Question

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

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

Answer 1

Answer: (C)

The given circuit is

Given: capacitor is charged to a potential difference of $10\, V \Rightarrow V_{0}=10\, V$
Therefore, charge on capacitor, $Q_{0}=C V_{0}=0.2\, \mu F \times 10\, V$
Now, $E_{c}=$ Energy stored in capacitor $=\frac{1}{2} C V_{0}^{2}$
$\Rightarrow E_{C}=\frac{1}{2} \times 0.2 \mu F \times(10 V )^{2}=10 \mu J =10 \times 10^{-6}\, J$
When inductor is connected in parallel with the capacitor in the circuit, the energy stored in capacitor when potential difference across capacitor is $V'=5 K$ is
$E_{C}'=\frac{1}{2} C V^{2}=\frac{1}{2} \times 0.2 \mu F \times(5 V )^{2}$
$=2.5 \mu\, J =2.5 \times 10^{-6}\, J$
Therefore, Energy stored in inductor $=E_{c}-E_{c}'=10 \times 10^{-6} J -2.5 \times 10^{-6}\, J$
$=7.5 \times 10^{-6}\, J$
Also, we know energy stored in inductor $=\frac{1}{2} L I^{2}$
$\Rightarrow \frac{1}{2} L I^{2}=7.5 \times 10^{-6}$
$\Rightarrow I^{2}=\frac{7.5 \times 10^{-6} \times 2}{0.5\, mH }=30 \times 10^{-3}$
$\Rightarrow I=\frac{\sqrt{3}}{10}=0.17\, A$