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anonymous
 one year ago
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
anonymous
 one year ago
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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Find power and energy delivered at t = .000625 seconds in watts and milliJoules

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0My calculations gave me: P(625microseconds) = 36.332 watts Work = 69 milliJoules

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2dw:1442225696953:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2you 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
 one year ago
Best ResponseYou've already chosen the best response.0Apologies 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.
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