anonymous
  • anonymous
The voltage and current at the terminals of this circuit are zero for t < 0 and for t ≥ 0 they are v = 54 e(^-1573t) Volts i = 5.6e(^-1573t) Amps
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Find power and energy delivered at t = .000625 seconds in watts and milliJoules
1 Attachment
anonymous
  • anonymous
My calculations gave me: P(625microseconds) = 36.332 watts Work = 69 milliJoules
IrishBoy123
  • IrishBoy123
|dw:1442225696953:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

IrishBoy123
  • IrishBoy123
you are using \(V = 54 e^{-1573t} \) and \( I = 5.6e^{-1573t}\) so \(P(t) = 54(5.6)e^{-2 \times 1573t}\) and \(W(t) = \int\limits_{0}^{0.000625} P(t) \ dt = \left[ \frac{54(5.6)}{-2 \times 1573}e^{-2 \times 1573t} \right]_{0}^{0.000625}\) is that your approach? because i'm getting different answers
anonymous
  • anonymous
Apologies late reply, so late... But you're approach is the correct approach. My mistake was that I took the instantaneous value of one of my variables before plugging it into the Work Integral.

Looking for something else?

Not the answer you are looking for? Search for more explanations.