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america8371
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In differential equations I was assigned a few circuit exercises. I'm having trouble understanding them, but I'm pretty sure that if can understand the first one then I can get the rest of them. The first exercise asks for time of capacitor charge and the others ask for things like capacitor maximum charge, steady state current, current/charge equations, etc.
Find the charge on the capacitor in an LRC series circuit at t=0.01s when L=0.05h, R=2[tex]\Omega[/tex], C=0.01f, E(t)=0V, q(0)=5C, and i(0)=0A. Determine the first time at which the charge on the capacitor is equal to zero. Please give answer to four decimal places.
a. t = 0.5095s
b. t = 0.9595s
c. t = 0.0509s
d. t = 0.5959s
e. t = 0.9059s
I'm not too sure but i think the correct diff. eq. to use is
L [tex]\frac{d^{2}q}{dt^{2}}[/tex]+R [tex]\frac{dq}{dt}[/tex]+ [tex]\frac{1}{C}[/tex] q = e(t)
I attempted to solve it by method of undetermined coefficients.
q''+40q'+2000q=0
r[tex]^{2}[/tex]+40r+2000=0 r=-20[tex]\pm[/tex]40i
q[tex]_{p}[/tex](t) = Ae[tex]^{-20t}[/tex]cos 40t + Be[tex]^{-20t}[/tex]sin 40t
etc.
It feels like I'm going in the wrong direction because it seem like there's more to this exercise than just solving the diff. eq.
Homework Statement
Find the charge on the capacitor in an LRC series circuit at t=0.01s when L=0.05h, R=2[tex]\Omega[/tex], C=0.01f, E(t)=0V, q(0)=5C, and i(0)=0A. Determine the first time at which the charge on the capacitor is equal to zero. Please give answer to four decimal places.
a. t = 0.5095s
b. t = 0.9595s
c. t = 0.0509s
d. t = 0.5959s
e. t = 0.9059s
Homework Equations
I'm not too sure but i think the correct diff. eq. to use is
L [tex]\frac{d^{2}q}{dt^{2}}[/tex]+R [tex]\frac{dq}{dt}[/tex]+ [tex]\frac{1}{C}[/tex] q = e(t)
The Attempt at a Solution
I attempted to solve it by method of undetermined coefficients.
q''+40q'+2000q=0
r[tex]^{2}[/tex]+40r+2000=0 r=-20[tex]\pm[/tex]40i
q[tex]_{p}[/tex](t) = Ae[tex]^{-20t}[/tex]cos 40t + Be[tex]^{-20t}[/tex]sin 40t
etc.
It feels like I'm going in the wrong direction because it seem like there's more to this exercise than just solving the diff. eq.