Solution manual fundamentals of electric circuits 3rd edition chapter06

... 6, Solution t −t −t t − e dt + = 12 t + e V = 12(t + e-t) – 12 v(t) = ∫ 12 o -2 v(2) = 12(2 + e ) – 12 = 13.624 V ( ) ( ) p = iv = [12 (t + e-t) – 12]6(1-e-t) p(2) = [12 (2 + e-2) – 12]6(1-e-2)... and b in the circuit of Fig 6.53 All capacitances are in μF Figure 6.53 Chapter 6, Solution 19 We combine 1 0-, 2 0-, and 3 0- μ F capacitors in parallel to get 60 μ F The 60 - μ F capacitor in series... x 1 0-6 x 3600 = 3.6mJ w6μF = 1/2 x x 1 0-6 x 400 = 1.2mJ w3μF = 1/2 x x 1 0-6 x 1600 = 2.4mJ PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc All rights reserved No part of this Manual