Question

an lc circuit has a capacitance of 40 μf and an inductance of 10 mh. at time t = 0, the charge on the capacitor is 12 μc, and the current is 15 ma. the maximum current is

32 ma.
19 ma.
39 ma.
24 ma.
15 ma.

Explanation:

Step1: Write the energy conservation formula for LC - circuit

The total energy \(U\) in an LC - circuit is the sum of the electrical energy in the capacitor \(U_E=\frac{q^{2}}{2C}\) and the magnetic energy in the inductor \(U_B = \frac{Li^{2}}{2}\). At any time \(t\), \(U=\frac{q^{2}}{2C}+\frac{Li^{2}}{2}\), and at maximum current \(i = I_{max}\), \(q = 0\) and \(U=\frac{LI_{max}^{2}}{2}\). So, \(\frac{q^{2}}{2C}+\frac{Li^{2}}{2}=\frac{LI_{max}^{2}}{2}\)

Step2: Substitute the given values

Given \(C = 40\times10^{- 6}\text{F}\), \(L=10\times10^{-3}\text{H}\), \(q = 12\times10^{-6}\text{C}\), \(i = 15\times10^{-3}\text{A}\)

First, calculate \(\frac{q^{2}}{2C}+\frac{Li^{2}}{2}\):

$$
LATEXBLOCK0
$$

Then, from \(\frac{LI_{max}^{2}}{2}=2.925\times10^{-6}\text{J}\), we can solve for \(I_{max}\):

$$
LATEXBLOCK1
$$

Answer:

\(24\text{mA}\)