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

a) 0.269 mC

b) 0.355 ms

c) 1.39W

Step-by-step 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=(dq)/(dt)=Q\omega cos(\omega t)\\\\for \ t= 0:\\\\I=Q\omega\\\\Q=(I)/(\omega)\\\\\omega=(1)/(√(LC))\\\\Q=I√(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=(q^2)/(2C)=(Q^2sin^2(\omega t))/(2C)`

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

`(du_c)/(dt)=(2\omega Q^2sin(\omega t)cos(\omega t))/(2C)=0\\\\(du_c)/(dt)=(\omega Q^2 sin(2\omega t))/(2C)=0\\\\sin(2\omega t)=0\\\\2\omega t= 2\pi\\\\t=(\pi)/(\omega)=\pi√(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:

`(du_c)/(dt)=(2Q^2sin(2(t)/(√(LC))))/(√(LC))`

`(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`

Answer 2

a) 2.693*10^-4 C

b) 8.875*10^-5 s

c) 2.96 W

Step-by-step 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