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:
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:
[tex]q=Qsin(\omega t)[/tex]
next, you calculate the current I by using the derivative of q:
[tex]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}[/tex] ( 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):
[tex]Q=(2.38A)(\sqrt{(4.24*10^{-3}H)(3.02*10^{-6}F)})=2.69*10^{-4}C=0.269mC[/tex]
b) to find the time t you use the following formula for the energy of the capacitor:
[tex]u_c=\frac{q^2}{2C}=\frac{Q^2sin^2(\omega t)}{2C}[/tex]
the maximum storage energy in the capacitor is obtained by derivating the energy:
[tex]\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[/tex]
hence, the time is 0.355 ms
c) The greatest rate is obtained for duc/dt evaluated in t=0.355 ms:
[tex]\frac{du_c}{dt}=\frac{2Q^2sin(2\frac{t}{\sqrt{LC}})}{\sqrt{LC}}[/tex]
[tex]\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[/tex]
