Just prior to t = 0, the capcitor (C = 1 x 10^-3 F) in the circuit below has a c
ID: 2107273 • Letter: J
Question
Just prior to t = 0, the capcitor (C = 1 x 10^-3 F) in the circuit below has a charge of 1 meuC. The inductor has inductance, L = 1 x 10^-6 H. At t = 0, a switch (not shown) is close, allowing current to flow around the circuit. What is the magnitude of the current when there is no charge on the capacitor has fallen to 0.5 meuC ? [Hint: Q(t) = Qo cos(t/(root (LC))] At the instant when the charge on the capacitor fallen to 0.5 meuC, what is the magnitude of the rate of change in current?
The answers are 2.7 x 10^-2 A and 500 A/s.
Could you explain how to solve?
Explanation / Answer
I=Io*cos(wt)/sqrt(LC) so di/dt=Io*w*sin(wt)/sqrt(LC) where w=1/sqrt(LC) do we find
.5=1*cos(wt)/sqrt(LC) so cos(wt)/sqrt(LC)=.5 so sin(wt)/sqrt(LC)=.866 so find di/dt=Io*w*sin(wt)/sqrt(LC)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.