_2$ is closed then the maximum current in the circuit will be:

• A. $1$ A
• B. 5A
• C. $2$ A
• D. None of these

Answer 1

The correct option is A $1$ A

$\begin{array}{c}Fromthefigure\\ voltageacrosscapacitor2\mu F\left(c_1\right)atsteadystate=10v\\ where{k}_{2}isclosed\\ Maxcurrentisobtainewhentotaleneegyof{c}_{1}istransferedtotheinductor\\ Hence,\\ \frac{1}{2}{I}^{2}L=\frac{1}{2}c{v}^2\\ I=\sqrt{\frac{c{v}^2}{2}}\\ =\sqrt{\frac{c}{L}}v\\ =w\sqrt{\frac{2\times {10}^{-5}}{0.2\times {10}^{-3}}}\\ =10\sqrt{10}\times {10}^{-3}\\ ={10}^{-1‘}\times 10=1A\end{array}$

hence, the option $A$ is the correct answer.