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16 October, 19:14

Consider a 50-mH inductor. a. Express the voltage across the inductor and then evaluate it at t = 0.25 s if iL (t) = 5e - 2t + 3te-2t - 2 A. Be certain to simplify your expression.

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  1. 16 October, 19:31
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    V=L (di/dt) where i is current, V=0.208

    Explanation:

    using expression iL (t) = 5e-2t+3te-2t-2 and L=0.05H (50/1000)

    V=0.05*d (5e-2t+3te-2t-2) / dt

    since there is no power of e, I'll assume the power to be 1

    V=0.05 * (-2+3e-2)

    at t=0.25

    V=0.15e-0.2

    V=0.208
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