analysis of circuit techniques nodal and mesh analysis

1. EE-103: Electrical Engineering
Circuit Analysis Techniques

2. Analysis by Inspection
Intuitive solution utilizing the element and circuit
laws
Repeated use of Ohm’s and Kirchhoff's laws
Uses simple techniques
 Series and parallel combination of resistors
 Voltage and current dividers
Start from the farthest branch

3. Nodal Analysis
Steps to Determine Node Voltages:
 Select a node as the reference node. Assign
voltage v1, v2, …vn-1 to the remaining n-1 nodes.
The voltages are referenced with respect to the
reference node.
 Apply KCL to each of the n-1 nonreference nodes.
Use Ohm’s law to express the branch currents in
terms of node voltages. Stick to Passive Sign
Convention.
 Solve the resulting simultaneous equations to
obtain the unknown node voltages.

4. Example
Typical circuit for nodal analysis

5. 322
2121
iiI
iiII


R
vv
i lowerhigher 

3
2
3
2
21
2
1
1
1
0
0
R
v
i
R
vv
i
R
v
i






3
2
2
21
2
2
21
1
1
21
R
v
R
vv
I
R
vv
R
v
II






6. Example
• Determine the voltage at the nodes

7. Example
• At node 1,
24
3
3
2131
1
vvvv
ii x






8. Example
• At node 2
4
0
82
23221
32







vvvvv
iiix

9. Example
• At node 3
2
)(2
84
2
213231
21
vvvvvv
iii x








10. Example
• In matrix form:








































0
0
3
8
3
8
9
4
3
8
1
8
7
2
1
4
1
2
1
4
3
3
2
1
v
v
v

11. Nodal Analysis with Voltage Sources
Case 1: The voltage source is connected between a
nonreferenced node and the reference node: The
nonreferenced node voltage is equal to the
magnitude of voltage source and the number of
unknown nonreferenced nodes is reduced by one
Case 2: The voltage source is connected between two
nonreferenced nodes: a generalized node (supernode)
is formed

12. Super Node
A supernode is formed by enclosing a
(dependent or independent) voltage source
connected between two nonreference nodes
and any elements connected in parallel with
it.

13. Nodal Analysis with Voltage Sources
5
6
0
8
0
42
32
323121
3241









vv
vvvvvv
iiii

14. Example
• For the circuit, find the node voltages
i1 i2

15. Example
• For the circuit, find the node voltages
i1 i2

16. Example
• For the circuit, find the node voltages
0
42
72
02172
21


vv
ii

17. Example
• For the circuit, find the node voltages
2
0
42
72
02172
21
21



vv
vv
ii

18. Example
• For the circuit, find the node voltages

22. Mesh Analysis
• Mesh analysis: another procedure for
analyzing circuits, applicable to planar circuit.
• A Mesh is a loop which does not contain any
other loops within it

23. (a) A Planar circuit with crossing branches,
(b) The same circuit redrawn with no crossing branches.

24. A nonplanar circuit.

25. • Steps to Determine Mesh Currents:
1. Assign mesh currents i1, i2, .., in to the n meshes.
2. Apply KVL to each of the n meshes. Use Ohm’s
law to express the voltages in terms of the mesh
currents.
3. Solve the resulting n simultaneous equations to
get the mesh currents.

26. A circuit with two meshes.

27. • Apply KVL to each mesh. For mesh 1,
• For mesh 2,
123131
213111
)(
0)(
ViRiRR
iiRiRV


223213
123222
)(
0)(
ViRRiR
iiRViR



28. • Solve for the mesh currents.
• Use i for a mesh current and I for a branch
current.
















2
1
2
1
323
331
V
V
i
i
RRR
RRR
2132211 ,, iiIiIiI 

29. Example
• Find the branch current I1, I2, and I3 using
mesh analysis.

30. Example
• For mesh 1,
• For mesh 2,
• We can find i1 and i2 by substitution method
or Cramer’s rule. Then,
123
010)(10515
21
211


ii
iii
12
010)(1046
21
1222


ii
iiii
2132211 ,, iiIiIiI 

31. Example 2
• Use mesh analysis to find the current I0

32. Example 2
• Apply KVL to each mesh. For mesh 1,
• For mesh 2,
126511
0)(12)(1024
321
3121


iii
iiii
02195
0)(10)(424
321
12322


iii
iiiii

33. Example 2
• For mesh 3,
• In matrix form eqs become
we can calculate i1, i2 and i3 by Cramer’s rule,
and find I0.
02
0)(4)(12)(4
,A,nodeAt
0)(4)(124
321
231321
210
23130




iii
iiiiii
iiI
iiiiI




























0
0
12
211
2195
6511
3
2
1
i
i
i

34.     021441211  AviiRiiRiR
    032612425  iiRiiRiR
    043823637  iiRiiRiR
    034814243  iiRiiRiR
Mesh Current Analysis

35. A Three Mesh Example

36. A Three Mesh Example
36
Follow each mesh
clockwise
Simplify
Solve the equations:
i1 = 3 A, i2 = 2 A, and i3 = 3 A.

37. Mesh Analysis with Current Sources
A circuit with a current source.

38. • Case 1
– Current source exist only in one mesh
– One mesh variable is reduced
• Case 2
– Current source exists between two meshes, a
super-mesh is obtained.
A51 i

39. Super Mesh
• a supermesh results when two meshes have a
(dependent , independent) current source in
common.

40. Properties of a Supermesh
1. The current is not completely ignored
– provides the constraint equation necessary to
solve for the mesh current.
2. A supermesh has no current of its own.
3. Several current sources in adjacency form a
bigger supermesh.

41. Find the three mesh currents in the
circuit below.

42. 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]
Rearranging,
i1 - 4 i2 + 4 i3 = 7 [1]
-i1 + 6 i2 - 3 i3 = 0 [2]
i1 - i3 = 7 [3]
Solving,
i1 = 9 A, i2 = 2.5 A, and i3 = 2 A.
Finally, we relate the currents in meshes 1 and 3:
i1 - i3 = 7 [3]

43. Example 3
• For the circuit, find i1 to i4 using mesh analysis.

44. 0102)(8
3
5
06)(842
434
432
21
24331




iii
iii
ii
iiiii