Bme211hw1

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Unformatted text preview: HW#1 Solutions P1.2-1 i( t) = d 4 1 - e -5t = 20 e -5t A dt ( ) P1.2-2 t t t t 4 4 q ( t ) = ) d + q ( 0 ) = 1 - e -5 d + 0 = d - e -5 d = 4 t + e -5t - C i( 4 4 4 0 0 0 0 5 5 ( ) P1.2-6 i = 600 A = 600 C s C s mg Silver deposited = 600 d 20 min 60 1.118 = 8.05 105 mg=805 g s min C P2.4-5 v1 = v 2 = vs = 150 V; R1 = 50 ; R2 = 25 v 1 and i1 adhere to the passive convention so i1 = v 1 150 = =3 A R 1 50 v2 150 =- = -6 A R2 25 v2 and i 2 do not adhere to the passive convention so i 2 = - The power absorbe...
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This note was uploaded on 03/07/2008 for the course BME 211 taught by Professor Cain during the Winter '07 term at University of Michigan.

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