... Chapter 1, Solution 14 q = ∫ idt = ∫ 10(1 - e -0 .5t )dt = 10(t + 2e -0 .5t ) (a) (b) = 10(1 + 2e -0 .5 − ) = 2.131 C p(t) = v(t)i(t) p(1) = 5cos2 ⋅ 10( 1- e-0.5) = (-2 .081)(3.935) = -8 .188 W Chapter... Figure 1.29 Chapter 1, Solution 18 p1 = 30 (-1 0) = -3 00 W p2 = 10(10) = 100 W p3 = 20(14) = 280 W p4 = 8 (-4 ) = -3 2 W p5 = 12 (-4 ) = -4 8 W PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc... Chapter 1, Solution 15 (a) q = ∫ idt = ∫ 3e ( -2 t ) − 2t dt = e = −1.5 e -4 − = 1.4725 C (b) 5di = −6e 2t ( 5) = −30e -2 t dt p = vi = − 90 e −4 t W v= (c) w = ∫ pdt = -9 0 ∫ e -4 t − 90 -4 t dt =