1. BYST Circuit -F2003: Nodal and Mesh Analysis 92 CPE220 Electric Circuit Analysis Chapter 3: Nodal and Mesh Analyses

2. BYST Circuit -F2003: Nodal and Mesh Analysis 93 Nodal Analysis 3. 1 Chapter 3 Nodal analysis provides a general procedure for analyzing electrical circuits using node voltages as the circuit variables. The given circuit commonly contains only current sources and KCL equation for each node is set up by expressing all the unknown currents coming through or leaving out of that node as a function of the node voltage. Then, the node voltages in the given circuit can be determined by solving those KCL equations. Nodal analysis can be used with either planar or non-planar circuits. Hence, it is a more general method compared with the mesh method which will be discussed later.

3. BYST Circuit -F2003: Nodal and Mesh Analysis 94 Recall that nodes are points at which two or more elements have a common connection. In the other words, nodes are the connected segments of conductor that remain when we remove the circuit elements. 9 A 16  8  12  3 A (a) A single node! (b) Figure 3.1(a) An electrical circuit. (b) Redraw the circuit with the circuit elements removed.

4. BYST Circuit -F2003: Nodal and Mesh Analysis 95 Fig. 3.1 illustrates nodes in an electrical circuit after all circuit elements are removed. Node and Branch Voltages reference node the highest number of branches A node voltage associated with a given node is defined to be the voltage difference between the given node and a reference node, which has been chosen from among the nodes. Usually, the reference node is the node to which the highest number of branches is connected. Hence, for a circuit with “n” essential nodes, there are “n–1” node voltages. Once the set of node voltages is determined, all the other voltages and currents can be obtained in a straightforward manner.

5. BYST Circuit -F2003: Nodal and Mesh Analysis 96 Identify a reference node and corresponding node voltages for the circuit in Fig. 3.1(a). Example 3.1 Solution: From the redraw circuit in Fig. 3.1(b), we select the bottom node (node 3) as the reference node. node 1 v1v1 v2v2 Reference node, ground, earth, “sea level” 9 A 16  8  12  3 A node 2 node 3 Ans.

6. BYST Circuit -F2003: Nodal and Mesh Analysis 97 branch voltage differencetwo node voltages A branch voltage is measured as the difference between two node voltages. As illustrated in Fig. 3.2, the voltage across the 16  resistor is not a node voltage. It is the branch voltage and is actually the difference voltage between two node voltages v 1 and v 2. In general, vjvj vkvk v jk = v j - v k (3.1) where v jk = the branch voltage in volts (V) v j and v k = the node voltages in volts (V)

7. BYST Circuit -F2003: Nodal and Mesh Analysis 98 Measurement of Branch and Node Voltages 9 A 16  42  12  3 A 12 v1v1v1v1 v2v2v2v2 64 V48 V 16 V voltmeter + – The voltage being measured is a branch voltage. Node voltage Figure 3.2Branch voltage v 12 in term of the node voltages v 1 and v 2.

8. BYST Circuit -F2003: Nodal and Mesh Analysis 99 Determine the relationship among the branch voltages and the node voltages in the following circuit. Example 3.2 Solution: There are three nodes. We choose node 3 as the reference node. Hence, only two node voltages which are v1 and v2. For branch voltages, there are three branch voltages: v 12, v 13, and v 23. However, the v1v1v1v1 v2v2v2v2 Ref. node 12 3 v 12 +- v 13 +- v 23 +-

9. BYST Circuit -F2003: Nodal and Mesh Analysis 100 branch voltages v 13 and v 23 are clearly equal to the node voltage v 1 and v 2, respectively. The branch voltage v 12 is the difference between v 1 and v 2. That is, v 13 = v 1 v 23 = v 2 v 12 = v 1 - v 2 Ans. Branch Currents Currents flowing through any resistors are defined as the branch currents. The value of each branch current can be determined by following the Ohm's law. For example, the current flowing through the resistor R in Fig. 3.3 can be determine as following:

10. BYST Circuit -F2003: Nodal and Mesh Analysis 101 vjvjvjvj vkvkvkvk Node j Node k Figure 3.3Branch currents flowing through the resistor R. The branch current flowing from node j to node k which can be calculated as: The branch current flowing from node k to node j which can be calculated as: (3.2) i jk = v j -v k R (3.3) i kj = v k -v j R

11. BYST Circuit -F2003: Nodal and Mesh Analysis 102 The rationale for nodal analysis is that once the node voltages are determined, all the other voltages and currents can be obtained in a simple manner. groundearth The reference node is chosen by the circuit analyst. In electronic circuits, we frequently choose the node to which lots of branches are connected. In power systems, we usually choose “ground” or “earth.” Basic Procedure for Nodal Analysis: 1. Select the node to which the highest number of branches is connected as the reference node. 2. Set up KCL equations for other nodes by expressing the unknown currents as a 3.1.1 Nodal Analysis by Examples

12. BYST Circuit -F2003: Nodal and Mesh Analysis 103 function of the node voltages measured with respect to (w.r.t.) the reference node. voltage sources 3. If the given circuit contains voltage sources, KCL equations of those two nodes connected by a voltage source are combined to eliminate the redundancy of KCL equations since the additional information is available through the node voltage. For example, if nodes j and k are connected by a voltage source, then v j - v k is already known. 4. Solve KCL equations to determine the node voltages.

13. BYST Circuit -F2003: Nodal and Mesh Analysis 104 Determine the node voltages v 1, v 2, and v 3 in the following circuit. Example 3.3 v1v1v1v1 v2v2v2v2 v3v3v3v3 Reference node Solution: KCL @ node 1 -8 -3 = v 1 -v 2 3 + v 1 -v 3 4 (3.4)

14. BYST Circuit -F2003: Nodal and Mesh Analysis 105 KCL @ node 2 = -(-3) = 3 v 2 -v 1 3 + v2v2 1 (3.4) v 2 -v 3 7 + KCL @ node 3 = -(-25) = 25 v 3 -v 2 7 + v3v3 5 (3.5) v 3 -v 1 4 + From Eq. 3.3, 3.4, and 3.5, we get v 1 = 5.412 V, v 2 = 7.736 V, and v 3 = 46.32 V Ans.

15. BYST Circuit -F2003: Nodal and Mesh Analysis 106 3.1.2 Nodal Analysis with Voltage Sources As mentioned earlier, nodal analysis usually involves with current sources. For electric circuits containing voltage sources, some special treatment is needed to eliminate the redundancy of KVL equations, since additional information is available on the node voltages. For example, if node j and node k are connected by the voltage source, the v j - v k is known. There are two possibilities of connecting the voltage source: reference nodenon-reference node A voltage source is connected between the reference node and a non-reference node. In this case, we simply set the

16. BYST Circuit -F2003: Nodal and Mesh Analysis 107 Supernodes supernode voltage source A supernode is a set of nodes connected to each other by voltage source, but not to the reference node by a path of voltage source. non-reference nodes supernode A voltage source is connected between the two non-reference nodes. In this case, the two non-reference nodes form a "supernode". Both KCL and KVL are applied to determine the node voltages. voltage at the non-reference node equal to the voltage of the voltage source. Referencenode vjvjvjvj (a) v j = 5 V

17. BYST Circuit -F2003: Nodal and Mesh Analysis 108 vjvjvjvj vkvkvkvk (b) v j - v k = 5 V supernode Figure 3.4(a) A voltage source is connected between the reference node and the non-reference node. (b) A voltage source is connected between two non-reference nodes. A supernode is formed. The voltage source in Fig. 3.4(a) is connected between node j and the reference node. Hence, the node voltage v j equal to 5 V. On the other hand, the voltage source in Fig. 3.4(b) is connected between two non- reference nodes (node j and k). Hence, node j and k form a supernode and the branch voltage v jk (v j - v k ) equal to 5 V.

18. BYST Circuit -F2003: Nodal and Mesh Analysis 109 Determine the node voltages v 1 in the following circuit. Example 3.4 v1v1v1v1 v3v3v3v3 Reference node v2v2v2v2Supernode Solution: If we choose the bottom node as the reference, then the voltage source will be connected between two non-reference nodes. Hence, node 2 and 3 form a supernode.

19. BYST Circuit -F2003: Nodal and Mesh Analysis 110 KCL @ node 1 -8 -3 = v 1 -v 2 3 + v 1 -v 3 4 (3.6) KCL @ supernode 2 & 3 3+25 = v 2 -v 1 3 + v2v2 1 (3.7) v 3 -v 1 4 + v3v3 5 + Since the branch voltage between node 2 and 3 must equal to 22 V. That is, v 3 - v 2 = 22 V. (3.8) From Eq. 3.6, 3.7, and 3.8, the solution for v 1 is 1.071 V. Ans.

20. BYST Circuit -F2003: Nodal and Mesh Analysis 111 Determine the values of the unknown node- to-reference voltages in the following circuit. Example 3.5 Solution: Only node 3 and 4 form a supernode since the independent voltage source 12 V is connected between node 1 and the reference node.Supernode

21. BYST Circuit -F2003: Nodal and Mesh Analysis 112 KCL @ node 2 v 1 = -12 = 14 v 2 -v 1 0.5 + v 2 -v 3 2 (3.9) KCL @ supernode 3&4 = 0.5v x v 3 -v 2 2 + v4v4 1 v 4 -v 1 2.5 + (3.10) v 3 - v 4 = 0.2v y Since v 4 - v 1 = v y (3.11) (3.12) v 2 - v 1 = v x (3.13) and (3.14) From all of above Eqs., we get v 1 = -12 V., v 2 =-4 V., v 3 = 0 V. and v 4 =-2 V. Ans.

22. BYST Circuit -F2003: Nodal and Mesh Analysis 113 Mesh Analysis 3. 2 mesh currents currents flowing around the meshes are independentmeshloop does not enclose any other loop within it For planar networks the mesh analysis provides another general procedure for analyzing electric circuits using mesh currents as the circuit variables. Basically, mesh analysis initially assumes that the currents flowing around the meshes are independent. A mesh is a circuit loop that does not enclose any other loop within it. In the other words, a mesh is the smallest circuit loop. A planar network is a network that can be drawn in a plane with no branches crossing one another. Fig. 3.5 illustrates examples of planar and nonplanar networks. Network in Fig. 3.5(a) and (b) are clearly a planar network and a nonplanar network, respectively. The network in Fig. 3.5(c) is

23. BYST Circuit -F2003: Nodal and Mesh Analysis 114 Figure 3.5Examples of planar and nonplanar networks. (a) and (c) are planar networks whereas (b) is a nonplanar network. drawn in such a way as to make it appear nonplanar. However, it is the planar network since it can be drawn in plan without any crossing branches. mesh current flows only around the perimeter of a mesh clockwise direction A mesh current is defined as a current that flows only around the perimeter of a mesh. It is traditionally labeled in the clockwise direction.

24. BYST Circuit -F2003: Nodal and Mesh Analysis 115 Basic Procedure for Mesh Analysis: 1. Assign all mesh currents. 2. Set up KVL equation for each mesh. Use Ohm's law to express the voltages in terms of the mesh currents. 3. If the given circuit contains current sources on the perimeter of any meshes. That is, two meshes share current source in common. Such meshes form a supermesh to eliminate the redundancy of KVL equations. 4. Solve KVL equations to determine the mesh currents. 3.2.1 Mesh Analysis by Examples

25. BYST Circuit -F2003: Nodal and Mesh Analysis 116 Determine the values of the mesh current i 1 and i 2 in the following circuit. Example 3.6 Solution: KVL in mesh 1: 6i 1 + 3(i 1 - i 2 ) = 42 (3.14) 9i 1 - 3i 2 = 42 KVL in mesh 2: 4i 2 + 3(i 2 - i 1 ) = 10 (3.15) -3i 1 + 7i 2 = 10

26. BYST Circuit -F2003: Nodal and Mesh Analysis 117 From Eq. 3.14 and 3.15, the mesh currents can be determined as: i 1 = 6 A. and i 2 = 4 A. Ans. Determine the values of the mesh current i 1, i 2 and i 3 in the following circuit. Example 3.7

27. BYST Circuit -F2003: Nodal and Mesh Analysis 118 Solution: KVL in mesh 1: (3.17) 3i 1 - i 2 - 2i 3 = 1 KVL in mesh 2: (3.18) -i 1 + 6i 2 - 3i 3 = 0 KVL in mesh 3: (3.19) -2i 1 - 3i 2 + 6i 3 = 6 From Eq. 3.17, 3.18 and 3.19, we get i 1 = 3 A., i 2 = 2 A. and i 3 = 3 A. Ans.

28. BYST Circuit -F2003: Nodal and Mesh Analysis 119 3.2.2 Mesh Analysis with Current Sources Current source on the perimeter of meshes can exist in two possible cases: A current source exists only in one mesh. Hence, the mesh current is known. For example, the mesh current i 2 in Fig. 3.6 equal to -5A. Figure 3.6A circuit with a current source existing only in one mesh.

29. BYST Circuit -F2003: Nodal and Mesh Analysis 120 supermesh A current source exists between two meshes. In this case, a supermesh is formed. That is, a supermesh results when two meshes have a dependent or independent current source in common. As shown in Fig. 3.7(a), two meshes share the current source 6A in common. A supermesh is formed by excluding the current source as shown in Fig. 3.7(b). Figure 3.7(a) A circuit with a current source existing between two meshes. (b) A supermesh is formed by excluding the current source. Exclude these elements (a) (b) x

30. BYST Circuit -F2003: Nodal and Mesh Analysis 121 A supermesh must, however, satisfy KVL similar to any other mesh. Thus, applying KVL to the supermesh in Fig. 3.7(b) yields 6i 1 + 14i 2 = 20 x At node x in Fig. 3.7(a), we apply KCL and get i 2 = i 1 + 6 For the circuit in Fig. 3.7(a), determine the values of the mesh current i 1 and i 2 using mesh analysis. Example 3.8 Solution: (3.20) (3.21) Solving Eq. 3.20 and 3.21, we get i 1 = -3.2 A. and i 2 = 2.8 A. Ans.

31. BYST Circuit -F2003: Nodal and Mesh Analysis 122 For the following circuit, determine the values of the mesh current i 1, i 2, and i 3 using mesh analysis. Example 3.9 Solution: Since mesh 1 and 3 share the current source 7A in common. They form a supermesh by excluding the branch where the current source is connected. Exclude these elements

32. BYST Circuit -F2003: Nodal and Mesh Analysis 123 KVL in supermesh: (3.22) i 1 - 4i 2 + 4i 3 = 7 KVL in mesh 2: (3.23) -i 1 + 6i 2 - 3i 3 = 0 @ the bottom node: (3.24) i 1 = i 3 + 7 From Eq. 3.22, 3.23 and 3.24, we get i 1 = 9 A., i 2 = 2.5 A. and i 3 = 2 A. Ans.

33. BYST Circuit -F2003: Nodal and Mesh Analysis 124 For the following circuit, determine the values of the mesh current i 1, i 2, and i 3 using mesh analysis. Example 3.10 Solution: KVL in mesh 2: (3.25) -i 1 + 6i 2 - 3i 3 = 0

34. BYST Circuit -F2003: Nodal and Mesh Analysis 125 From the given circuit, we get (3.26) i 1 = 15 and (3.27) = i 3 - i 1 1 9 vxvx Since (3.28) v x = 3(i 3 - i 2 ) Substitute Eq. 3.28 into 3.27, we get (3.29) - i 1 + + i 3 = 0 1 3 i2i2 2 3 From Eq. 3.25, 3.26 and 3.29, we get i 1 = 15 A., i 2 = 11 A. and i 3 = 17 A. Ans.