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Step-by-Step Solution
Step 1: Identify the Physical Quantities
• Capacitance of the capacitor, $C = 0.2\,\mu F = 0.2 \times 10^{-6}\,F$.
• Initial voltage across the capacitor, $V_{\text{initial}} = 10\,V$.
• Self-inductance of the inductor, $L = 0.5\,mH = 0.5 \times 10^{-3}\,H$.
• Voltage across the capacitor at the moment of interest, $V_{\text{final}} = 5\,V$.
• Current through the inductor at that same moment, $I = ?$
Step 2: Write Down the Energy Conservation Principle
When the charged capacitor is connected to the inductor, the total energy at the start (which is purely in the capacitor) converts partly into the energy in the capacitor (at the new voltage) and partly into the energy in the inductor (due to the current). Therefore:
$ \frac{1}{2} C V_{\text{initial}}^{2} \;=\; \frac{1}{2} C V_{\text{final}}^{2} \;+\; \frac{1}{2} L I^{2} $
Step 3: Substitute the Known Values
Initial energy in the capacitor:
$ U_{\text{initial}} \;=\; \frac{1}{2} \times 0.2 \times 10^{-6} \times (10)^{2}\;J. $
Final energy in the capacitor plus energy in the inductor:
$ U_{\text{final}} \;=\; \frac{1}{2} \times 0.2 \times 10^{-6} \times (5)^{2} \;+\; \frac{1}{2} \times 0.5 \times 10^{-3} \times I^{2}\;J. $
Step 4: Simplify the Expressions and Solve for I
Calculate Initial Capacitor Energy:
$ \frac{1}{2} \times 0.2 \times 10^{-6} \times 100 \;=\; 10^{-5}\;J. $
Calculate Final Capacitor Energy:
$ \frac{1}{2} \times 0.2 \times 10^{-6} \times 25 \;=\; 2.5 \times 10^{-6}\;J. $
Set Up the Energy Conservation Equation:
$ 10^{-5} \;=\; 2.5 \times 10^{-6} \;+\; \frac{1}{2} \times 0.5 \times 10^{-3} \times I^{2}. $
Isolate the Term with I2:
$ 10^{-5} - 2.5 \times 10^{-6} \;=\; \frac{1}{2} \times 0.5 \times 10^{-3} \times I^{2}. $
Compute the Left Side and Factor Out Constants:
Left side $= 7.5 \times 10^{-6}$.
Right side $= \frac{1}{2} \times 0.5 \times 10^{-3} = 0.25 \times 10^{-3} = 2.5 \times 10^{-4}$.
So,
$$
7.5 \times 10^{-6} = (2.5 \times 10^{-4}) \, I^{2}.
$$
Solve for I:
\[
I^{2} = \frac{7.5 \times 10^{-6}}{2.5 \times 10^{-4}} \;=\; 0.03,
\]
\[
I = \sqrt{0.03} \;\approx\; 0.17\,A.
\]
Step 5: State the Final Answer
$\boxed{0.17\text{ A}}$
