Question

An ideal capacitor of capacitance $0.2\mu F$ is charged to a potential difference of $10V.$ The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inductance $0.5mH.$ The current at a time when the potential difference across the capacitor is $5V,$ is :

• A. $0.34A$
• B. $0.25A$
• C. $0.17A$
• D. $0.15A$

Answer 1

Answer: (C)

The given circuit is

Given: capacitor is charged to a potential difference of $10V\Rightarrow V_0=10V$
Therefore, charge on capacitor, $Q_0=C{V}_0=0.2\mu F\times 10V$
Now, $E_c=$ Energy stored in capacitor $=\frac{1}{2}C{V}_0^2$
$\Rightarrow E_C=\frac{1}{2}\times 0.2\mu F\times (10V{)}^2=10\mu J=10\times 10^{-6}J$
When inductor is connected in parallel with the capacitor in the circuit, the energy stored in capacitor when potential difference across capacitor is $V^{\prime }=5K$ is
$E_C^{\prime }=\frac{1}{2}C{V}^2=\frac{1}{2}\times 0.2\mu F\times (5V{)}^2$
$=2.5\mu J=2.5\times 10^{-6}J$
Therefore, Energy stored in inductor $=E_c-E_c^{\prime }=10\times 10^{-6}J-2.5\times 10^{-6}J$
$=7.5\times 10^{-6}J$
Also, we know energy stored in inductor $=\frac{1}{2}L{I}^2$
$\Rightarrow \frac{1}{2}L{I}^2=7.5\times 10^{-6}$
$\Rightarrow I^2=\frac{7.5\times 10^{-6}\times 2}{0.5mH}=30\times 10^{-3}$
$\Rightarrow I=\frac{\sqrt{3}}{10}=0.17A$