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

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2II3Sv b=R3VS− v= bRApplying KCL to node (a) and (b) :II00+ I1+ IS5+ I2+ I1= 0− ISolving the system we obtain:3= 020− vb20 − v+R0+ 1 220− vb54 − v+R0+ 1 3v = 24 V and = 1ΩbbbR 0.+ 4 = 020 − v+2bvb−4= 0________________________________________________________________________Solution:Problem 2.34Known quantities:NOTE: Typo in Problem Statement for units of R 3Schematic of the circuit shown in Figure P2.34 with resistorsand voltage source V S= 12 V.Find:a) The mesh currents i a, i b, i cb) The current through each resistor.Analysis:Applying KVL to mesh (a), mesh (b) and mesh (c):iaR0+ ( ia− ib) R1= 0( ia− ib) R1− ibR2+ ( ic− ib)V = ( i − i ) RSSolving the system we obtain:iiiabcc= 2 A= 6 A= 8 Ab3IIIIRRRR0123= i= i= i= iabbc− i− iR= 2 Aa= 6 Ab3= 0= 4 A= 2 AR0212R3= Ω,R = 1Ω,R = 4 / 3Ω,= 6Ω2ia+ ( ia− ib) = 04( ia− ib) − ib+ 6( ic− ib)36( i − i ) = 12cb(positive in the direction of i(positive in the direction of i(positive in the direction of i(positive in the direction of iabbc= 0________________________________________________________________________))))2.26