Nodal and Mesh Circuit

3. a. Describe the Nodal Analysis Including
Steps of Determination of Node Voltages
(Use 3 nodes and a reference Node. Also
show 1 voltage source and 2 current
sources).

4. 1.
Steps To Solve A Circuit Using
Nodal Analysis

6. Now Lets Solve A
Circuit Using Nodal
Analysis

7. A3
2
48
24 8
A5
V12
0V
1V 2V 3V6i
1i
2i
3i
4i
5i
6i

8. Solution
 1............1647
284
35
35
321
31211
321






vvv
vvvvv
iii
KCL at Node 02KCL at Node 01
542 iii 
 2.....027
428
321
32221





vvv
vvvvv

9.  3.....36724
3
428
12
3
321
32313
536









vvv
vvvvv
iii

10. Now We Got 3
Equations We Need To Solve
Them To Find The Values Of
Nodal Voltages

11. Now Multipling 7 with equation no 2 and adding this resultant equation with
equation no 1 We get ,
014497
1647
321
321


vvv
vvv
 4......161848 32  vv
Now Multipling 4 with equation no 2 and adding this resultant
equation with equation no 3 We get

12.  5....361530 32  vv
Now Multipling 15 with equation no 4 and
multipling 18 with equation no 5 and Than
Subtracting equation 5 from equation 4 we get
36724
08284
321
321


vvv
vvv

13. 93.4
888180
2
2


v
v
vv
v
93.4
888180
2
2



14. vv
v
267.12
161893.448
3
3


 
vv
v
10
16267.12493.47
1
1


vvvvvvAnswer 267.12;93.4;10: 321 

15. Thus We Can Solve Any
Given Circuit With Nodal
Analysis
Got It ?

16. B. Describe mesh analysis including steps of
determination of 3 loop currents(at least 1
voltage sources in each loop and 1 common
to loop 1 &2 or 2&3)

17. MESH ANALYSIS
 It is only applicable to planar circuits (a circuit that can be drawn on a plane with
no branches crossing each other).
 Mesh analysis applies KVL to find unknown currents.
 A mesh is a loop that does not contain any other loops.
 The current through a mesh is known as the mesh current.
 Assume for simplicity that the circuit contains only voltage sources.

18. Steps of Mesh Analysis
 Identify Mesh (Loops).
 Assign Current to each Mesh.
 Apply KVL around each Loop to get an Equation in terms of the Loop Currents.
 Solve the Resulting System of Linear Equations.

19. +
_
+
_
+
_
+ - + -
+ -
- + - +
- + - +
+
-
50v
50v 50v
30Ω
40Ω 40Ω
20Ω
10Ω 10Ω
MESH 1 MESH 2
MESH 3
-
+
Identify Mesh (Loops)

20.  Identify Mesh (Loops).
 Assign Current to each Mesh.
 Apply KVL around each Loop to get an Equation in terms of the Loop Currents.
 Solve the Resulting System of Linear Equations.
Steps of Mesh Analysis

21. +
_
+
_
+
_
+ - + -
+ -
- + - +
- + - +
50v
50v 50v
𝐼1 𝐼2
𝐼3
30Ω
40Ω 40Ω
+
-
20
Ω
10Ω 10Ω
-
+
Assign Current to each Mesh

22.  Identify Mesh (Loops).
 Assign Current to each Mesh.
 Apply KVL around each Loop to get an Equation in terms of the Loop Currents.
 Solve the Resulting System of Linear Equations.
Steps of Mesh Analysis

23. KVL Around Mesh Loop 1
)1........(..........5427
50204070
02040)201040(50
321
231
231



iii
iii
iii
+
_
+
_
+
_
+ -
- +
50v
50v 50v
𝐼1 𝐼2
𝐼3
40Ω
40Ω
+
-
20
Ω
10Ω 10Ω
30Ω

24. )2.........(..........5472
5427
50402070
040207050
04020)401020(50
321
312
312
312
312





iii
iii
iii
iii
iii
KVL Around Mesh Loop 2
+
_
+
_
+
_
+ -
- +
50v 50v
𝐼1 𝐼2
𝐼3
30Ω
40Ω 40Ω
20
Ω
-
+
10Ω 10Ω

25. KVL Around Mesh Loop 3
)3(....................51144
501104040
0404011050
04040)304040(50
321
321
213
213




iii
iii
iii
iii
+
_
+
_
+
_
+ -
- + - +
50v
𝐼1 𝐼2
𝐼3
30Ω
40Ω 40Ω
20
Ω
10Ω 10Ω

26. Steps of Mesh Analysis
 Identify Mesh (Loops).
 Assign Current to Each Mesh.
 Apply KVL around each Loop to get an Equation in terms of the Loop Currents.
 Solve the resulting System of Linear Equations.

27.  Using Equation (1) and (3);  Using Equation (1) and (2);
)4(..........51918
51144
108414
)3(2)1(
31
321
321




ii
iii
iii
)5(..........253645
108144
35281449
)2(27)1(
31
321
321




ii
iii
iii
Solving the Equations

28. The Final Equations
Now, with Three equations, we can use one of several methods to mathematically
solve for the unknown currents i1 , i2 and i3.
)1........(..........5427 321  iii
)3(....................51144 321  iii
)2.........(..........5472 321  iii

29.  Using Equation (5) and (4);
Ai
ii
ii
425.1
365)1936()3618(
1925)1936()1945(
36)4(19)5(
1
31
31




 Using value of i1 From Equation (4) we get i3 =>
Ai 086.13 
 Using value of i1& i3 From Equation (1) we get i2 =>
Ai 314.02 

30. Any questions?