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

To find the maximum current in the LC circuit, we can use the principle of conservation of energy. Initially, the energy is stored in the capacitor, and it is given by the formula:

\(E_{\text{cap}} = \frac{1}{2} C V^2\) 

where \(C = 100 \, \mu \text{F} = 100 \times 10^{-6} \, \text{F}\) and \(V = 12 \, \text{V}\).

Plugging in the values, we get:

\(E_{\text{cap}} = \frac{1}{2} \times 100 \times 10^{-6} \times (12)^2 = \frac{1}{2} \times 100 \times 10^{-6} \times 144\)

\(E_{\text{cap}} = 0.0072 \, \text{J}\)

In an LC circuit, the maximum energy stored in the capacitor (at its maximum charge) will be equal to the maximum energy stored in the inductor when the capacitor is fully discharged, i.e., when the current through the inductor is maximum. The energy in the inductor is given by:

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

Equating the energies, we have:

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

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

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

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

\(I_{\text{max}}^2 = \frac{0.0144}{6.4 \times 10^{-3}}\)

\(I_{\text{max}}^2 = 2.25\)

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

Thus, the maximum current in the circuit is 1.5 A, which matches the correct answer option: \(1.5 \, \text{A}\).

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