CHAPTER 4. Techniques of Circuit Analysis

CHAPTER 4 Techniques of Circuit Analysis

4.1 Terminology Planar circuits those circuits that can be drawn on a plane with no crossing branches. Figure 4.1 (a) A planar circuit. (b) The same circuit redrawn to verify that it is planar. Figure 4.1 (a) A planar circuit. (b) The same circuit redrawn to verify that it is planar.

Describing a Circuit The Vocabulary Figure 4.2 A nonplanar circuit.

Example 4.1 For the circuit in Fig. 4.3, identify a) all nodes. b) all essential nodes. c) all branches. d) all essential branches. e) all meshes. f) two paths that are not loops or essential branches. g) two loops that are not meshes. Figure 4.3 A circuit illustrating nodes, branches, meshes, paths, and loops.

Example 4.1

The Systematic Approach An Illustration The circuit has four essential nodes and six essential branches, Figure 4.4 The circuit shown in Fig. 4.3 with six unknown branch currents defined. Figure 4.4 The circuit shown in Fig. 4.3 with six unknown branch currents defined.

We use the nodes b, c, and e to get Using the other three meshes gives

Rearranging Eqs. 4.1 and 4.2 to facilitate their solution yields the set Note that summing the current at the nth node (g in this example) gives

4.2 Introduction to the Node-Voltage Method The optimum choice of the reference node (if one exists) becomes apparent after you have gained some experience using this method. In Fig. 4.5, the lower node connects the most branches, so we use it as the reference node. We flag the chosen reference node with the symbol, as in Fig. 4.6. Figure 4.6 The circuit shown in Fig. 4.5 with a reference node and the node voltages. Figure 4.5 A circuit used to illustrate the node-voltage method of circuit analysis.

A node voltage is defined as the voltage rise from the reference node to a nonreference node. For this circuit, we must define two node voltages, which are denoted v 1 and v 2 in Fig. 4.6 Figure 4.6 The circuit shown in Fig. 4.5 with a reference node and the node voltages.

Figure 4.7 Computation of the branch current i.

The sum of the three currents leaving node 1 must equal zero; therefore the node-voltage equation derived at node 1 is The node-voltage equation derived at node 2 is Solving for v 1 and v 2 yields

Example 4.2 a) Use the node-voltage method of circuit analysis to find the branch currents i a, i b, and i c in the circuit shown in Fig. 4.8. b) Find the power associated with each source, and state whether the source is delivering or absorbing power. Figure 4.8 The circuit for Example 4.2.

Example 4.2 Figure 4.9 The circuit shown in Fig. 4.8 with a reference node and the unknown node voltage v 1.

Example 4.2

4.3 The Node-Voltage Method and Dependent Sources If the circuit contains dependent sources, the nodevoltage equations must be supplemented with the constraint equations imposed by the presence of the dependent sources.

Example 4.3 Use the node-voltage method to find the power dissipated in the 5 Ω resistor in the circuit shown in Fig. 4.10. Figure 4.10 The circuit for Example 4.3.

Example 4.3

Example 4.3 Figure 4.11 The circuit shown in Fig. 4.10, with a reference node and the node voltages.

4.4 The Node-Voltage Method: Some Special Cases When a voltage source is the only element between two essential nodes, the node-voltage method is simplified. But the 100 V source constrains the voltage between node 1 and the reference node to 100 V. This means that there is only one unknown node voltage (v 2 ). Figure 4.12 A circuit with a known node voltage.

But v 1 = 100 V, so Eq. 4.7 can be solved for :

Figure 4.13 A circuit with a dependent voltage source connected between nodes. Figure 4.14 The circuit shown in Fig. 4.13. with the selected node voltages defined.

Figure 4.14 shows the redrawn circuit, with the reference node flagged and the node voltages defined. Also, we introduce the current i because we cannot express the current in the dependent voltage source branch as a function of the node voltages v 1 and v 2. Thus, at node 2 and at node 3 We eliminate i simply by adding Eqs. 4.9 and 4.10 to get

The Concept of a Supernode When a voltage source is between two essential nodes, we can combine those nodes to form a supernode. Figure 4.15 Considering nodes 2 and 3 to be a supernode.

Using Eqs. 4.13 and 4.14 and v 1 = 50 V reduces Eq. 4.12 to From Eqs. 4.13 and 4.14:

Node-Voltage Analysis of the Amplifier Circuit The circuit has four essential nodes: Nodes a and d are connected by an independent voltage source as are nodes b and c. Figure 4.16 The transistor amplifier circuit shown Figure in Fig. 2.24. 4.16 The transistor amplifier circuit shown in Fig. 2.24. Figure Figure 4.17 4.17 The circuit The shown circuit in shown Fig. 4.16, in with Fig. voltages 4.16, and with the voltages supernode and identified. the supernode identified.

Using d as the reference node, combine nodes b and c into a supernode. We now eliminate both v c and i B from Eq. 4.15 by noting that Substituting Eqs. 4.16 and 4.17 into Eq. 4.15 yields Solving Eq. 4.18 for v b yields

4.5 Introduction to the Mesh-Current Method Fig. 4.18 contains seven essential branches where the current is unknown and four essential nodes. Therefore, to solve it via the mesh-current method, we must write four [7-(4-1)] mesh-current equations. A mesh current is the current that exists only in the perimeter of a mesh. Figure 4.18 The circuit Figure 4.18 The circuit shown in Fig. 4.1(b), with the shown in Fig. 4.1(b), with mesh currents defined. the mesh currents defined.

The evolution of the mesh current technique Applying Kirchhoff s current law to the upper node and Kirchhoff s voltage law around the two meshes generates the following set of equations: Figure 4.19 A circuit used to illustrate development of the mesh-current method of circuit analysis.

Apply Kirchhoff s voltage law around the two meshes Figure 4.20 Mesh currents i a and i b.

Collecting the coefficients of i a and i b in Eqs. 4.25 and 4.26 gives

Example 4.4 a) Use the mesh-current method to determine the power associated with each voltage source in the circuit shown in Fig. 4.21. b) Calculate the voltage v o across 8 Ω the resistor. Figure 4.21 The circuit for Example 4.4.

Example 4.4 Figure 4.22 The three mesh currents used to analyze the circuit shown in Fig. 4.21.

Example 4.4

4.6 The Mesh-Current Method and dependent Sources If the circuit contains dependent sources, the meshcurrent equations must be supplemented by the appropriate constraint equations.

Example 4.5 Use the mesh-current method of circuit analysis to determine the power dissipated in the resistor 4 Ω in the circuit shown in Fig. 4.23. Figure 4.23 The circuit for Example 4.5.

Example 4.5 Figure 4.24 The circuit shown in Fig. 4.23 with the three mesh currents.

Example 4.5

4.7 The Mesh-Current Method: Some Special Cases When a branch includes a current source, the meshcurrent method requires some additional manipulations. Figure 4.25 A circuit illustrating mesh analysis when Figure 4.25 A circuit illustrating mesh a branch contains an independent current source. analysis when a branch contains an independent current source.

Thus, for mesh a: and for mesh c: We now add Eqs. 4.36 and 4.37 to eliminate and obtain Summing voltages around mesh b gives

We reduce Eqs. 4.38 and 4.39 to two equations and two unknowns by using the constraint that We leave to you the verification that, when Eq. 4.40 is combined with Eqs. 4.38 and 4.39, the solutions for the three mesh currents are

The Concept of a Supermesh To create a supermesh, we mentally remove the current source from the circuit by simply avoiding this branch when writing the mesh-current equations. Figure 4.26 The circuit shown in Fig. 4.25, illustrating Figure 4.26 The circuit shown in Fig. 4.25, the concept of a supermesh. illustrating the concept of a supermesh.

When we sum the voltages around the supermesh (denoted by the dashed line), we obtain the equation which reduces to Note that Eqs. 4.42 and 4.38 are identical.

Mesh-Current Analysis of the Amplifier Circuit In Fig. 4.27 using the concept of the supermesh, we redraw the circuit as shown in Fig. 4.28. Figure 4.27 The circuit shown in Fig. Figure 2.244.27 with The the circuit mesh shown currents in Fig. 2.24 i a, with the mesh currents ia, ib, and ic i. b, and i c. Figure 4.28 The circuit shown in Fig. 4.27, depicting the supermesh Figure 4.28 The circuit shown in Fig. 4.27, depicting created the supermesh by the created presence by the presence of the of the dependent current source. dependent current source.

We now sum the voltages around the supermesh in terms of the mesh currents i a, i b and i c to obtain The mesh b equation is The constraint imposed by the dependent current source is The branch current controlling the dependent current source, expressed as a function of the mesh currents, is

From Eqs. 4.45 and 4.46 We now use Eq. 4.47 to eliminate i c from Eqs. 4.43 and 4.44:

You should verify that the solution of Eqs. 4.48 and 4.49 for i a and i b gives We also leave you to verify that, when Eqs. 4.50 and 4.51 are used to find i B, the result is the same as that given by Eq. 2.25.

4.8 The Node-Voltage Method Versus the Mesh-Current Method Example 4.6 Find the power dissipated in the 300 Ω resistor in the circuit shown in Fig. 4.29. Figure 4.29 The circuit for Example 4.6.

Example 4.6 Figure 4.30 The circuit shown in Fig. 4.29, with the five mesh currents.

Example 4.6

Example 4.6 Figure 4.31 The circuit shown in Fig. 4.29, with a reference node.

Example 4.6

Example 4.6 Figure 4.32 The circuit shown in Fig. 4.29 with an alternative reference node.

Example 4.6

Example 4.7 Find the voltage v o in the circuit shown in Fig. 4.33. Figure 4.33 The circuit for Example 4.7.

Example 4.7

Example 4.7 Figure 4.34 The circuit shown in Fig. 4.33 with the three mesh currents. Figure 4.35 The circuit shown in Fig. 4.33 with node voltages.

Example 4.7

4.9 Source Transformations Figure 4.36 Source transformations.

Example 4.8 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). Figure 4.37 The circuit for Example 4.8.

Example 4.8

Example 4.8 Figure 4.38 Step-by-step simplification of the circuit shown in Fig. 4.37.

Example 4.8

Example 4.8 Figure 4.39 Equivalent circuits containing a resistance in parallel with a voltage source or in series with a current source.

Example 4.9 a) Use source transformations to find the voltage v o 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. Figure 4.40 The circuit for Example 4.9.

Example 4.9 Figure 4.41 A simplified version of the circuit shown in Fig. 4.40.

Example 4.9 Figure 4.42 The circuit shown in Fig. 4.41 after a source transformation. Figure 4.43 The circuit shown in Fig. 4.42 after combining sources and resistors.

Example 4.9

4.10 Thévenin and Norton Equivalents Thévenin equivalent circuit is an independent voltage source in series with a resistor. Figure 4.44 (a) A general circuit. (b) The Thévenin equivalent circuit.

Finding a Thévenin Equivalent Figure 4.45 A circuit used to illustrate a Thévenin equivalent.

Figure 4.46 The circuit shown in Fig. 4.45 with terminals a and b short-circuited.

Figure 4.47 The Thévenin equivalent of the circuit shown in Fig. 4.45.

The Norton Equivalent A Norton equivalent circuit consists of an independent current source in parallel with the Norton equivalent resistance. Figure 4.48 Step-by-step derivation of the Thévenin and Norton equivalents of the circuit shown in Fig. 4.45.

Figure 4.48 Step-by-step derivation of the Thévenin and Norton equivalents of the circuit shown in Fig. 4.45.

Example 4.10 Find the Thévenin equivalent for the circuit containing dependent sources shown in Fig. 4.49. Figure 4.49 A circuit used to illustrate a Thévenin equivalent when the circuit contains dependent sources.

Example 4.10

Example 4.10 Figure 4.50 The circuit shown in Fig. 4.49 with terminals a and b shortcircuited.

Example 4.10

Example 4.10 Figure 4.51 The Thévenin equivalent for the circuit shown in Fig. 4.49.

4.11 More on Deriving a Thévenin Equivalent Figure 4.52 A circuit used to illustrate a Thévenin equivalent. Figure 4.53 The circuit shown in Fig. 4.52 after deactivation of the independent sources.

Example 4.11 Find the Thévenin resistance R TH for the circuit in Fig. 4.49, using the alternative method described.

Example 4.11 Figure 4.54 An alternative method for computing the Thévenin resistance.

Example 4.11

Using the Thévenin Equivalent in the Amplifier Circuit Figure 4.55 The application of a Thévenin equivalent in circuit analysis. Figure 4.56 A modified version of the circuit shown in Fig. 4.55.

Now we replace the circuit made up of V CC, R 1 and R 2 with a Thévenin equivalent, with respect to the terminals b,d. The Thévenin voltage and resistance are With the Thévenin equivalent, the circuit in Fig. 4.56 becomes the one shown in Fig. 4.57.

Figure 4.57 The circuit shown in Fig. 4.56 modified by a Thévenin equivalent.

We now derive an equation for i B simply by summing the voltages around the left mesh. In writing this mesh equation, we recognize that i E = (1 + b)i B. Thus, from which

4.12 Maximum Power Transfer Figure 4.58 A circuit describing maximum power transfer. Figure 4.59 A circuit used to determine the value of R L for maximum power transfer.

Example 4.12 a) For the circuit shown in Fig. 4.60, find the value of R L that results in maximum power being transferred to R L. Figure 4.60 The circuit for Example 4.12. b) Calculate the maximum power that can be delivered to R L. c) When R L is adjusted for maximum power transfer, what percentage of the power delivered by the 360 V source reaches R L?

Example 4.12 Figure 4.61 Reduction of the circuit shown in Fig. 4.60 by means of a Thévenin equivalent.

Example 4.12

4.13 Superposition An individual response is the result of an independent source acting alone. The principle is applicable to any linear system. Superposition is applied in both the analysis and the design of circuits.

Figure 4.62 A circuit used to illustrate superposition. Figure 4.63 The circuit shown in Fig. 4.62 with the current source deactivated.

Figure 4.64 The circuit shown in Fig. 4.62 with the voltage source deactivated. Figure 4.65 The circuit shown in Fig. 4.64 showing the node voltages υ 3 and υ 4.

Example 4.13 Use the principle of superposition to find v o in the circuit shown in Fig. 4.66. Figure 4.66 The circuit for Example 4.13.

Example 4.13 Figure 4.67 The circuit shown in Fig. 4.66 with the 5 A source deactivated.

Example 4.13

Example 4.13 Figure 4.68 The circuit shown in Fig. 4.66 with the 10 V source deactivated.

Example 4.13

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.

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. 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.

Summary Several new circuit simplification techniques were introduced in this chapter: Source transformations allow us to exchange a voltage 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 equivalence holds provided that

Summary Thévenin equivalents and Norton equivalents allow us to simplify a circuit comprised of sources and resistors into an equivalent circuit consisting of a voltage source and a series resistor (Thévenin) 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 Thévenin voltage (V TH ) is the open-circuit voltage across the terminals of the original circuit, (2) the Thévenin resistance (R TH ) is the ratio of the Thévenin voltage to the shortcircuit current across the terminals of the original circuit; and (3) the Norton equivalent is obtained by performing a source transformation on a Thévenin equivalent.

Summary 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 R L = R TH, the Thévenin resistance as seen from the resistor R L. The equation for the maximum power transferred is

Summary 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 determine the voltages and currents that exist when all independent sources are active. Dependent sources are never deactivated when applying superposition.