Question

An oscillating LC circuit consists of a 75 mH inductor and a 1.2 uF capacitor. If the maximum charge to the capacitor is 2.7μC. The maximum current in the circuit will be _____mA.

Hint:

Energy conservation in LC circuits ensures smooth oscillations between capacitor and inductor.

Answer 1

Correct Answer: 9

To determine the maximum current in an oscillating LC circuit, we start with the known parameters: inductance \(L = 75 \text{ mH} = 75 \times 10^{-3} \text{ H}\), capacitance \(C = 1.2 \text{ μF} = 1.2 \times 10^{-6} \text{ F}\), and the maximum charge \(Q = 2.7 \text{ μC} = 2.7 \times 10^{-6} \text{ C}\). 

The energy in the LC circuit is conserved and the maximum energy stored in the capacitor is equal to the energy stored in the inductor at maximum current. This can be expressed as:

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

The maximum charge is related to the maximum voltage across the capacitor by \(Q = CV\). So the maximum voltage \(V\) can be given by:

\(V = \frac{Q}{C}\)

Substituting \(Q = 2.7 \times 10^{-6}\) C and \(C = 1.2 \times 10^{-6}\) F,

\(V = \frac{2.7 \times 10^{-6}}{1.2 \times 10^{-6}} = 2.25 \text{ V}\)

Now substituting \(V\) into the energy equation, we write:

\(\frac{1}{2}C \left(\frac{Q}{C}\right)^2 = \frac{1}{2}LI_{\text{max}}^2\)

This simplifies to:

\(\frac{Q^2}{2C} = \frac{1}{2}LI_{\text{max}}^2\)

Solving for \(I_{\text{max}}\),

\(I_{\text{max}} = \sqrt{\frac{Q^2}{LC}}\)

Inserting the values,

\(I_{\text{max}} = \sqrt{\frac{(2.7 \times 10^{-6})^2}{75 \times 10^{-3} \times 1.2 \times 10^{-6}}}\)

\(I_{\text{max}} = \sqrt{\frac{7.29 \times 10^{-12}}{90 \times 10^{-9}}}\)

\(I_{\text{max}} = \sqrt{8.1 \times 10^{-5}}\)

\(I_{\text{max}} = 0.009 \text{ A} = 9 \text{ mA}\)

The computed maximum current is 9 mA, which correctly fits within the given range of 9,9.