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

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2Therefore, Req= 4 + 7.4 = 11.4 Ω._______________________________________________________________________Solution:Problem 2.39Known quantities:Schematic of the circuit shown in Figure P2.39 with source voltage,V s12VR = Ω,R = 2Ω,R = 50Ω,R = 8Ω,R = 10Ω,R = 12Ω,= 6Ω.0412345R6Find:= ; and resistances,The equivalent resistance of the circuit seen by the source; the current i through the resistance R2.Analysis:Starting from the right side, we can combine the three parallel resistors, namely the 10 Ω resistor, the 12 Ωresistor and the 6 Ω resistor:R1 1parallel1 1 1= + +10 12 6−20( Ω ) R = Ωparallel7Then, we can combine the two resistors in series, namely the 8 Ω and the 2.86 Ω resistor:Then, we can combine the three parallel resistors, namely the 4 Ω resistor, the 50 Ω resistor and the 10.86Ω resistor:2.30