ECE-320 CH02.pdf - Unix.eng.ua.edu

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 22( 1.2A) 14. W2P1 = R1i= 10Ω⋅= 4 , therefore the minimum power rating for R1is 16 W.________________________________________________________________________Solution:Problem 2.24Known quantities:Schematic of the circuit shown in Figure P2.24 with resistors,R = Ω,R = 10Ω,R = 5Ω,= 7Ω.252341RFind:a) The currents i 1and i2b) The power delivered by the 3-A current source and the 12-V voltage sourcec) The total power dissipated by the circuit.Analysis:a) KCL at node 1 requires that:v 1R 2+v 1Solving for v1we haveTherefore,-12 VR3( 4 + R )- 3 A = 03R2v1= 3 = 18 VR + R2v118i1= − = − = −1.8 AR21012 − v16i2= = − = −1.2 AR 53b) The power delivered by the 3-A source is:P3-A = (v3-A)(3)3Thus, we can compute the voltage across the 3-A source asv3-A = R + v = 3⋅25 + 18 93 VThus,31 1=P3-A = (93)(3) = 279 W.Similarly, the power supplied by the 12-V source is:P12-V = (12)(I12-V)We have I12-V =12R4+ i 2= 514.3 mA, thus:P12-V = (12)(I12-V) = 6.17 Wc) Since the power dissipated equals the total power supplied:2.19