McGraw-Hill © 2013 The McGraw-Hill Companies Inc. All rights reserved. Electricity Principles & Applications Eighth Edition Chapter 6 Complex-Circuit Analysis (student version) Richard J. Fowler McGraw-Hill © 2013 The McGraw-Hill Companies Inc. All rights reserved

McGraw-Hill © 2013 The McGraw-Hill Companies Inc. All rights reserved. INTRODUCTION Superpositon Theorem (Page 156) Simultaneous Equations (Page 136) Voltage Sources (Page 160) Thevenin’s Theorem (Page 161) Norton’s Theorem (Page 169) 6 - 2

McGraw-Hill © 2013 The McGraw-Hill Companies Inc. All rights reserved. Dear Student: This presentation is arranged in segments. Each segment is preceded by a Concept Preview slide and is followed by a Concept Review slide. When you reach a Concept Review slide, you can return to the beginning of that segment by clicking on the Repeat Segment button. This will allow you to view that segment again, if you want to

McGraw-Hill © 2013 The McGraw-Hill Companies Inc. All rights reserved. Concept Preview Two or more sources are required to use the superposition theorem. (Page 156) Superposition theorem requires only series-parallel rules and procedures to determine values in a complex circuits. (Page 157) Superposition theorem will solve for all currents and voltages in the circuit. (Page 157) 6 - 4

McGraw-Hill © 2013 The McGraw-Hill Companies Inc. All rights reserved. Superposition Technique 10  30  110 V 20  110 V 20  10  30  110 V Replace B 2 with a shortand calculate the currents. 110 V 10  30  110 V 20  B2B2 B1B1 5 A 3 A 2 A Replace B 1 with a shortand calculate the currents. 3 A 4 A 1 A Algebraically add the two currents for each resistor. 8 A 7 A1 A 110 V Rty u 1 10 V Rty u 6 - 5

McGraw-Hill © 2013 The McGraw-Hill Companies Inc. All rights reserved. Verification of Superposition Results Using Circuit Measurements. The dark bars on the ammeters are the negative terminals. Notice that both the directions and magnitudes of the currents agree with those obtained by the superposition method

McGraw-Hill © 2013 The McGraw-Hill Companies Inc. All rights reserved. Simultaneous equations A procedure for determining the values of I 1 and I 2 using two independent equations is shown below. The independent equations are: #1 8.5V = 5  (I 1 ) - 1  (I 2 ) #2 2.5V = -1  (I 1 ) + 3  (I 2 ) Multiply equation # 2 by 5 and add the results to equation #1. #2 [2.5V = -1  (I 1 ) + 3  (I 2 )] x 5 yields 12.5V = -5  (I 1 ) + 15  (I 2 ) #1 8.5V = 5  (I 1 ) - 1  (I 2 ) 21V = 14  (I 2 ) Now solve for I 2 which yields I 2 = 21V / 14  = 1.5A Finally, using this value of I 2 in either equation #1 or #2, solve for I 1 which yields I 1 = (8.5V + 1.5V) / 5  = 2A In summary the values are: I 1 = 2A and I 2 = 1.5A 6 - 7

McGraw-Hill © 2013 The McGraw-Hill Companies Inc. All rights reserved. Concept Review Two or more sources are required to use the superposition theorem. Superposition theorem requires only series-parallel rules and procedures to determine values in a complex circuits. Superposition theorem will solve for all currents and voltages in the circuit. Repeat Segment Simultaneous equations techniques can be used to solve loop equations

McGraw-Hill © 2013 The McGraw-Hill Companies Inc. All rights reserved. Concept Preview Thevenin’s theorem reduces a complex circuit to an equivalent-circuit voltage source with a load. (Page 161) Norton’s theorem reduces a complex circuit to an equivalent-circuit current source with a load. (Page 169 Norton’s theorem and Thevenin’s theorem may not solve for all values of current and voltage. (Page 171) 6 - 9

McGraw-Hill © 2013 The McGraw-Hill Companies Inc. All rights reserved. Voltage Sources (Page 160) Loaded V 15 V Loaded V 13 V No load V 15 V No load 15 V V Ideal Source Voltage does not change when loaded. Equivalent-Circuit Source Voltage changes when loaded

McGraw-Hill © 2013 The McGraw-Hill Companies Inc. All rights reserved. Thevenizing A Complex Circuit (Page 161) 110 V 10  110 V 20  B2B2 B1B1 30  R2R2 R2R2 V TH = 36.7 V R TH = 6.7  Remove R 2 and calculate V TH. e Replace B 1 and B 2 with shortsand calculate R TH. 110 V 20  10  30  110 V R2R2 1 A 30 V 8 A 7 A Return to the original circuit and calculate the other currents. R 2 30  Draw the Thevenin equivalent circuit. Load the circuit with R 2. Calculate the load V and I. 30 V 1 A R TH 6.7  V TH 36.7 V

McGraw-Hill © 2013 The McGraw-Hill Companies Inc. All rights reserved. Applying Norton’s Theorem to a Complex Circuit (Page 169) Select a resistor to be the load. Replace the load with a short. Calculate the current through the short. This is I N. Remove the short from the load terminals. Replace all sources with shorts. Calculate the resistance between the load terminals. This is R N. Use I N and R N for the Norton circuit. Connect the load to the Norton circuit. Calculate V load and I load

McGraw-Hill © 2013 The McGraw-Hill Companies Inc. All rights reserved. R TH 8  V TH 16 V - + R N 8  I N 2 A R TH 8  V TH 16 V - + R TH 8  V TH 16 V - + R N 8  I N 2 A R N 8  I N 2 A R L 24  R L 24  2 A A A 16 V V V The open-circuit voltages are equal. The short-circuit currents are equal. When equally loaded, 12 V 0.5 A 12 V 0.5 A the voltages and currents are equal. Equivalency Of Norton and Thevenin Circuits Page (172)

McGraw-Hill © 2013 The McGraw-Hill Companies Inc. All rights reserved. Complex-Circuits Quiz Which theorem(s) require(s) algebraically adding the currents for each resistor? ____ Which theorem(s) require(s) shorting all voltage sources at the same time? ____ Which theorem(s) use(s) an equivalent- circuit current source? ____ Which theorem(s) use(s) an equivalent- circuit voltage source? ____ Which theorem(s) determine(s) the current for all resistors in the circuit? ____ The voltage of a(n) ____ voltage source is independent of the load current. Superposition Thevenin Norton Thevenin Superposition ideal Solutions by simultaneous equations require ____ equations. independent

McGraw-Hill © 2013 The McGraw-Hill Companies Inc. All rights reserved. Concept Review Repeat Segment Thevenin’s theorem reduces a complex circuit to an equivalent-circuit voltage source with a load. Norton’s theorem reduces a complex circuit to an equivalent-circuit current source with a load. Norton’s theorem and Thevenin’s theorem may not solve for all values of current and voltage

McGraw-Hill © 2013 The McGraw-Hill Companies Inc. All rights reserved. REVIEW Superpositon Theorem Simultaneous Equations Voltage Sources Thevenin’s Theorem Norton’s Theorem