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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

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  1. anonymous
    • one year ago
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    Find power and energy delivered at t = .000625 seconds in watts and milliJoules

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  2. anonymous
    • one year ago
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    My calculations gave me: P(625microseconds) = 36.332 watts Work = 69 milliJoules

  3. IrishBoy123
    • one year ago
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    |dw:1442225696953:dw|

  4. IrishBoy123
    • one year ago
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    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

  5. anonymous
    • one year ago
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    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.

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