Question

In circuit shown in the figure charge on the capacitor is $Q=100\mu \text{C}$. If the switch is closed at $t=0$, the charge on the capacitor reduces to $50\mu \text{C}$ and the current in the circuit is $\frac{n}{10}\text{A}$. The value of $n$ is (Integer only)

Answer 1

Let the current flowing in the circuit is $i$.

From energy conservation principle,

$\frac{q^2}{2C}+\frac{1}{2}L{i}^2=\frac{Q^2}{2C}$

$\Rightarrow i=\frac{\sqrt{Q^2-q^2}}{\sqrt{LC}}$

Substituting the values,

$i=\frac{\sqrt{(100\times {10}^{-6}{)}^2-(50\times {10}^{-6}{)}^2}}{\sqrt{6\times {10}^{-3}\times 5\times {10}^{-6}}}$

$=\frac{{10}^{-6}}{{10}^{-4}}\times \frac{\sqrt{7500}}{\sqrt{3}}$

$={10}^{-2}\times 50=\frac{5}{10}\text{A}$

Given, $i=\frac{n}{10}$

$\therefore n=5$