# Basic Electronics Ninth Edition Basic Electronics Ninth Edition ©2002 The McGraw-Hill Companies Grob Schultz.

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Basic Electronics Ninth Edition Basic Electronics Ninth Edition ©2002 The McGraw-Hill Companies Grob Schultz

Basic Electronics Ninth Edition Basic Electronics Ninth Edition ©2003 The McGraw-Hill Companies 10 CHAPTER Network Theorems

Topics Covered in Chapter 10  Superposition Method  Thevenin’s Theorem  Norton’s Theorem

Topics Covered in Chapter 10  Conversion of Voltage and Current Sources  Millman’s Theorem   and Y Conversions

Superposition Theorem In a linear, bilateral network that has more than one source, the current and voltage in any part of the network can be found by adding algebraically the effect of each source separately. This analysis is done by:  shorting each voltage source in turn  opening each current source in turn

Superposition Method Applied R1R1 R2R2 R3R3 V1V1 V2V2 100  20  10  15 V13 V R1R1 R2R2 R3R3 V1V1 100  20  10  15 V V 2 shorted R EQ = 106.7 , I T = 0.141 A and I R 3 = 0.094 A

Superposition Method Applied R1R1 R2R2 R3R3 V1V1 V2V2 100  20  10  15 V13 V R EQ = 29.09 , I T = 0.447 A and I R 3 = 0.406 A R1R1 R2R2 R3R3 V2V2 100  20  10  13 V V 1 shorted

Superposition Method Applied R1R1 R2R2 V1V1 V2V2 100  20  15 V13 V Adding the currents gives I R 3 = 0.5 A R EQ = 106.7 , I T = 0.141 A and I R 3 = 0.094 A R EQ = 29.09 , I T = 0.447 A and I R 3 = 0.406 A With V 2 shorted With V 1 shorted 0.094 A0.406 A

Checking the Superposition Solution R1R1 R2R2 R3R3 V1V1 V2V2 100  20  10  15 V13 V With 0.5 A flowing in R 3, the voltage across R 3 must be 5 V (Ohm’s Law). The voltage across R 1 must therefore be 10 volts (KVL) and the voltage across R 2 must be 8 volts (KVL). Solving for the currents in R 1 and R 2 will verify that the solution agrees with KCL. 0.5 A I R 1 = 0.1 A and I R 2 = 0.4 A I R 3 = 0.1 A + 0.4 A = 0.5 A

Thevenin’s Theorem Any network with two open terminals can be replaced by a single voltage source ( V TH ) and a series resistance ( R TH ) connected to the open terminals. A component can be removed to produce the open terminals.

A Wheatstone bridge can be Thevenized. 30  40  R1R1 R2R2 60  40  R3R3 R4R4 A B 45 V R5R5 Problem: Find the voltage drop across R 5. The bridge is unbalanced and Thevenin’s theorem is a good choice. R 5 will be removed in this procedure making A and B the Thevenin terminals.

Determining Thevenin Resistance and Voltage R TH is determined by shorting the voltage source and calculating the circuit’s total resistance as seen from open terminals A and B. V TH is determined by calculating the voltage between open terminals A and B.

Applying Thevenin’s Theorem 30  40  R1R1 R2R2 60  40  R3R3 R4R4 A B 45 V R5R5 R 5 is removed.The voltage source is shorted. R TH is the resistance from A to B. R TH = 40  30  40  60  40  R TH AB

Applying Thevenin’s Theorem 30  40  R1R1 R2R2 60  40  R3R3 R4R4 A B 45 V R5R5 Choose a reference point. V TH is the voltage from A to B. V TH = 7.5 V Remove R 5. V A = 60 90 x 45 = 30 V V B = 40 80 x 45 = 22.5 V V AB = 30 – 22.5 = 7.5 V

Applying Thevenin’s Theorem R TH 7.5 V V TH 40  A B The Thevenin equivalent circuit 40  R5R5 Connect R 5 to the equivalent circuit. V R 5 = 40 80 x 7.5 = 3.75 V

Norton’s Theorem Any network with two terminals can be replaced by a single current source and parallel resistance connected across the terminals. The two terminals are usually labeled something such as A and B The Norton current is usually labeled I N The Norton resistance is usually labeled R N

Determining Norton Current and Voltage I N is determined by calculating the current through a short placed across terminals A and B. R N is determined by shorting the voltage source and calculating the circuit’s total resistance as seen from open terminals A and B (same procedure as for R TH ).

A Wheatstone bridge can be Nortonized. 30  40  R1R1 R2R2 60  40  R3R3 R4R4 A B 45 V R5R5 Replace R 5 with a short and determine I N. R EQ = 41.14  I T = 1.094 A I R 2 = 0.6563 A I R 1 = 0.4689 A I N = 0.1874 A Apply the current divider. Apply KCL. ITIT ITIT IR2IR2 IR1IR1 ININ R N = R TH

The Norton Equivalent Circuit RNRN ININ 40  A B R5R5 0.1874 A Connect R 5. Apply the current divider. Use Ohm’s Law. The circle is a symbol for a current source. It provides 0.1874 A total flow, regardless of what is connected across it. With no load, all of the current will flow in R N. When shorted, all of the current will flow in the short. I R 5 = 0.0937 A V R 5 = 3.75 V

Conversion of Real Sources R I 40  A B 0.1874 A R 7.5 V V 40  A B All real voltage sources have a series resistance. All real current sources have a parallel resistance. Use Ohm’s Law to convert from one to the other. The resistance stays the same. 7.5 V 40  0.1874 A x 40 

Millman’s Theorem The common voltage across parallel branches with different voltage sources can be determined by:

Delta-to-Wye Conversion A delta (  ) circuit can be converted to a wye (Y) equivalent circuit by: R1R1 R2R2 R3R3 RARA RBRB RCRC

A Wheatstone bridge can be simplified. 30  40  R1R1 R2R2 60  40  R3R3 R4R4 45 V R5R5 It’s possible to convert this delta section to an equivalent wye. R EQ = 41.74  I T = 1.078 A Apply Ohm’s Law. V A = 28.12 and V B = 24.37 V AB = 3.75 V 0.4687 A 0.6093 A AB 60  40  45 V 10.91  14.55  AB Apply current divider here.

Wye-to-delta Conversion A wye (Y) circuit can be converted to a delta (  ) equivalent circuit by: R1R1 R2R2 R3R3 RARA RBRB RCRC