- #1
Potatochip911
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- 3
Homework Statement
Determine ##q(t=0)##, ##i(t=0)##, the phase difference ##\phi##, angular frequency ##w## and the time constant ##\tau## from the graph of the capacitor waveform below:(pulse voltage source)
For this circuit we have ##q(t)=Ae^{-\frac{t}{2\tau}}\cos(\omega t+\phi)## therefore
##v(t)=\frac{A}{C}e^{-\frac{t}{2\tau}}\cos(\omega t+\phi)##
Homework Equations
##i=\frac{dq}{dt}##
The Attempt at a Solution
For ##q(t=0)## I get ##q(t=0)=A\cos(\phi)##, for ##i(t)=\frac{dq}{dt}=-A(\frac{t}{2\tau}e^{-\frac{t}{2\tau}}\cos(\omega t+\phi)+\omega\sin(\omega t +\phi)e^{-\frac{t}{2\tau}})## therefore
##i(t=0)=-A\omega\sin(\phi)##
For the phase difference since at ##t=0## the function is starting at 0 and we have a cosine the phase is ##\phi=-\frac{\pi}{2}##
To find ##\omega## I used the fact that a fourth of the period corresponds to ##\pi/2## so taking a time difference that satisfies this ##t_1=0## and ##t_2=0.533\mu_s## we have ##\omega t_2-\omega t_1=\omega(t_2-t_1)=\pi/2\Longrightarrow \omega=\frac{\pi}{2t_2}## (Since ##t_1=0##),
This gave ##\omega=2.9rad/\mu_s##
Now I'm stuck trying to calculate ##\tau##, I wanted to sub into my voltage formula ##t=2\tau## then I would have ##v(t)=\frac{A}{C}e^{-1}\cos(2\omega \tau-\pi/2)## since I wanted to look graphically for the time where the amplitude became ##\approx \frac{1}{e}## of the original but I'm not sure what to do with the ##\cos## term.