Question

n an oscillating LC circuit, L = 3.76 mH and C = 3.13 μF. At t = 0 the charge on the capacitor is zero and the current is 2.95 A. (a) What is the maximum charge that will appear on the capacitor? (b) At what earliest time t > 0 is the rate at which energy is stored in the capacitor greatest, and (c) what is that greatest rate?

Answer 1

Part a)

`Q = 320 \mu C`

Part b)

`t = 8.52 * 10^(-5) s`

Part c)

Rate of energy = 301.5 J/s

Step-by-step explanation:

Part a)

Since energy is always conserved in LC oscillating system

So here for maximum charge stored in the capacitor is equal to the magnetic field energy stored in inductor

`(1)/(2)Li^2 = (Q^2)/(2C)`

now we have

`Q = √(LC) i`

`Q = \sqrt{(3.76 * 10^(-3))(3.13 * 10^(-6))} (2.95)`

`Q = 320 \mu C`

Part b)

Energy stored in the capacitor is given as

`U = (q^2)/(2C)`

now rate of energy stored is given as

`(dU)/(dt) = (q)/(C)(dq)/(dt)`

so here we also know that

`q = Q sin(\omega t)`

`(dq)/(dt) = Q\omega cos(\omega t)`

now from above equation

`(dU)/(dt) = (Qsin(\omega t))/(C) (Q\omega cos\omega t)`

so maximum rate of energy will be given when

`sin\omega t = cos\omega t`

`\omega t = (\pi)/(4)`

`t = (\pi)/(4)√(LC)`

`t = 8.52 * 10^(-5) s`

Part c)

Greatest rate of energy is given as

`(dU)/(dt) = (Q^2\omega)/(C)`

`(dU)/(dt) = \frac{(320 \mu C)^2 \sqrt{(1)/((3.76 mH)(3.13 \mu C))}}{3.13 \mu C}`

`(dU)/(dt) = 301.5 J/s`