Question:
In an oscillating LC circuit, \(L = 1.6\,mH\), \(C = 4\,\mu F\). If the maximum charge is \(4 \times 10^{-6}\,C\), then the maximum current is:
In an oscillating LC circuit, \(L = 1.6\,mH\), \(C = 4\,\mu F\). If the maximum charge is \(4 \times 10^{-6}\,C\), then the maximum current is:
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In LC circuits: \(I_{max} = Q_{max}\omega\) and \(\omega = 1/\sqrt{LC}\).
Updated On: Jun 19, 2026
- 75 mA
- 12.5 mA
- 125 mA
- 50 mA
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The Correct Option is D
Solution and Explanation
Step 1: Formula for maximum current in LC oscillation.
\[ I_{max} = Q_{max}\,\omega \] where \[ \omega = \frac{1}{\sqrt{LC}} \]
Step 2: Convert values.
\[ L = 1.6 \times 10^{-3}\,H,\quad C = 4 \times 10^{-6}\,F \]
Step 3: Compute \(LC\).
\[ LC = 6.4 \times 10^{-9} \]
Step 4: Compute \(\omega\).
\[ \sqrt{LC} = 8 \times 10^{-5} \Rightarrow \omega = 1.25 \times 10^{4} \]
Step 5: Compute current.
\[ I_{max} = (4 \times 10^{-6})(1.25 \times 10^{4}) = 5 \times 10^{-2}\,A \]
Step 6: Final result.
\[ I_{max} = 50\,mA \]
Final Answer: \[ \boxed{50\,mA} \]
\[ I_{max} = Q_{max}\,\omega \] where \[ \omega = \frac{1}{\sqrt{LC}} \]
Step 2: Convert values.
\[ L = 1.6 \times 10^{-3}\,H,\quad C = 4 \times 10^{-6}\,F \]
Step 3: Compute \(LC\).
\[ LC = 6.4 \times 10^{-9} \]
Step 4: Compute \(\omega\).
\[ \sqrt{LC} = 8 \times 10^{-5} \Rightarrow \omega = 1.25 \times 10^{4} \]
Step 5: Compute current.
\[ I_{max} = (4 \times 10^{-6})(1.25 \times 10^{4}) = 5 \times 10^{-2}\,A \]
Step 6: Final result.
\[ I_{max} = 50\,mA \]
Final Answer: \[ \boxed{50\,mA} \]
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