Question

In the circuit shown below, the switch is connected to position $1$ for a very long time. What is the maximum current flowing through the circuit, when the switch is taken to position $2$?

• A. $1.5\text{A}$
• B. $2.5\text{A}$
• C. $3.5\text{A}$
• D. $4.5\text{A}$

Answer 1

The correct option is A $1.5\text{A}$


When the switch is connected to position $1$ for a very long time, the potential difference across capacitor will become $6\text{V}$.

So, total energy stored in the capacitor,

$U_T=\frac{1}{2}C{V}^2=\frac{1}{2}\times 50\times {10}^{-6}\times 6^2=9\times {10}^{-4}\text{J}$.

For the current in the circuit to be maximum, all the energy must be stored in the inductor and the inductor starts charging the capacitor.

$\Rightarrow \frac{1}{2}L{i}_{\text{max}}^2=9\times {10}^{-4}$

$\Rightarrow \frac{1}{2}\times 0.8\times {10}^{-3}\times i_{\text{max}}^2=9\times {10}^{-4}$

$\Rightarrow i_{\text{max}}=1.5\text{A}$

Hence, option $\left(A\right)$ is the correct answer.