EE 2120 Chapter : Prob1 36s1
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Problem 1. 36 after t = 0, the current entering the positive terminal of a ashlight bulb is given by i(t) = 2(1 e 10t) (a), and the voltage across the bulb is (t) = 12e 10t (v). Determine the maximum power level delivered to the ashlight. = 24(e 10t e 20t). to nd pmax, we take the derivative of p(t) and equate it to zero: d p dt d dt. = 24( 10)e 10t 24( 20)e 20t = 0, which simpli es to. Dividing by e 10t gives or e 10t 2e 20t = 0. Taking the natural log of both sides gives or ln(e 10t) = ln(cid:18) 1. Using this value of t in the expression for p(t) gives pmax = 24(e 10 0. 0693 e 20 0. 0693)
