Question

A charged capacitor is connected to an inductor to form an LC circuit with a frequency of oscillation f = 1.6 Hz. At time t = 0 the capacitor is fully charged. At a given instant later the charge on the capacitor is measured to be Q = 3 μC and the current in the circuit is equal to 75 μA. What is the maximum charge of the capacitor?

Answer 1

Answer:

$Q = 8.61 \times 10^{-4} C$

Explanation:

Since in LC oscillation there is no energy loss

so here we can say that

initial total energy of capacitor = energy stored in capacitor + energy stored in inductor at any instant of time

so we can say

$\frac{Q^2}{2C} = \frac{q^2}{2C} + \frac{1}{2}Li^2$

now we have

$q = 3\mu C$

$i = 75 \mu A$

now we have

$Q^2 = q^2 + (LC) i^2$

we also know that

$2\pi f = \frac{1}{\sqrt{LC}}$

$2\pi(1.6) = \frac{1}{\sqrt{LC}}$

$LC = 9.89 \times 10^{-3}$

now from above equation

$Q^2 = (3\mu C)^2 + (9.89 \times 10^{-3})(75 \mu A)$

$Q = 8.61 \times 10^{-4} C$