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ET 162 Circuit Analysis Series and Parallel Networks Electrical and Telecommunication Engineering Technology Professor Jang Acknowledgement I want to express my gratitude to Prentice Hall giving me the permission to use instructor’s material for developing this module. I would like to thank the Department of Electrical and Telecommunications Engineering Technology of NYCCT for giving me support to commence and complete this module. I hope this module is helpful to enhance our students’ academic performance. OUTLINES Introduction to Series-Parallel Networks Reduce and Return Approach Block Diagram Approach Descriptive Examples Ladder Networks Key Words: Series-Parallel Network, Block Diagram, Ladder Network ET162 Circuit Analysis – Series and Parallel Networks Boylestad 2 Series-Parallel Networks – Reduce and Return Approach Series-parallel networks are networks that contain both series and parallel circuit configurations For many single-source, series-parallel networks, the analysis is one that works back to the source, determines the source current, and then finds its way to the desired unknown. FIGURE 7.1 Introducing the reduce and return approach. ET162 Circuit Analysis – Series and parallel networks Boylestad 3 Series-Parallel Networks Block Diagram Approach The block diagram approach will be employed throughout to emphasize the fact that combinations of elements, not simply single resistive elements, can be in series or parallel. FIGURE 7.2 Introducing the block diagram approach. ET162 Circuit Analysis – Series and parallel networks Boylestad 4 Ex. 7-1 If each block of Fig.7.3 were a single resistive element, the network of Fig. 7.4 might result. FIGURE 7.4 FIGURE 7.3 6k I s 1 1 I s 9mA 3 mA 6k 12k 3 3 12k I s 2 2 IC I s 9mA 6 mA 12k 6k 3 3 IB ET162 Circuit Analysis – Series and parallel networks Boylestad 4 Ex. 7-2 It is also possible that the blocks A, B, and C of Fig. 7.2 contain the elements and configurations in Fig. 7.5. Working with each region: A : RA 4 B : RB R2 // R3 R2 // 3 R 4 2 N 2 C : RC R4 R5 R4,5 2 FIGURE 7.5 R 2 RB // C 1 N 2 RT RA RB // C 4 1 5 ET162 Circuit Analysis FIGURE 7.6 E 10 V Is 2A RT 5 Boylestad 6 I A Is 2 A I A Is 2 A I B IC 1 A 2 2 2 IB I R2 I R3 0.5 A 2 FIGURE 7.6 VA I A RA 2 A4 8 V VB I B RB 1 A2 2 V VC VB 2 V ET162 Circuit Analysis – Series and parallel networks Boylestad 7 Ex. 7-3 Another possible variation of Fig. 7.2 appears in Fig. 7.7. RA R1// 2 9 6 9 6 54 3.6 15 RB R3 R4 // 5 9 3 4 9 3 4 2 6 RC 3 FIGURE 7.7 FIGURE 7.8 RT RA RB // C 6 3 3.6 6 3 3.6 2 5.6 E 16.8V Is 3 A RT 5.6 ET162 Circuit Analysis – Series and parallel networks I A Is 3 A 8 Applying the current divider rule yields RC I A 3 3 A IB 1 A RC RB 3 6 By Kirchhoff ' s current law, IC I A I B 3 A 1 A 2 A By Ohm ' s law, VA I A RA 3 A3.6 10.8 V V I R V I R 2 A3 6 V R2 I A I1 R2 R1 6 3 A 6 9 1.2 A I 2 I A I1 3 A 1 .2 A 1.8 A Series-Parallel Networks - Descriptive Examples Ex. 7-4 Find the current I4 and the voltage V2 for the network of Fig. 7.2 . E E 12V I4 1.5 A RB R4 8 RD R2 // R3 3 // 6 2 RD E V2 RD RC 2 12V 4V FIGURE 7.9 2 4 FIGURE 7.10 ET162 Circuit Analysis – Series and parallel networks FIGURE 7.11 Boylestad 10 Ex. 7-5 Find the indicated currents and the voltages for the network of Fig. 7.12 . FIGURE 7.12 R 6 R1// 2 3 N 2 3 2 6 RA R1// 2 // 3 1.2 3 2 5 8 12 96 RB R4 // 5 4.8 8 12 20 ET162 Circuit Analysis – Series and parallel networks Boylestad FIGURE 7.13 11 RT R1// 2 // 3 R4 // 5 1 .2 4 .8 6 E 24 V Is 4A RT 6 FIGURE 7.13 V1 I s R1// 2 // 3 4 A1.2 4.8V V2 I s R4 // 5 4 A4.8 19.2V V5 19.2 V I4 2.4 A R4 8 ET162 Circuit Analysis – Series and parallel networks V2 V1 4.8V I2 0.8 A R2 Boylestad R2 6 12 Ex. 7-6 a. Find the voltages V1, V2, and Vab for the network of Fig. 7.14. b. Calculate the source current Is. a. FIGURE 7.14 FIGURE 7.15 Applying the voltage divider rule yields Applying Kirchhoff ' s voltage law around the indicated loop of Fig. 512V R1 E V1 7.5 V V1 V3 Vab 0 R1 R2 5 3 Vab V3 V1 9V 7.5V 15 .V 612V R3 E V3 9V R3 R4 6 2 ET162 Circuit Analysis – Series and parallel networks Boylestad 13 b. By Ohm' s law, V1 7.5V I1 15 . A R1 5 V3 9V I3 15 . A R3 6 ET162 Circuit Analysis – Series and parallel networks Applying Kirchhoff’s current law, Is = I1 + I3 = 1.5A + 1.5A = 3A Boylestad 14 Ex. 7-7 For the network of Fig. 7.16, determine the voltages V1 and V2 and current I. FIGURE 7.17 FIGURE 7.16 V2 = – E1 = – 6V Applying KVL to the loop E1 – V1 + E2 = 0 V1 = E2 + E1 =18V + 6V = 24V Applying KCL to node a yields I I1 I 2 I 3 V1 E1 E1 R1 R4 R2 R3 24V 6V 6V 6 6 12 4 A 1 A 0.5 A 55 . A ET162 Circuit Analysis – Series and parallel networks Boylestad 15 Ex. 7-9 Calculate the indicated currents and voltage of Fig.7.17. FIGURE 7.17. E 72V 72V I5 3mA R(1, 2,3) // 4 R5 12k 12k 24k V7 9 kΩ I6 R7 //(8,9) E R7 //(8,9) R6 V7 R7 //(8,9) 4.5k 72V 324V 4.5k 12k 16.5 19.6V 19.6V 4.35mA 4.5k Is = I5 + I6 = 3 mA +4.35 mA = 7.35 mA ET162 Circuit Analysis – Series and parallel networks Boylestad 16 Ex. 7-10 This example demonstrates the power of Kirchhoff’s voltage law by determining the voltages V1, V2, and V3 for the network of Fig.7.18. FIGURE 7.17. FIGURE 7.18. For the path 1, E1 V1 E 3 0 V1 E1 E 3 20V 8V 12 V For the path 2, E 2 V1 V2 0 V2 E 2 V1 5V 12V 7 V For the path 3, V3 V2 E 3 0 V3 E 3 V2 8V ( 7V ) 15 V ET162 Circuit Analysis – Series and parallel networks Boylestad 17 Series-Parallel Networks – Ladder Networks A three-section ladder appears in Fig. 7.19. FIGURE 7.19. Ladder network. ET162 Circuit Analysis – Series and parallel networks Boylestad 18 FIGURE 7.20. RT 5 3 8 240V E Is 30 A RT 8 FIGURE 7.21. ET162 Circuit Analysis – Series and parallel networks Boylestad 19 I1 I s I s 30 A I3 15 A 2 2 6 I 3 6 I6 15 A 10 A 6 3 9 V6 I 6 R6 10 A2 20 V ET162 Circuit Analysis – Series and parallel networks Boylestad 20