2. Chapter 4 Basic Nodal and Mesh Analysis Engineering Circuit Analysis Sixth Edition W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin Copyright © 2002 McGraw-Hill, Inc. All Rights Reserved. User Note: Run View Show under the Slide Show menu to enable slide selection. Fig. 4.1 “Obtain values for the unknown voltages …” Fig. 4.5(a) The circuit of Example 4.2 with a 22-V source... Fig. 4.7 “Determine the node-to-reference voltages …” Fig. 4.9 Examples of planar and nonplanar networks... Fig. 4.10 (a) The set of branches identified by the heavy lines… Fig. 4.12 “Determine the two mesh currents, i 1 and i 2, in …” Fig. 4.16 “Find the three mesh currents in the circuit below.” Fig. 4.19 Circuit from Practice Problem 4.8.

3. Nodal Analysis

4. Fig. 4.1 (a) A simple three- node circuit. (b) Redrawn circuit to emphasize nodes. (c) Reference node selected and voltages assigned. (d) Shorthand voltage references. If desired, an appropriate ground symbol may be substituted for “Ref.” W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition. Copyright ©2002 McGraw-Hill. All rights reserved. Obtain values for the unknown voltages across the elements in the circuit below. At node 1 At node 2

5. Fig. 4.5 (a) The circuit of Example 4.2 with a 22-V source in place of the 7-  resistor. (b) Expanded view of the region defined as a supernode; KCL requires that all currents flowing into the region must sum to zero, or we would pile up or run out of electrons. W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition. Copyright ©2002 McGraw-Hill. All rights reserved. (a) The circuit of Example 4.2 with a 22-V source in place of the 7-  resistor. (b) Expanded view of the region defined as a supernode; KCL requires that all currents flowing into the region must sum to zero, or we would pile up or run out of electrons. At node 1: At the “supernode:”

6. Fig. 4.7 “Determine the node-to-reference voltages in the circuit below.” W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition. Copyright ©2002 McGraw-Hill. All rights reserved. Determine the node-to-reference voltages in the circuit below.

7. Fig. 4.9 Examples of planar and nonplanar networks; crossed wires without a solid dot are not in physical contact with each other. W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition. Copyright ©2002 McGraw-Hill. All rights reserved. Examples of planar and nonplanar networks; crossed wires without a solid dot are not in physical contact with each other.

8. Fig. 4.10 (a) The set of branches identified by the heavy lines is neither a path nor a loop. (b) The set of branches here is not a path, since it can be traversed only by passing through the central node twice. (c) This path is a loop but not a mesh, since it encloses other loops. (d) This path is also a loop but not a mesh. (e, f) Each of these paths is both a loop and a mesh. W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition. Copyright ©2002 McGraw-Hill. All rights reserved. (a) The set of branches identified by the heavy lines is neither a path nor a loop. (b) The set of branches here is not a path, since it can be traversed only by passing through the central node twice. (c) This path is a loop but not a mesh, since it encloses other loops. (d) This path is also a loop but not a mesh. (e, f) Each of these paths is both a loop and a mesh.

9. Mesh Current Analysis

10. Fig. 4.12 “Determine the two mesh currents, i 1 and i 2, in the circuit below.” W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition. Copyright ©2002 McGraw-Hill. All rights reserved. Determine the two mesh currents, i 1 and i 2, in the circuit below. For the left-hand mesh, -42 + 6 i 1 + 3 ( i 1 - i 2 ) = 0 For the right-hand mesh, 3 ( i 2 - i 1 ) + 4 i 2 - 10 = 0 Solving, we find that i 1 = 6 A and i 2 = 4 A. (The current flowing downward through the 3-  resistor is therefore i 1 - i 2 = 2 A. )

11. Fig. 4.16 “Find the three mesh currents in the circuit below.” W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition. Copyright ©2002 McGraw-Hill. All rights reserved. Find the three mesh currents in the circuit below. Creating a “supermesh” from meshes 1 and 3: -7 + 1 ( i 1 - i 2 ) + 3 ( i 3 - i 2 ) + 1 i 3 = 0 [1] Around mesh 2: 1 ( i 2 - i 1 ) + 2 i 2 + 3 ( i 2 - i 3 ) = 0[2] Rearranging, i 1 - 4 i 2 + 4 i 3 = 7[1] -i 1 + 6 i 2 - 3 i 3 = 0[2] i 1 - i 3 = 7[3] Solving, i 1 = 9 A, i 2 = 2.5 A, and i 3 = 2 A. Finally, we relate the currents in meshes 1 and 3: i 1 - i 3 = 7[3]

12. Fig. 4.19 Circuit from Practice Problem 4.8. W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition. Copyright ©2002 McGraw-Hill. All rights reserved. Find the voltage v 3 in the circuit below.