Chapter 4
Basic Nodal and Mesh
Analysis
Fig. 4.1
“Obtain values for the unknown voltages …”
Fig. 4.5
(a) The circuit of Example 4.2 with a 22-V source ...
User Note:
Fig. 4.7
“Determine the node-to-reference voltages …”
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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, i1 and i2, in …”
Fig. 4.16
“Find the three mesh currents in the circuit below.”
Fig. 4.19
Circuit from Practice Problem 4.8.
Engineering Circuit Analysis Sixth Edition
W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin
Copyright © 2002 McGraw-Hill, Inc. All Rights Reserved.
Nodal Analysis
v1  v2
v2  v1
i1 
and i2 
R
R
Obtain values for the unknown voltages across the
elements in the circuit below.
At node 1
At node 2
v1
v1  v2

 3.1
2
5
v2
v v
 2 1  - (-1.4)
1
5
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-W
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:
v1  v2 v1  v3
83 

3
4
At the “supernode:”
3  25 
v2  v1 v3  v1 v3 v2

 
3
4
5 1
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.
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.
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.
W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition.
Copyright ©2002 McGraw-Hill. All rights reserved.
Mesh Current Analysis




V1  I1  I 2 R and V2  I 2  I1 R
Determine the two mesh currents, i1 and i2, in the circuit below.
For the left-hand mesh,
-42 + 6 i1 + 3 ( i1 - i2 ) = 0
For the right-hand mesh,
3 ( i2 - i1 ) + 4 i2 - 10 = 0
Solving, we find that i1 = 6 A and i2 = 4 A.
(The current flowing downward through
the 3-W resistor is therefore i1 - i2 = 2 A. )
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 ( i1 - i2 ) + 3 ( i3 - i2 ) + 1 i3 = 0
[1]
Around mesh 2:
1 ( i2 - i1 ) + 2 i2 + 3 ( i2 - i3 ) = 0
[2]
Finally, we relate the currents in meshes 1 and 3:
i1 - i3 = 7
[3]
Rearranging,
i1 - 4 i2 + 4 i3 = 7
[1]
-i1 + 6 i2 - 3 i3 = 0
[2]
i1
[3]
- i3 = 7
Solving,
i1 = 9 A, i2 = 2.5 A, and i3 = 2 A.
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 v3 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.