Question

In an oscillating LC circuit, L = 4.24 mH and C = 3.02 μF. At t = 0 the charge on the capacitor is zero and the current is 2.38 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:

a) 2.693*10^-4 C

b) 8.875*10^-5 s

c) 2.96 W

Explanation:

Given that

Inductance of the circuit, L = 4.24 mH

Capacitance of the circuit, C = 3.02 μF

Current in the circuit, I = 2.38 A

See attachment for calculations

Answer 2

Answer:

a) 0.269 mC

b) 0.355 ms

c) 1.39W

Explanation:

a) To find the charge off the capacitor you start by using the following expression for the charge in the capacitor:

$q=Qsin(\omega t)$

next, you calculate the current I by using the derivative of q:

$I=\frac{dq}{dt}=Q\omega cos(\omega t)\\\\for \ t= 0:\\\\I=Q\omega\\\\Q=\frac{I}{\omega}\\\\\omega=\frac{1}{\sqrt{LC}}\\\\Q=I\sqrt{LC}$ ( 1 )

L: inductance = 4.24*10^{-3}H

C: capacitance = 3.02*10^{-6}F

I: current = 2.38 A

you replace the values of the parameters in (1):

$Q=(2.38A)(\sqrt{(4.24*10^{-3}H)(3.02*10^{-6}F)})=2.69*10^{-4}C=0.269mC$

b) to find the time t you use the following formula for the energy of the capacitor:

$u_c=\frac{q^2}{2C}=\frac{Q^2sin^2(\omega t)}{2C}$

the maximum storage energy in the capacitor is obtained by derivating the energy:

$\frac{du_c}{dt}=\frac{2\omega Q^2sin(\omega t)cos(\omega t)}{2C}=0\\\\\frac{du_c}{dt}=\frac{\omega Q^2 sin(2\omega t)}{2C}=0\\\\sin(2\omega t)=0\\\\2\omega t= 2\pi\\\\t=\frac{\pi}{\omega}=\pi\sqrt{LC}=\pi\sqrt{(4.24*10^{-3}H)(3.02*10^{-6}F)}=3.55*10^{-4}s=0.355\ ms$

hence, the time is 0.355 ms

c) The greatest rate is obtained for duc/dt evaluated in t=0.355 ms:

$\frac{du_c}{dt}=\frac{2Q^2sin(2\frac{t}{\sqrt{LC}})}{\sqrt{LC}}$

$\frac{du_c}{dt}=\frac{(2.69*10^{-4}C)^2sin(2\frac{3.55*10^{-4}s}{\sqrt{(4.24*10^{-3}H)(3.02*10^{-6}C)}})}{2(3.02*10^{-6}C)\sqrt{(4.24*10^{-3}H)(3.02*10^{-6}C)}}=-1.39W$