Question

A capacitor of capacitance 100 μF is charged to a potential of 12 V and connected to a 6.4 mH inductor to produce oscillations. The maximum current in the circuit would be:

• A. 2 A
• B. 1.2 A
• C. 2.4 A
• D. 1.5 A

Answer 1

Answer: (D)

1.5 A

Answer 2

Step 1: Using Energy Conservation in LC Circuit - The energy stored in the capacitor initially is equal to the maximum energy stored in the inductor:

$\frac{1}{2}CV^2 = \frac{1}{2}LI_{\text{max}}^2$

- Solving for $I_{\text{max}}$:

$I_{\text{max}} = V \sqrt{\frac{C}{L}}$

Step 2: Substitute Given Values - $C = 100 \times 10^{-6} \, \text{F}$, $L = 6.4 \times 10^{-3} \, \text{H}$, $V = 12 \, \text{V}$

$I_{\text{max}} = 12 \sqrt{\frac{100 \times 10^{-6}}{6.4 \times 10^{-3}}}$

Step 3: Calculate $I_{\text{max}}$

$I_{\text{max}} = 12 \times \sqrt{\frac{1}{8}} = 12 \times \frac{1}{\sqrt{8}} = \frac{12}{\sqrt{8}} = 1.5 \, \text{A}$

So, the correct answer is: 1.5 A