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

The Correct Option is D

The maximum current in the LC circuit can be determined using energy conservation. Initially, the capacitor stores energy, calculated as \(E_{\text{cap}} = \frac{1}{2} C V^2\). With \(C = 100 \, \mu \text{F} = 100 \times 10^{-6} \, \text{F}\) and \(V = 12 \, \text{V}\), the initial capacitor energy is:

\(E_{\text{cap}} = \frac{1}{2} \times 100 \times 10^{-6} \times (12)^2 = 0.0072 \, \text{J}\)

In an LC circuit, this maximum capacitor energy equals the maximum inductor energy when the capacitor is discharged. The inductor energy is given by \(E_{\text{ind}} = \frac{1}{2} L I_{\text{max}}^2\). Equating these energies:

\(\frac{1}{2} L I_{\text{max}}^2 = 0.0072\)

Given \(L = 6.4 \, \text{mH} = 6.4 \times 10^{-3} \, \text{H}\), we solve for \(I_{\text{max}}\):

\(\frac{1}{2} \times 6.4 \times 10^{-3} \times I_{\text{max}}^2 = 0.0072\)

\(I_{\text{max}}^2 = \frac{0.0072 \times 2}{6.4 \times 10^{-3}} = 2.25\)

\(I_{\text{max}} = \sqrt{2.25} = 1.5 \, \text{A}\)

Therefore, the maximum current in the circuit is 1.5 A.