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:

Initially, the capacitor is charged by the battery and carries a potential difference of $$V_0 = 10\text{ V}$$.

Its capacitance is given as $$C = 0.2\;\mu\text{F} = 0.2 \times 10^{-6}\text{ F} = 2 \times 10^{-7}\text{ F}$$.

The charge stored at this instant is obtained from the basic relation $$q_0 = C V_0.$$

Substituting the values, we have $$q_0 = (2 \times 10^{-7}\text{ F})(10\text{ V}) = 2 \times 10^{-6}\text{ C}.$$

After the battery is disconnected and the capacitor is connected to an ideal inductor, the system becomes an ideal LC circuit. For such a circuit, the charge on the capacitor varies harmonically as

$$q(t) = q_0 \cos(\omega t),$$

where $$\omega$$ is the natural angular frequency of the LC circuit. The expression for $$\omega$$ is

$$\omega = \frac{1}{\sqrt{LC}}.$$

The given inductance is $$L = 0.5\;\text{mH} = 0.5 \times 10^{-3}\text{ H} = 5 \times 10^{-4}\text{ H}.$$

Now computing the product $$LC$$:

$$LC = (5 \times 10^{-4}\text{ H})(2 \times 10^{-7}\text{ F}) = 1 \times 10^{-10}\text{ H}\cdot\text{F}.$$

Taking the square root,

$$\sqrt{LC} = \sqrt{1 \times 10^{-10}} = 1 \times 10^{-5}.$$

Hence,

$$\omega = \frac{1}{1 \times 10^{-5}}\; \text{rad s}^{-1} = 1 \times 10^{5}\; \text{rad s}^{-1}.$$

The potential difference across the capacitor at any time is

$$V_C(t) = \frac{q(t)}{C} = \frac{q_0}{C}\cos(\omega t) = V_0 \cos(\omega t).$$

We are told that at the required instant the capacitor voltage has fallen to $$V_C = 5\text{ V}.$$ Using the above relation,

$$V_0 \cos(\omega t) = 5.$$

Substituting $$V_0 = 10\text{ V},$$

$$10 \cos(\omega t) = 5 \;\;\Longrightarrow\;\; \cos(\omega t) = 0.5.$$

We recognise $$\cos^{-1}(0.5) = \frac{\pi}{3},$$ so we can take

$$\omega t = \frac{\pi}{3}\;(= 60^\circ)$$

for the first time when the voltage becomes 5 V.

The instantaneous current in an LC circuit is the time-derivative of the charge:

$$i(t) = \frac{dq}{dt} = -q_0\,\omega \sin(\omega t).$$

We need only the magnitude, so

$$|i| = q_0 \omega |\sin(\omega t)|.$$

The sine corresponding to $$\omega t = \pi/3$$ is

$$\sin\!\left(\frac{\pi}{3}\right) = \frac{\sqrt3}{2} \approx 0.866.$$

Now substitute each known quantity:

First, the product $$q_0 \omega$$ is

$$q_0 \omega = (2 \times 10^{-6}\text{ C})(1 \times 10^{5}\text{ s}^{-1}) = 2 \times 10^{-1}\text{ A} = 0.20\text{ A}.$$

Then

$$|i| = 0.20\text{ A} \times 0.866 = 0.1732\text{ A}.$$

Rounding to two significant figures,

$$|i| \approx 0.17\text{ A}.$$

Hence, the correct answer is Option A.