1. The KCL equations for nodes 1 through 4 follow
Example: Write equations for all nodes.
The KCL equations for nodes 1 through
4 follow

2. For node 1
I1 + I2 – I5 = 0

3. For node 2
-I2 +I3 -50I2=0

4. For node 3
-I1 + 50I2 + I4 = 0

5. For node 4
I5 – I3 – I4 = 0

6. Example
For node A 10mA is entering the node and the source current It is leaving and 60mA is also entering so,

7. So for node A
It -60mA – 10mA= 0

8. 40mA
10mA A B
60mA
20mA It
For node B
60mA – 40mA – 20mA = 0
0 = 0

9. Example
By KCL the equation for node A
-12mA + 4mA + I =0

10. Example
The equation of currents at node A and B
by KCL
For node A
-I1+I2+ 3mA=0

11. I2 I1
4mA
12mA
3mA B A
For node B
-12mA + 4mA +I1=0

12. Example
Writing equation for node A .
For node A
- 10 Ix+ Ix+44 mA – 12 mA = 0

13. Example
For node A equation will be by KCL
Ix+ 10Ix – 44mA = 0

14. Example
We want to calculate the values of I1 and I2
For node A
4mA + 8mA- I1 = 0
I1 = 12mA

15. Now for I2
For node B
-8mA + 2 mA + I2=0
I2 = 6mA

16. Example
We want to calculate the values of I1 , I2 ,
and I3, so we will use node analysis.

17. I1 C A B I2
8mA I3
2mA
For node A
-I1 – I2 +8mA= 0
-I1 – I2 = -8mA

18. I1 C A B I2
8mA I3
2mA
For node B
I2 + I3 +4mA=0
I2 + I3 = - 4mA

19. I1 C A B I2
8mA I3
2mA
For node C
-I3 + 2mA – 8mA = 0
I3 = - 6mA

20. Putting the value of I3 in equation of node B I2 - 6mA = -4mA I2 = 2mAC A B I2
8mA I3
2mA
Putting the value of I3 in equation of node B
I2 - 6mA = -4mA
I2 = 2mA

21. Putting the value of I2 in equation of node A -I1 – 2mA = -8mA C A B I2
8mA I3
2mA
Putting the value of I2 in equation of
node A
-I1 – 2mA = -8mA
I1 = 6mA

22. Example
We want to write the equations for nodes A, B, C and D.

23. 4mA A
5mA
2mA I1 I2
3mA C D B
8mA I3
For node A
-5mA + 8mA + 4mA =0

24. 4mA A
5mA
2mA I1 I2
3mA C D B
8mA I3
For node B
I1 – I2 + 5mA =0

25. For node C -I1 – 2mA + 3mA – 8mA =0 I3 A 8mA 5mA 4mA I1 3mA B C D I2

26. 4mA A
5mA
2mA I1 I2
3mA C D B
8mA I3
For node D
-4mA – 3mA + I3=0

27. Example
At node 1
(V1/12k) + ((V1 – V2)/10k) = 6mA
5V1 +6V1 – 6V2 = (60k)(6mA)
11V1 – 6V2 = 360

28. (V2/3k) +(V2/6k) + ((V2- V1)/10k) =0 10V2 +5V2+3V2-3V1=0 18V2 -3V1 =0Io
3k Io
At node 2
(V2/3k) +(V2/6k) + ((V2- V1)/10k) =0
10V2 +5V2+3V2-3V1=0
18V2 -3V1 =0

29. Equating equation of node1 and node2 33V1 -18V2 =1080 -3V1 + 18V2 =0
3k Io
Equating equation of node1 and node2
33V1 -18V2 =1080
-3V1 + 18V2 =0
30V1 =1080, V1 = 36volts

30. Io
3k Io
18V2 -3V1 =0
6V2 – V1=0
6V2 -36 =0
... V2 =36/6=6volts

31. Io
3k Io
6k 
3k
Io =V2/6k
=6/6k
=1mA