1. Network Theorems Topics Covered in Chapter 8 8-1: Kirchhoff’s Current Law (KCL) 8-2: Kirchhoff’s Voltage Law (KVL) 8-3: Superposition Theorem 8-4: Thevenin’s Theorem Chapter 8 8 © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

2. 8-2 Topics Covered in Chapter 8 8-5: Thevenizing a Bridge Circuit 8-6: Voltage Sources 8-7: Maximum Power, Transfer Theorem

3. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-3 8-1: Kirchhoff’s Current Law (KCL)  The sum of currents entering any point in a circuit is equal to the sum of currents leaving that point.  Otherwise, charge would accumulate at the point, reducing or obstructing the conducting path.  Kirchhoff’s Current Law may also be stated as I IN = I OUT Fig. 8-1: Current I C out from point P equals 5A + 3A into P.

4. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-4 8-1: Kirchhoff’s Current Law (KCL) Fig. 8-2: Series-parallel circuit illustrating Kirchhoff’s laws.

5. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-5 8-1: Kirchhoff’s Current Law (KCL) The 6-A I T into point C divides into the 2-A I 3 and 4-A I 4-5 I 4-5 is the current through R 4 and R 5 I T − I 3 − I 4-5 = 0 6A − 2A − 4A = 0 At either point C or point D, the sum of the 2-A and the 4-A branch currents must equal the 6A line current. Therefore, I in = I out

6. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-6 8-2: Kirchhoff’s Voltage Law (KVL)  Loop Equations  A loop is a closed path.  This approach uses the algebraic equations for the voltage around the loops of a circuit to determine the branch currents.  Use the IR drops and KVL to write the loop equations.  A loop equation specifies the voltages around the loop.

7. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-7 8-2: Kirchhoff’s Voltage Law (KVL)  Loop Equations  ΣV = V T means the sum of the IR voltage drops must equal the applied voltage. This is another way of stating Kirchhoff’s Voltage Law.

8. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-8 8-2: Kirchhoff’s Voltage Law (KVL) Fig. 8-2: Series-parallel circuit illustrating Kirchhoff’s laws.

9. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-9 8-2: Kirchhoff’s Voltage Law (KVL) In Figure 8-2, for the inside loop with the source V T, going counterclockwise from point B, 90V + 120V + 30V = 240V If 240V were on the left side of the equation, this term would have a negative sign. The loop equations show that KVL is a practical statement that the sum of the voltage drops must equal the applied voltage.

10. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-10 8-2: Kirchhoff’s Voltage Law (KVL)  The algebraic sum of the voltage rises and IR voltage drops in any closed path must total zero. For the loop CEFDC without source the equation is −V 4 − V 5 + V 3 = 0 −40V − 80V + 120V = 0 0 = 0 Fig. 8-2: Series-parallel circuit illustrating Kirchhoff’s laws.

11. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-11 8-3: Superposition Theorem  The superposition theorem extends the use of Ohm’s Law to circuits with multiple sources.  In order to apply the superposition theorem to a network, certain conditions must be met: 1. All the components must be linear, meaning that the current is proportional to the applied voltage.

12. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-12 8-3: Superposition Theorem 2. All the components must be bilateral, meaning that the current is the same amount for opposite polarities of the source voltage. 3. Passive components may be used. These are components such as resistors, capacitors, and inductors, that do not amplify or rectify. 4. Active components may not be used. Active components include transistors, semiconductor diodes, and electron tubes. Such components are never bilateral and seldom linear.

13. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-13 8-3: Superposition Theorem  In a linear, bilateral network that has more than one source, the current or 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.

14. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-14 8-3: Superposition Theorem Fig. 8-5: Superposition theorem applied to a voltage divider with two sources V 1 and V 2. (a) Actual circuit with +13 V from point P to chassis ground. (b) V 1 alone producing +16 V at P. (c) V 2 alone producing −3 V at P.

15. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-15 8-3: Superposition Theorem 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

16. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-16 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 8-3: Superposition Theorem (Applied)

17. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-17 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 A 0.406 A 8-3: Superposition Theorem (Applied)

18. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-18 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 8-3: Superposition Method (Check)

19. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-19 8-4: Thevenin’s Theorem  Thevenin’s theorem simplifies the process of solving for the unknown values of voltage and current in a network by reducing the network to an equivalent series circuit connected to any pair of network terminals.  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.

20. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-20 8-4: Thevenin’s Theorem Fig. 8-7: Application of Thevenin’s theorem. (a) Actual circuit with terminals A and B across R L. (b) Disconnect R L to find that V AB is 24V. (c) Short-circuit V to find that R AB is 2Ω.

21. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-21 8-4: Thevenin’s Theorem Fig. 8-7 (d) Thevenin equivalent circuit. (e) Reconnect R L at terminals A and B to find that V L is 12V.

22. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-22 8-4: Thevenin’s Theorem  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.

23. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-23 8-4: Thevenin’s Theorem Fig. 8-8: Thevenizing the circuit of Fig. 10-3 but with a 4-Ω R 3 in series with the A terminal. (a) V AB is still 24V. (b) Now the R AB is 2 + 4 = 6 Ω. (c) Thevenin equivalent circuit. Note that R 3 does not change the value of V AB produced by the source V, but R 3 does increase the value of R TH.

24. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-24 8-5: Thevenizing a Bridge Circuit  A Wheatstone Bridge Can Be Thevenized.  Problem: Find the voltage drop across R L.  The bridge is unbalanced and Thevenin’s theorem is a good choice.  R L will be removed in this procedure making A and B the Thevenin terminals. Fig. 8-9: Thevenizing a bridge circuit. (a) Original circuit with terminals A and B across middle resistor R L.

25. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-25 8-5: Thevenizing a Bridge Circuit Fig. 8-9(b) Disconnect R L to find V AB of −8 V. (c) With source V short-circuited, R AB is 2 + 2.4 = 4.4 Ω. V AB = −20 −(−12) = −8V R AB = R TA + R TB = 2 + 2.4 = 4.4 Ω

26. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-26 8-5: Thevenizing a Bridge Circuit Fig. 8-9(d) Thevenin equivalent with R L reconnected to terminals A and B.

27. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-27 8-6: Voltage Sources  An ideal dc voltage source produces a load voltage that is constant.  An example is a perfect battery, one whose internal resistance is zero.  An ideal dc voltage source produces a constant load voltage, regardless of how small or large the load resistance is.  With an ideal voltage source, only load current changes when the load resistance changes.

28. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-28 8-6: Voltage Sources Second Approximation  An ideal voltage source cannot exist in nature.  When the load resistance approaches zero, the load current approaches infinity.  No real voltage source can produce infinite current because a real voltage source always has some internal resistance.  The second approximation of a dc voltage source includes this internal resistance.

29. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-29 8-6: Voltage Sources Second Approximation  The source (internal) resistance can be ignored when it is at least 100 times smaller than the load resistance.  Any source that satisfies this condition is referred to as a stiff voltage source.  Using the second approximation for an ac voltage source is valid only at low frequencies.  At high frequencies, factors such as lead inductance and stray capacitance come into play.

30. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-30 8-7: Maximum Power, Transfer Theorem  When transferring power from a voltage source to a load, it is often necessary to transfer maximum power to the load.  Maximum power is transferred to the load when the load resistance (R L ) equals the internal resistance of the generator (r i ).

31. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-31 8-7: Maximum Power, Transfer Theorem In Figure 8-13, when R L equals r i, the load and generator are matched. When they are matched, the generator then produces maximum power in R L. Figure 8-13

32. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8-32 8-7: Maximum Power, Transfer Theorem  If maximum voltage is desired, the load should have as high a resistance as possible.  Efficiency increases as R L in Figure 8-13 (previous slide) increases.  Efficiency increases because there is less current, resulting in less power lost in r i.  Matching the load and generator resistances is desirable when the load requires maximum power, rather than maximum voltage or efficiency, assuming that the match does not result in excessive current.