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

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2Analysis:The same current flows through R1 and R3. Therefore, they are connected in series. Similarly, R2 and R4are connected in series.SPECIFY THE ASSUMED POLARITY OF THE VOLTAGE BETWEEN NODES A AND B. THISWILL HAVE TO BE A WILD GUESS AT THIS POINT.Specify the polarities of the voltage across R3 and R4 which will be determined using voltage division. Theactual polarities are not difficult to determine. Do so.VD :VD :KVL :VVR3R4VSR3=R1+ R3=VSR4=R + R=−VR32+ VAB4+ V[ 12 V ][ 6.8 kΩ]11+6.8 kΩ−6[ 12 V ][ 0.22×10 kΩ]−6( 220 + 0.22×10 ) kΩR4= 0∴VAB= V= 4.584 VR3−VR4= 1.20×10= 4.584VThe voltage is negative indicating that the polarity of VABis opposite of that specified.A solution is not complete unless the assumed positive direction of a current or assumed positive polarity ofa voltage IS SPECIFIED ON THE CIRCUIT.________________________________________________________________________Solution:Problem 2.50Known quantities:Schematic of the circuit shown in Figure P2.49 with source voltage,R = .2kΩ,R = 18kΩ,R = 4.7kΩ,R = 3. 3kΩ12234−8V≈ 0V s= 5V; and resistances,Find:The voltage between nodes A and B.Analysis:The same current flows through R1 and R3. Therefore, they are connected in series. Similarly, R2 and R4are connected in series.SPECIFY THE ASSUMED POLARITY OF THE VOLTAGE BETWEEN NODES A AND B. THISWILL HAVE TO BE A WILD GUESS AT THIS POINT.Specify the polarities of the voltage across R3 and R4 which will be determined using voltage division. Theactual polarities are not difficult to determine. Do so.VD :VD :KVL :VR3VSR3=R + R[ 5V][ 4.7 KΩ]== 3.406V2.2KΩ + 4.7 KΩ[ 5V][ 3.3 KΩ]VSR4VR4= == 0.917 VR2+ R418KΩ + 3.3 KΩ−V+ V + V = 0 V = V −V= 2.489VR31AB3R4ABA solution is not complete unless the assumed positive direction of a current or assumed positive polarity ofa voltage IS SPECIFIED ON THE CIRCUIT.________________________________________________________________________R3R42.38