106 Techniques of Circuit Analysis I/ASSESSMENT PROBLEMS Objective 2—Understand and be able to use the mesh-current method 4.10 Use the mesh-current method to find the power dissipated in the 2 ft resistor in the circuit shown. 30 V Answer: 72 W. 4.11 Use the mesh-current method to find the mesh current / a in the circuit shown. 75 VI Answer: 15 A. 4.12 Use the mesh-current method to find the power dissipated in the 1 ft resistor in the cir- cuit shown. 16 A ^vw 4 Wv 4 10 A e 20 'a <\t, 15 n in •A^V- 2V* 2A e iovT*j 2 a Answer: 36 W. NOTE: Also try Chapter Problems 4.42, 4.44, 4.48, and 4.51. 6V 4,8 The Node-Voltage Method Versus the Mesh-Current Method The greatest advantage of both the node-voltage and mesh-current meth- ods is that they reduce the number of simultaneous equations that must be manipulated. They also require the analyst to be quite systematic in terms of organizing and writing these equations. It is natural to ask, then, "When is the node-voltage method preferred to the mesh-current method and vice versa?" As you might suspect, there is no clear-cut answer. Asking a number of questions, however, may help you identify the more efficient method before plunging into the solution process: • Does one of the methods result in fewer simultaneous equations to solve? • Does the circuit contain supernodes? If so, using the node-voltage method will permit you to reduce the number of equations to be solved. 4.8 The Node-Voltage Method Versus the Mesh-Current Method 107 • Does the circuit contain supermeshes? If so, using the mesh-current method will permit you to reduce the number of equations to be solved. • Will solving some portion of the circuit give the requested solution? If so, which method is most efficient for solving just the pertinent portion of the circuit? Perhaps the most important observation is that, for any situation, some time spent thinking about the problem in relation to the various analytical approaches available is time well spent. Examples 4.6 and 4.7 illustrate the process of deciding between the node-voltage and mesh-current methods. Example 4.6 Understanding the Node-Voltage Method Versus Mesh-Current Method Find the power dissipated in the 300 Ct resistor in the circuit shown in Fig. 4.29. 300 ft J 1 - -Wv 150 ft 100ft -AAA^ 250 ft -AVv t 50 / A 500 ft 256 V £200 ft 400 ft k 128 V Figure 4.29 A The circuit for Example 4.6. Solution To find the power dissipated in the 300 H resistor, we need to find either the current in the resistor or the voltage across it. The mesh-current method yields the current in the resistor; this approach requires solving five simultaneous mesh equations, as depicted in Fig. 4.30. In writing the five equa- tions, we must include the constraint / A = — i b . Before going further, let's also look at the circuit in terms of the node-voltage method. Note that, once we know the node voltages, we can calculate either the current in the 300 il resistor or the voltage across it. The circuit has four essential nodes, and therefore only three node-voltage equations are required to describe the circuit. Because of the dependent volt- age source between two essential nodes, we have to sum the currents at only two nodes. Hence the prob- lem is reduced to writing two node-voltage equations and a constraint equation. Because the node-voltage method requires only three simultaneous equations, it is the more attractive approach. Once the decision to use the node-voltage method has been made, the next step is to select a reference node. Two essential nodes in the circuit in Fig. 4.29 merit consideration. The first is the refer- ence node in Fig. 4.31. If this node is selected, one of the unknown node voltages is the voltage across the 300 CL resistor, namely, v 2 in Fig. 4.31. Once we know this voltage, we calculate the power in the 300 f! resistor by using the expression Psaon = «1/300. 300 ft -*' A -AAA. 150ft 100ft 250 ft -AAA-—f—-WV- 500 ft : f20uftSO 50 *A)f .+. Figure 4.30 A The circuit shown in Fig. 4.29, with the five mesh currents. 300 ft J s vvV 150 ft <^ 500 ft 400 ft £ 128 V Figure 4.31 A The circuit shown in Fig. 4.29, with a reference node. Note that, in addition to selecting the reference node, we defined the three node voltages V\, v 2 , and u 3 and indicated that nodes 1 and 3 form a super- node, because they are connected by a dependent voltage source. It is understood that a node voltage is a rise from the reference node; therefore, in Fig. 4.31, we have not placed the node voltage polarity refer- ences on the circuit diagram. 108 Techniques of Circuit Analysis The second node that merits consideration as the reference node is the lower node in the circuit, as shown in Fig. 4.32. It is attractive because it has the most branches connected to it, and the node- voltage equations are thus easier to write. However, to find either the current in the 300 11 resistor or the voltage across it requires an additional calcula- tion once we know the node voltages v a and v c . For example, the current in the 300 H resistor is (v c - v a )/300, whereas the voltage across the resis- tor is v r - v». 300 ft l* 128 V Figure 4.32 A The circuit shown in Fig. 4.29 with an alternative reference node. t? 3 + 256 + — = 0. 150 Att? 2 , v 2 v 2 - Vi v 2 - VT. VI + 128 - v-x ' + _•_ + + — — = 0. 300 250 400 500 From the supernode, the constraint equation is v 2 v 3 = vi - 50/ A = v, - — Set 2 (Fig 4.32) At%, v a v a - 256 v. d - v b v a - v c _ 200 150 Atv c , 100 300 We compare these two possible reference nodes by means of the following sets of equations. The first set pertains to the circuit shown in Fig. 4.31, and the second set is based on the circuit shown in Fig. 4.32. Set 1 (Fig 4.31) At the supernode, V\ V\ — v 2 v 3 V3 — v 2 v 3 — (v 2 + 128) KX) + 250 + 200 + 400 500 v c v c + 128 v c — v b v c - v a 400 500 250 300 From the supernode, the constraint equation is 50(v c - v a ) v c - v a v b = 50/ A 300 You should verify that the solution of either set leads to a power calculation of 16.57 W dissipated in the 300 O, resistor. Example 4.7 Companng the Node-Voltage and Mesh-Current Methods Find the voltage v 0 in the circuit shown in Fig. 4.33. Solution At first glance, the node-voltage method looks appealing, because we may define the unknown voltage as a node voltage by choosing the lower ter- minal of the dependent current source as the refer- ence node. The circuit has four essential nodes and two voltage-controlled dependent sources, so the node-voltage method requires manipulation of three node-voltage equations and two constraint equations. Let's now turn to the mesh-current method for finding v 0 . The circuit contains three meshes, and we can use the leftmost one to calculate v 0 . If we let / a denote the leftmost mesh current, then v 0 = 193 — 10/' a . The presence of the two current sources reduces the problem to manipulating a sin- gle supermesh equation and two constraint equa- tions. Hence the mesh-current method is the more attractive technique here. 4ft 0.8¾ 6 ft 7.5 ft 8ft Figure 4.33 • The circuit for Example 4.7. 4.9 Source Transformations 109 411 2.512 193 VI r ~ - "A ,, W AC )^D^ -AMr- -VvV -Wr 6 n 7.5 n 8 a Figure 4.34 A The circuit shown in Fig. 4.33 with the three mesh currents. 4 0 2.5 ft *« 2 ft -'vw f -vw- Ml93V^/t>0.4^ (T)o.5A 60 V^/v and the constraint equations are i\j — / a = 0.4i; A = 0.8/ c ; v 9 = — 7.5/ b ; and / c — / b = 0.5. We use the constraint equations to write the super- mesh equation in terms of / a : 160 = 80*' a , or / a = 2 A, v a = 193 - 20 = 173 V. The node-voltage equations are 0.8 v (l 7.5 ft "h Figure 4.35 A The circuit shown in Fig. 4.33 with node voltages. To help you compare the two approaches, we summarize both methods.The mesh-current equa- tions are based on the circuit shown in Fig. 4.34, and the node-voltage equations are based on the circuit shown in Fig. 4.35. The supermesh equation is 193 = 104 + 10 'b + 10/'c + 0.8v 0 , v„ - 193 10 A 2.5 0. 2.5 10 ^ b t 0 5 1 % + °' 8Ve ~ Va ~ 0 7.5 " 10 The constraint equations are v 9 = -v b , v A = "« a - (v b + 0.8¾¾) 1 10 2. We use the constraint equations to reduce the node- voltage equations to three simultaneous equations involving v () , u a , and v b . You should verify that the node-voltage approach also gives v 0 = 173 V. ^ASSESSMENT PROBLEMS Objective 3—Deciding between the node-voltage and mesh-current methods 4.13 Find the power delivered by the 2 A current source in the circuit shown. 4.14 Find the power delivered by the 4 A current source in the circuit shown. 4A 20 V 128 V Answer: 70 W NOTE: Also try Chapter Problems 4.52 and 4.53. Answer: 40 W. 4.9 Source Transformations Even though the node-voltage and mesh-current methods are powerful tech- niques for solving circuits, we are still interested in methods that can be used to simplify circuits. Series-parallel reductions and A-to-Y transformations are 110 Techniques of Circuit Analysis -•a -•b (b) Figure 4.36 A Source transformations. already on our list of simplifying techniques. We begin expanding this list with source transformations. A source transformation, shown in Fig. 4.36, allows a voltage source in series with a resistor to be replaced by a current source in parallel with the same resistor or vice versa. The double-headed arrow emphasizes that a source transformation is bilateral; that is, we can start with either configuration and derive the other. We need to find the relationship between v s and i s that guarantees the two configurations in Fig. 4.36 are equivalent with respect to nodes a,b. Equivalence is achieved if any resistor R L experiences the same current flow, and thus the same voltage drop, whether connected between nodes a,b in Fig. 4.36(a) or Fig. 4.36(b). Suppose R f is connected between nodes a,b in Fig. 4.36(a). Using Ohm's law, the current in R L is tL (4.52) R + R L Now suppose the same resistor R L is connected between nodes a,b in Fig. 4.36(b). Using current division, the current in R, is R '/. (4.53) if the two circuits in Fig. 4.36 are equivalent, these resistor currents must be the same. Equating the right-hand sides of Eqs. 4.52 and 4.53 and simplifying. (4.54) When Eq. 4.54 is satisfied for the circuits in Fig. 4.36, the current in R L is the same for both circuits in the figure for all values of R L . If the current through R L is the same in both circuits, then the voltage drop across R { is the same in both circuits, and the circuits are equivalent at nodes a,b. If the polarity of v s is reversed, the orientation of i s must be reversed to maintain equivalence. Example 4.8 illustrates the usefulness of making source transforma- tions to simplify a circuit-analysis problem. Example 4.8 Using Source Transformations to Solve a Circuit a) For the circuit shown in Fig. 4.37, find the power associated with the 6 V source. b) State whether the 6 V source is absorbing or delivering the power calculated in (a). Solution a) If we study the circuit shown in Fig. 4.37, know- ing that the power associated with the 6 V source is of interest, several approaches come to mind. The circuit has four essential nodes and six essential branches where the current is unknown. Thus we can find the current in the branch containing the 6 V source by solving either three [6-(4-1)] mesh-current equa- tions or three [4-1] node-voltage equations. Choosing the mesh-current approach involves 6 V §30 n ?20£l 40 V Figure 4.37 • The circuit for Example 4.8. solving for the mesh current that corresponds to the branch current in the 6 V source. Choosing the node-voltage approach involves solving for the voltage across the 30 O resistor, from which the branch current in the 6 V source can be calculated. But by focusing on just one branch current, we can first simplify the circuit by using source transformations. 4.9 Source Transformations 111 We must reduce the circuit in a way that pre- serves the identity of the branch containing the 6 V source. We have no reason to preserve the identity of the branch containing the 40 V source. Beginning with this branch, we can transform the 40 V source in series with the 5 ft resistor into an 8 A current source in parallel with a5fi resistor, as shown in Fig. 4.38(a). 4 n 6 ii -f 'WV 32 V (a) First step 4 11 (b) Second step 412 12 O 19.2 V (c) Third step Figure 4.38 A Step-by-step simplification of the circuit shown in Fig. 4.37. Next, we can replace the parallel combination of the 20 ft and 5 ft resistors with a 4 ft resistor. This 4 ft resistor is in parallel with the 8 A source and therefore can be replaced with a 32 V source in series with a 4 ft resistor, as shown in Fig. 4.38(b).The 32 V source is in series with 20 ft of resistance and, hence, can be replaced by a cur- rent source of 1.6 A in parallel with 20 ft, as shown in Fig. 4.38(c). The 20 ft and 30 ft parallel resis- tors can be reduced to a single 12 ft resistor. The parallel combination of the 1.6 A current source (d) Fourth step and the 12 ft resistor transforms into a voltage source of 19.2 V in series with 12 ft. Figure 4.38(d) shows the result of this last transformation. The current in the direction of the voltage drop across the 6 V source is (19.2 - 6)/16, or 0.825 A. Therefore the power associated with the 6 V source is p 6V = (0.825)(6) = 4.95 W. b) The voltage source is absorbing power. A question that arises from use of the source transformation depicted in Fig. 4.38 is, "What happens if there is a resistance R p in parallel with the voltage source or a resistance R s in series with the current source?" In both cases, the resistance has no effect on the equivalent circuit that pre- dicts behavior with respect to terminals a,b. Figure 4.39 summarizes this observation. The two circuits depicted in Fig. 4.39(a) are equivalent with respect to terminals a,b because they produce the same voltage and current in any resistor R L inserted between nodes a,b. The same can be said for the cir- cuits in Fig. 4.39(b). Example 4.9 illustrates an application of the equiva- lent circuits depicted in Fig. 4.39. R -wv—»a »« \R, -•b (a) (b) Figure 4.39 • Equivalent circuits containing a resistance in parallel with a voltage source or in series with a current source. 112 Techniques of Circuit Analysis Example 4.9 Using Special Source Transformation Techniques a) Use source transformations to find the voltage v () in the circuit shown in Fig. 4.40. b) Find the power developed by the 250 V voltage source. c) Find the power developed by the 8 A current source. b) The current supplied by the 250 V source equals the current in the 125 ft resistor plus the current in the 25 ft resistor. Thus 250 250 - 20 „ „ '< = l25 + -^- =1UA - 25 ft 250 V Figure 4.40 A The circuit for Example 4.9. Solution Therefore the power developed by the voltage source is /> 25() v(developed) = (250)(11.2) = 2800 W. c) To find the power developed by the 8 A current source, we first find the voltage across the source. If we let v s represent the voltage across the source, positive at the upper terminal of the source, we obtain a) We begin by removing the 125 ft and 10 ft resis- tors, because the 125 ft resistor is connected across the 250 V voltage source and the 10 ft resistor is connected in series with the 8 A current source. We also combine the series-connected resistors into a single resistance of 20 ft. Figure 4.41 shows the sim- plified circuit. v s + 8(10) = v 0 = 20, or v s = -60 V, and the power developed by the 8 A source is 480 W. Note that the 125 ft and 10 ft resistors do not affect the value of v 0 but do affect the power calculations. 25 ft 250 V Figure 4.41 • A simplified version of the circuit shown in Fig. 4.40. Figure 4.42 • The circuit shown in Fig. 4.41 after a source transformation. We now use a source transformation to replace the 250 V source and 25 ft resistor with a 10 A source in parallel with the 25 ft resistor, as shown in Fig. 4.42. We can now simplify the circuit shown in Fig. 4.42 by using Kirchhoffs current law to com- bine the parallel current sources into a single source. The parallel resistors combine into a single resistor. Figure 4.43 shows the result. Hence v„ = 20 V. + D„5 io ft Figure 4.43 A The circuit shown in Fig. 4.42 after combining sources and resistors. 4.10 Thevenin and Norton Equivalents 113 /"ASSESSMENT PROBLEM Objective 4—Understand source transformation 4.15 a) Use a series of source transformations to find the voltage v in the circuit shown. b) How much power does the 120 V source deliver to the circuit? Answer: (a) 48 V; (b) 374.4 W. NOTE: Also try Chapter Problems 4.59 and 4.60. 20 a , j 60V -r (t ) 36A 1120 V L 1.6 a 4.10 Thevenin and Norton Equivalents At times in circuit analysis, we want to concentrate on what happens at a specific pair of terminals. For example, when we plug a toaster into an outlet, we are interested primarily in the voltage and current at the ter- minals of the toaster. We have little or no interest in the effect that con- necting the toaster has on voltages or currents elsewhere in the circuit supplying the outlet. We can expand this interest in terminal behavior to a set of appliances, each requiring a different amount of power. We then are interested in how the voltage and current delivered at the outlet change as we change appliances. In other words, we want to focus on the behavior of the circuit supplying the outlet, but only at the out- let terminals. Thevenin and Norton equivalents are circuit simplification techniques that focus on terminal behavior and thus are extremely valuable aids in analysis. Although here we discuss them as they pertain to resistive cir- cuits, Thevenin and Norton equivalent circuits may be used to represent any circuit made up of linear elements. We can best describe a Thevenin equivalent circuit by reference to Fig. 4.44, which represents any circuit made up of sources (both inde- pendent and dependent) and resistors. The letters a and b denote the pair of terminals of interest. Figure 4.44(b) shows the Thevenin equiva- lent. Thus, a Thevenin equivalent circuit is an independent voltage source V Th in series with a resistor R Th , which replaces an interconnec- tion of sources and resistors. This series combination of V Th and R T h is equivalent to the original circuit in the sense that, if we connect the same load across the terminals a,b of each circuit, we get the same volt- age and current at the terminals of the load. This equivalence holds for all possible values of load resistance. To represent the original circuit by its Thevenin equivalent, we must be able to determine the Thevenin voltage V Jh and the Thevenin resist- ance R lh . First, we note that if the load resistance is infinitely large, we have an open-circuit condition. The open-circuit voltage at the terminals a,b in the circuit shown in Fig. 4.44(b) is Vj h . By hypothesis, this must be • a A resistive network containing independent and dependent sources (a) (b) Figure 4.44 A (a) A general circuit, (b) The Thevenin equivalent circuit. 114 Techniques of Circuit Analysis the same as the open-circuit voltage at the terminals a,b in the original circuit. Therefore, to calculate the Thevenin voltage V Th , we simply calcu- late the open-circuit voltage in the original circuit. Reducing the load resistance to zero gives us a short-circuit condition. If we place a short circuit across the terminals a,b of the Thevenin equiva- lent circuit, the short-circuit current directed from a to b is /„- = Tli Th (4.55) By hypothesis, this short-circuit current must be identical to the short-circuit current that exists in a short circuit placed across the terminals a,b of the original network. From Eq. 4.55, R Th - V-, ih (4.56) Thus the Thevenin resistance is the ratio of the open-circuit voltage to the short-circuit current. 40 ^vw—• a + + 2sy Cz) 2 °^f 3A 0 ) V] ^ Finding a Thevenin Equivalent To find the Tlievenin equivalent of the circuit shown in Fig. 4.45, we first calculate the open-circuit voltage of v. db . Note that when the terminals a,b are open, there is no current in the 4 O resistor. Therefore the open-circuit voltage v. db is identical to the voltage across the 3 A current source, labeled V\. We find the voltage by solving a single node-voltage equation. Choosing the lower node as the reference node, we get Figure 4.45 • A circuit used to illustrate a Thevenin equivalent. Vi - 25 V] — + — - 3 = 0. 5 20 (4.57) Solving for V\ yields v x = 32 V. (4.58) 25 V Figure 4.46 • The circuit shown in Fig. 4.45 with terminals a and b short-circuited. Hence the Thevenin voltage for the circuit is 32 V. The next step is to place a short circuit across the terminals and calcu- late the resulting short-circuit current. Figure 4.46 shows the circuit with the short in place. Note that the short-circuit current is in the direction of the open-circuit voltage drop across the terminals a,b. If the short-circuit current is in the direction of the open-circuit voltage rise across the termi- nals, a minus sign must be inserted in Eq. 4.56. The short-circuit current (/ sc ) is found easily once v 2 is known. Therefore the problem reduces to finding v 2 with the short in place. Again, if we use the lower node as the reference node, the equation for v 2 becomes v 2 - 25 ih v-y (4.59) 4.10 Thevenin and Norton Equivalents 115 Solving Eq. 4.59 for v 2 gives th = 16 V. (4.60) Hence, the short-circuit current is 16 A A <ac = "J" = 4 A - (4.61) We now find the Thevenin resistance by substituting the numerical results from Eqs. 4.58 and 4.61 into Eq. 4.56: R Th = V Th 32 4 8H. (4.62) Figure 4.47 shows the Thevenin equivalent for the circuit shown in Fig. 4.45. You should verify that, if a 24 Cl resistor is connected across the ter- minals a,b in Fig. 4.45, the voltage across the resistor will be 24 V and the current in the resistor will be 1 A, as would be the case with the Thevenin circuit in Fig. 4.47. This same equivalence between the circuit in Figs. 4.45 and 4.47 holds for any resistor value connected between nodes a.b. The Norton Equivalent A Norton equivalent circuit consists of an independent current source in parallel with the Norton equivalent resistance, We can derive it from a Thevenin equivalent circuit simply by making a source transforma- tion. Thus the Norton current equals the short-circuit current at the terminals of interest, and the Norton resistance is identical to the Thevenin resistance. 32 V SO 'VW • a Figure 4.47 • The Thevenin equivalent of the circuit shown in Fig. 4.45. 4 O -wv—• a J25V J20I1 ( f J3A -•b Step 1: Source transformation t Step 2: Parallel sources and parallel resistors combined • 4ft •—-Wv—e a 8A 4 0 -•b Step 3: Source transformation; series resistors combined, producing the Thevenin equivalent circuit 8ft -VW- 32 V Using Source Transformations Sometimes we can make effective use of source transformations to derive a Thevenin or Norton equivalent circuit. For example, we can derive the Thevenin and Norton equivalents of the circuit shown in Fig. 4.45 by making the series of source transformations shown in Fig. 4.48. This technique is most useful when the network contains only independent sources. The presence of dependent sources requires retaining the identity of the controlling voltages and/or currents, and this constraint usually prohibits continued reduction of the circuit by source transformations. We discuss the problem of finding the Thevenin equivalent when a circuit contains dependent sources in Example 4.10. Step 4: Source transformation, producing the Norton equivalent circuit 4A 8 O Figure 4.48 • Step-by-step derivation of the Thevenin and Norton equivalents of the circuit shown in Fig. 4.45. . values of R L . If the current through R L is the same in both circuits, then the voltage drop across R { is the same in both circuits, and the circuits are equivalent at nodes a,b. If the. though the node-voltage and mesh-current methods are powerful tech- niques for solving circuits, we are still interested in methods that can be used to simplify circuits. Series-parallel