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

Answer:

Part a)

$Q = 320 \mu C$

Part b)

$t = 8.52 \times 10^{-5} s$

Part c)

Rate of energy = 301.5 J/s

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

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

now we have

$Q = \sqrt{LC} i$

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

$Q = 320 \mu C$

Part b)

Energy stored in the capacitor is given as

$U = \frac{q^2}{2C}$

now rate of energy stored is given as

$\frac{dU}{dt} = \frac{q}{C}\frac{dq}{dt}$

so here we also know that

$q = Q sin(\omega t)$

$\frac{dq}{dt} = Q\omega cos(\omega t)$

now from above equation

$\frac{dU}{dt} = \frac{Qsin(\omega t)}{C} (Q\omega cos\omega t)$

so maximum rate of energy will be given when

$sin\omega t = cos\omega t$

$\omega t = \frac{\pi}{4}$

$t = \frac{\pi}{4}\sqrt{LC}$

$t = 8.52 \times 10^{-5} s$

Part c)

Greatest rate of energy is given as

$\frac{dU}{dt} = \frac{Q^2\omega}{C}$

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

$\frac{dU}{dt} = 301.5 J/s$