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

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G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2Assumptions:Resistors are available in 18 - ¼- ½-, and 1-W ratings.Analysis:R 4300 2v out=V = 110⋅= 64.30 1 2160014300R + R + R + R + R1= 1. 45kΩV110VI === 15 maR0+ R1+ R21600 Ω + 1450 Ω + 4300 Ω21P0 = I R0= 360 mW PR= W0221P1 = I R1= 326.25 mW PR= W122P2 = I R2= 967.5mW PR 2= 1 W________________________________________________________________________Solution:Problem 2.23Known quantities:Schematic of the circuit shown in Figure P2.23 with source voltage,R o= Ω,R 1= 10Ω,R = 2Ω.82Find:a) The equivalent resistance seen by the sourceb) The current ic) The power delivered by the sourced) The voltages v1and v2e) The minimum power rating required for R1Analysis:a) The equivalent resistance seen by the source isR eqb) Applying KVL:= R + R + = 8 + 10 + 2 = 20Ω0 1R2V 24VV − Req i = 0 , therefore i = = = 1.2A20ΩR eqv = 24V; and resistances,c)P source= Vi = 24 V ⋅1.2A= 28. 8Wd) Applying Ohm's law:e)v = R i = 10Ω ⋅1.2A12 V , and v = R i = 2Ω⋅1.2A2. 4 V1 1=2 2=2.18
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
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ECE-320 CH02.pdf - Unix.eng.ua.edu

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