126 Techniques of Circuit Analysis dv 2 = R 3 R 4 {R y R,I, 2 - [R 2 (R 3 + R 4 ) + R 3 R 4 ]I gl } dR, " [(R l + R 2 )(R 3 + R 4 ) + R 3 R 4 ] 2 (4.98) Figure 4.69 A Circuit used to introduce sensitivity analysis. We now consider an example with actual component values to illustrate the use of Eqs. 4.97 and 4.98. EXAMPLE Assume the nominal values of the components in the circuit in Fig. 4.69 are: R l = 25 H; R 2 = 5 ft; R 3 = 50 ft; R 4 = 75 ft; I g] = 12 A and I g2 = 16 A. Use sensitivity analysis to predict the values of v x and v 2 if the value of R^ is different by 10% from its nominal value. Solution From Eqs. 4.95 and 4.96 we find the nominal values of v v and v 2 . Thus 25(3750(16) - [5(125) + 3750112} and 30(125) + 3750 3750[30(16) - 25(12)] v 2 = . = 90 V 30(125) + 3750 (4.100) Now from Eqs. 4.97 and 4.98 we can find the sensitivity of V\ and v 2 to changes in R t . Hence dv { [3750 + 5(125)] - (3750(16) - [3750 + 5(125)]12} dRi [(30)(125) + 3750]' 12 ' (4.101) and dv 2 _ 3750(3750(16) - [5(125) + 3750J12}] tf/c7 " (7500) 2 = 0.5 V/a (4.102) How do we use the results given by Eqs. 4.101 and 4.102? Assume that Ri is 10% less than its nominal value, that is, R\ — 22.5 ft. Then Ai?! = -2.5 ft and Eq. 4.101 predicts Av x will be Av i = (^V 2 - 5 ) = -1-4583 V. Therefore, if R^ is 10% less than its nominal value, our analysis predicts that v\ will be Vi. = 25 - 1.4583 = 23.5417 V. (4.103) Similarly for Eq. 4.102 we have Av 2 = 0.5(-2.5) = -1.25 V, v 2 = 90 - 1.25 = 88.75 V. (4.104) We attempt to confirm the results in Eqs. 4.103 and 4.104 by substituting the value R^ = 22.5 ft into Eqs. 4.95 and 4.96. When we do, the results are v x = 23.4780 V, (4.105) v 2 = 88.6960 V. (4.106) Why is there a difference between the values predicted from the sensitivity analysis and the exact values computed by substituting for R^ in the equa- tions for V\ and v 2 l We can see from Eqs. 4.97 and 4.98 that the sensitivity of Vi and v 2 with respect to R^ is a function of R lt because R^ appears in the denominator of both Eqs. 4.97 and 4.98. This means that as Ri changes, the sensitivities change and hence we cannot expect Eqs. 4.97 and 4.98 to give exact results for large changes in /¾. Note that for a 10% change in R u the percent error between the predicted and exact values of v x and v 2 is small. Specifically, the percent error in v { = 0.2713% and the percent error in v 2 = 0.0676%. From this example, we can see that a tremendous amount of work is involved if we are to determine the sensitivity of v^ and v 2 to changes in the remaining component values, namely R 2f R 3t R 4f I gU and I g2 . Fortunately, PSpice has a sensitivity function that will perform sensitivity analysis for us. The sensitivity function in PSpice calculates two types of sensitivity. The first is known as the one-unit sensitivity, and the second is known as the 1% sensitivity. In the example circuit, a one-unit change in a resistor would change its value by 1 ft and a one-unit change in a current source would change its value by 1 A. In contrast, 1% sensitiv- ity analysis determines the effect of changing resistors or sources by 1% of their nominal values. The result of PSpice sensitivity analysis of the circuit in Fig. 4.69 is shown in Table 4.2. Because we are analyzing a linear circuit, we can use superposition to predict values of v\ and v 2 if more than one component's value changes. For example, let us assume /?, decreases to 24 ft and R 2 decreases to 4 ft. From Table 4.2 we can combine the unit sensitivity of V\ to changes in R { and R 2 to get Auj Av\ A#[ + 1R~ 2 = 0.5833 - 5.417 = -4.8337 V/ft, Techniques of Circuit Analysis Similarly, At?? Av 2 + AS, ' AR 2 0.5+ 6.5 = 7.0 V/a. Thus if both R { and R 2 decreased by 1II we would predict Vi = 25 + 4.8227 = 29.8337 V, v 2 = 90 - 7 = 83 V. TABLE 4.2 PSpice Sensitivity Analysis Results Element Element Name Value Element! Sensitivity (Volts/Unit) (a) DC Sensitivities of Node Voltage VI Rl 25 R2 5 R3 50 R4 75 IG1 12 IG2 16 (b) Sensitivities of Output V2 Rl 25 R2 5 R3 50 R4 75 IG1 12 IG2 16 0.5833 -5.417 0.45 0.2 -14.58 12.5 0.5 6.5 0.54 0.24 -12.5 15 Normalized Sensitivity (Volts/Percent) 0.1458 -0.2708 0.225 0.15 -1.75 2 0.125 0.325 0.27 0.18 -1.5 2.4 If we substitute R { = 24 fl and R 2 = 4 ft into Eqs. 4.95 and 4.96 we get vi = 29.793 V, v 2 = 82.759 V. In both cases our predictions are within a fraction of a volt of the actual node voltage values. Circuit designers use the results of sensitivity analysis to determine which component value variation has the greatest impact on the output of the circuit. As we can see from the PSpice sensitivity analysis in Table 4.2, the node voltages Dj and v 2 are much more sensitive to changes in R 2 than to changes in R u Specifically, V\ is (5.417/0.5833) or approximately 9 times more sensitive to changes in R 2 than to changes in R x and v 2 is (6.5/0.5) or 13 times more sensitive to changes in R 2 than to changes in Ri, Hence in the example circuit, the tolerance on R 2 must be more strin- gent than the tolerance on R^ if it is important to keep V\ and v 2 close to their nominal values. NOTE: Assess your understanding of this Practical Perspective by trying Chapter Problems 4.105-4.107. Summary 129 Summary • For the topics in this chapter, mastery of some basic terms, and the concepts they represent, is necessary. Those terms are node, essential node, path, branch, essential branch, mesh, and planar circuit. Table 4.1 provides definitions and examples of these terms. (See page 91.) • Two new circuit analysis techniques were introduced in this chapter: • The node-voltage method works with both planar and nonplanar circuits. A reference node is chosen from among the essential nodes. Voltage variables are assigned at the remaining essential nodes, and Kirchhoff s current law is used to write one equation per voltage variable. The number of equations is n e — 1, where n e is the number of essential nodes. (See page 93.) • The mesh-current method works only with planar circuits. Mesh currents are assigned to each mesh, and Kirchhoff's voltage law is used to write one equation per mesh. The number of equations is b — (n — 1), where b is the number of branches in which the current is unknown, and n is the number of nodes. The mesh currents are used to find the branch currents. (See page 99.) • Several new circuit simplification techniques were introduced in this chapter: • Source transformations allow us to exchange a volt- age source (v s ) and a series resistor (R) for a current source (i s ) and a parallel resistor (R) and vice versa. The combinations must be equivalent in terms of their terminal voltage and current. Terminal equiva- lence holds provided that s R (See page 109.) • Thevenin equivalents and Norton equivalents allow us to simplify a circuit comprised of sources and resis- tors into an equivalent circuit consisting of a voltage source and a series resistor (Thevenin) or a current source and a parallel resistor (Norton). The simplified circuit and the original circuit must be equivalent in terms of their terminal voltage and current. Thus keep in mind that (1) the Thevenin voltage (Kiii) is the open-circuit voltage across the terminals of the original circuit, (2) the Thevenin resistance (i? T h) is the ratio of the Thevenin voltage to the short-circuit current across the terminals of the original circuit; and (3) the Norton equivalent is obtained by per- forming a source transformation on a Thevenin equivalent. (See page 113.) • Maximum power transfer is a technique for calculating the maximum value of p that can be delivered to a load, R L . Maximum power transfer occurs when Ri = Rjh, the Thevenin resistance as seen from the resistor R L . The equation for the maximum power transferred is (See page 120.) • In a circuit with multiple independent sources, superposition allows us to activate one source at a time and sum the resulting voltages and currents to deter- mine the voltages and currents that exist when all inde- pendent sources are active. Dependent sources are never deactivated when applying superposition. (See page 122.) 130 Techniques of Circuit Analysis Problems Section 4.1 4.1 For the circuit shown in Fig. P4.1, state the numerical value of the number of (a) branches, (b) branches where the current is unknown, (c) essential branches, (d) essential branches where the current is unknown, (e) nodes, (f) essential nodes, and (g) meshes. Figure P4.1 4.4 Assume the current i g in the circuit in Fig. P4.4 is known. The resistors R^ - R 5 are also known. a) How many unknown currents are there? b) How many independent equations can be writ- ten using Kirchhoff s current law (KCL)? c) Write an independent set of KCL equations. d) How many independent equations can be derived from Kirchhoff s voltage law (KVL)? e) Write a set of independent KVL equations. Figure P4.4 V L 4.2 a) If only the essential nodes and branches are identified in the circuit in Fig. P4.1, how many simultaneous equations are needed to describe the circuit? b) How many of these equations can be derived using Kirchhoff s current law? c) How many must be derived using Kirchhoffs voltage law? d) What two meshes should be avoided in applying the voltage law? 4.3 a) How many separate parts does the circuit in Fig. P4.3 have? b) How many nodes? c) How many branches are there? d) Assume that the lower node in each part of the circuit is joined by a single conductor. Repeat the calculations in (a)-(c). Figure P4.3 4.5 A current leaving a node is defined as positive. a) Sum the currents at each node in the circuit shown in Fig. P4.4. b) Show that any one of the equations in (a) can be derived from the remaining three equations. Section 4.2 4.6 Use the node-voltage method to find Uj and v 2 in PSPICE the circuit in Fig. P4.6. Figure P4.6 144 V son 4.7 Use the node-voltage method to find how much PSPICE power the 2 A source extracts from the circuit in HULTISIM pjg p4J Figure P4.7 2AI so a 45 V Problems 131 4.8 Use the node-voltage method to find v x and v 2 in PSPICE the circuit shown in Fig. P4.8. MULTISIM Figure P4.8 8ft -VvV- <>i|40ft kmVL u 2 f 120ft C\ J1A 4.9 Use the node-voltage method to find v () in the cir- PSPICE cuit in Fig. P4.9. Figure P4.9 2011 24 V 8012 -A^ m 40 mA 4.10 a) Find the power developed by the 40 mA current PSPKE source in the circuit in Fig. P4.9. b) Find the power developed by the 24 V voltage source in the circuit in Fig. P4.9. c) Verify that the total power developed equals the total power dissipated. 4.11 A 50 O resistor is connected in series with the PSPICE 40 mA current source in the circuit in Fig. P4.9. MULTISIM N ^,. . a) Find v a . b) Find the power developed by the 40 mA current source. c) Find the power developed by the 24 V voltage source. d) Verify that the total power developed equals the total power dissipated. e) What effect will any finite resistance connected in series with the 40 mA current source have on the value of v a l 4.12 The circuit shown in Fig. P4.12 is a dc model of a PSPICE residential power distribution circuit. a) Use the node-voltage method to find the branch currents i { — / 6 . b) Test your solution for the branch currents by showing that the total power dissipated equals the total power developed. Figure P4.12 125 V 125 V ':, 4.13 a) Use the node-voltage method to find the PSPKE branch currents / a - / e in the circuit shown in MULTISIM Fi g.p4 >13 . b) Find the total power developed in the circuit. Figure P4.13 128 V 320 V 4.14 Use the node-voltage method to find the total power PSPICE dissipated in the circuit in Fig. P4.14. MULTISIM Figure P4.14 40 V 132 Techniques of Circuit Analysis 4.15 a) Use the node-voltage method to find v h v 2 , and !™. v 3 in the circuit in Fig. P4.15. MULTISIM b) How much power does the 40 V voltage source deliver to the circuit? Figure P4.15 40 V 4.16 a) Use the node-voltage method to show that the PSPICE output voltage v 0 in the circuit in Fig. P4.16 is equal to the average value of the source voltages. b) Find v 0 if v { = 100 V, v 2 = 80 V, and v 3 = -60 V. Figure P4.16 4.19 Use the node-voltage method to calculate the PSPICE power delivered by the dependent voltage source in WLTISIM ., . ^. „, n the circuit in Fig. P4.19. Figure P4.19 160 V ion 3011 ('„ I loo a 150 L 20 a ^vw- Section 4.3 4.17 a) Use the node-voltage method to find the total power developed in the circuit in Fig. P4.17. MULTISIM b) Check your answer by finding the total power absorbed in the circuit. Figure P4.17 84/ A PSPICE MULTISIM 4.18 a) Use the node-voltage method to find v„ in the circuit in Fig. P4.18. b) Find the power absorbed by the dependent source. c) Find the total power developed by the independ- ent sources. Figure P4.18 20 a 3A 80 V 4.20 a) Find the node voltages V\, «2, and v$ in the cir- PSPICE cu it in Fig. P4.20. •IULTISIM b) Find the total power dissipated in the circuit. Figure P4.20 5 n 10 a m VA f VvV- f\5/ ( , iv*20ft y 2 ?40O yyk (^/) 96 V Section 4.4 4.21 Use the node-voltage method to find i 0 in the cir- PSPICE cuit in Fig. P4.21. MULTISIM Figure P4.21 2kft 5kft 20 V 6 30kll lkft PSPICE MULTISIM 4.22 a) Use the node-voltage method to find the branch currents ij, i 2 , and /3 in the circuit in Fig. P4.22. b) Check your solution for ij, i 2 , and i 3 by showing that the power dissipated in the circuit equals the power developed. Figure P4.22 30 V 80 V Problems 133 4.23 a) Use the node-voltage method to find the power dissipated in the 2 (2 resistor in the circuit in Fig. P4.23. b) Find the power supplied by the 230 V source. Figure P4.23 230 v C-y in in in: la- in l n in sn 2n :5 n 4.24 Use the node-voltage method to find the value of v 0 KM « in the circuit in Fig. P4.24. MUITISIM Figure P4.26 25 V 55 n 4.27 Use the node-voltage method to find v 0 in the cir- PSPICE cuit in Fig. P4.27. MULTISIM Figure P4.27 15 V Figure P4.24 50 V 4.25 Use the node-voltage method to find the value of v a PSPICE in the circuit in Fig. P4.25. «!ULTISIM Figure P4.25 4.26 a) Use the node-voltage method to find v 0 and PSPICE the power delivered by the 2 A current source in the circuit in Fig. P4.26. Use node a as the reference node. b) Repeat part (a), but use node b as the refer- ence node. c) Compare the choice of reference node in (a) and (b). Which is better, and why? 4.28 Use the node-voltage method to find the power devel- oped by the 20 V source in the circuit in Fig. P4.28. PSPICE MULTISIM Figure P4.28 80n (1)3.125 v A 4.29 Assume you are a project engineer and one of your PSPICE MULTISIM staff is assigned to analyze the circuit shown in Fig. P4.29. The reference node and node numbers given on the figure were assigned by the analyst. Her solution gives the values of v 3 and v 4 as 108 V and 81.6 V, respectively. Test these values by checking the total power developed in the circuit against the total power dis- sipated. Do you agree with the solution submitted by the analyst? Figure P4.29 20 n 1)1.75¾ 134 Techniques of Circuit Analysis 4.30 Show that when Eqs. 4.16,4.17, and 4.19 are solved for i B , the result is identical to Eq. 2.25. Section 4.5 4.31 Solve Problem 4.12 using the mesh-current method. 4.32 Solve Problem 4.13 using the mesh-current method. 4.33 a) Use the mesh-current method to find the branch currents L, //„ and L in the circuit in Fig. P4.33. MULTISIM b b) Repeat (a) if the polarity of the 60 V source is reversed. Figure P4.33 60 V to a 20 V 4.34 a) Use the mesh-current method to find the total power developed in the circuit in Fig. P4.34. MULTISIM b) Check your answer by showing that the total power developed equals the total power dissipated. Figure P4.34 460 V 4.35 Solve Problem 4.21 using the mesh-current method. 4.36 Solve Problem 4.23 using the mesh-current method. Section 4.6 4.37 a) Use the mesh-current method to find v 0 in the PSPICE circuit in Fig. P4.37. MULTISIM b) Find the power delivered by the dependent source. Figure P4.37 io a 7 a 4.38 Use the mesh-current method to find the power dissi- PSPICE pate( j in the 20 fl resistor in the circuit in Fig. P4.38. MULTISIM Figure P4.38 5 0 3 a •AW 135 V © 4ft 20 a 10 i. 2 a la -"VW 4.39 Use the mesh-current method to find the power delivered by the dependent voltage source in the JLTISIM circuit seen in Fig. P4.39. Figure P4.39 660 V 25 a 20 /,, 4.40 Use the mesh-current method to find the power >SPICE developed in the dependent voltage source in the JLTISIM . . _. _ . ._ circuit in Fig. P4.40. Figure P4.40 30 V Section 4.7 4.41 Solve Problem 4.8 using the mesh-current method. 4.42 a) Use the mesh-currcnt method to find how much power the 4 A current source delivers to the cir- cuit in Fig. P4.42. b) Find the total power delivered to the circuit. c) Check your calculations by showing that the total power developed in the circuit equals the total power dissipated Problems 135 Figure P4.42 120 V 5 0 80 V 4.43 Solve Problem 4.20 using the mesh-current method. 4.44 a) Use the mesh-current method to solve for i± in PSPICE the circuit in Fig. P4.44. MULTISIM b) Find the power delivered by the independent current source. c) Find the power delivered by the dependent volt- age source. 4.47 Solve Problem 4.22 using the mesh-current method. 4.48 Use the mesh-current method to find the total PSPICE power dissipated in the circuit in Fig. P4.48. MULTISIM Figure P4.48 20 V 90 V 4.49 a) Assume the 20 V source in the circuit in Fig. P4.48 is changed to 60 V. Find the total power dissipated in the circuit. b) Repeat (a) if the 6 A current source is replaced by a short circuit. c) Explain why the answers to (a) and (b) are the same. Figure P4.44 ikn 150 k 4.45 Use the mesh-current method to find the total power PSPICE developed in the circuit in Fig. P4.45. MULTISIM Figure P4.45 20 A 6.5 / A 4.46 a) Use the mesh-current method to determine PSPICE which sources in the circuit in Fig. P4.46 are gen- IULTISIM crating power. b) Find the total power dissipated in the circuit. Figure P4.46 2ft 4.50 a) Find the branch currents / a - i c for the circuit nna shown in Fig. P4.50. MULTISIM b) Check your answers by showing that the total power generated equals the total power dissipated. Figure P4.50 4/ d -i c 19 A 4.51 a) Use the mesh-current method to find the branch PSPICE MULTISIM currents in i a — i e in the circuit in Fig. P4.51. b) Check your solution by showing that the total power developed in the circuit equals the total power dissipated. Figure P4.51 100ft