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

The Correct Option is C

To solve this problem, we need to understand the dynamics of an LC circuit (an inductor and capacitor connected together). In such a circuit, the energy interchanges between the magnetic field of the inductor and the electric field of the capacitor.

Step-by-Step Solution

1. Initially, the capacitor is charged with a potential difference of 10 \, V. The total energy stored in the capacitor at this time is given by the formula: E = \frac{1}{2} C V^2, where C = 0.2 \, \mu F = 0.2 \times 10^{-6} \, F and V = 10 \, V.
   So, E = \frac{1}{2} \times 0.2 \times 10^{-6} \times 10^2 = 10 \times 10^{-6} \, J.
2. After the battery is disconnected and the inductor is connected, the potential difference across the capacitor is given as 5 \, V at some time t.
3. The energy stored in the capacitor at this new potential is: E_{cap} = \frac{1}{2} C V'^2, where V' = 5 \, V.
   Thus, E_{cap} = \frac{1}{2} \times 0.2 \times 10^{-6} \times 5^2 = 2.5 \times 10^{-6} \, J.
4. At this moment, part of the initial energy is stored in the inductor as magnetic energy: E_{inductor} = E_{initial} - E_{cap} = (10 \times 10^{-6} - 2.5 \times 10^{-6}) \, J = 7.5 \times 10^{-6} \, J
5. The energy stored in an inductor is also given by: E_{inductor} = \frac{1}{2} L I^2
   where L = 0.5 \, mH = 0.5 \times 10^{-3} \, H.
6. Setting the energies equal to solve for the current I: \frac{1}{2} \times 0.5 \times 10^{-3} \times I^2 = 7.5 \times 10^{-6}
   This simplifies to: 0.25 \times 10^{-3} \times I^2 = 7.5 \times 10^{-6}.
7. Solving for I, we get: I^2 = \frac{7.5 \times 10^{-6}}{0.25 \times 10^{-3}}
   I^2 = \frac{7.5}{0.25} \times 10^{-3} = 30 \times 10^{-3}
   I = \sqrt{30} \times 10^{-3} = \sqrt{3} \times 10^{-1} = 0.1732 \, A
8. Since we are given options, rounding 0.1732 \, A to 0.17 \, A, which matches the correct answer in the options provided.

Therefore, the correct answer is 0.17 \, A.