1. Example The voltage between node A and B is 4V
the source voltage can be calculated as
V= (R/Rt) Vs
4=(8/12) Vs
Vs=12 (4/8)= 6Volts.

2. Example
12V
We want to calculate the source current.The voltage across 3k resistor is 12V. So the current (I) flowing through it, will be
I =V/R
=12/3k= 4mA

3. 12V
The same current is following through the series combination of 3k resistor and 9k resistor. 3k is in series with 9k and 2k is in series with 4k, so

4. Now by current division rule the source current will be
I= (6k/(12k +6k)) Is
Is= (18k/6k) 4mA =12mA

5. Example 4V
We want to calculate the source voltage. The voltage across 2k resistor is 4V, so the current flowing through it will be
I=V/R =(4/2k) =2mA

6. 4V
The same current is following through the series combination of 2k and 4k. So the voltage across 4k resistor will be
V= IR=(2m) x(4k)=8 Volts.

7. So the total voltage across 2k and 4k resistor will be V= 4V +8V=12V

8. 4k is in series with 2k these may be combined as =4k+2k=6k.4V
4k is in series with 2k these may be combined as
=4k+2k=6k.

9. 6k is parallel with 6k resistor so 6k||6k = (6x6)/(6+6)=3k

10. 3k
The voltage across 3k resistor is 12k so the source voltage is
V=(3/(3+9)) Vs
Vs=12 (12/3)=48V

11. We want to calculate the voltage across 4k resistor.
12k||4k = (12 x4)/(12+4)=3k.

12. 9k is in series with 3k their combined effect will be = 9k +3k =12k

13. 12k is parallel with 6k resistor so 12k||6k = (12 x 6)/18 = 72/18
= 72/18
=4k

14. 12k is parallel with 4k resistor so
12k||4k = (12 x4)/(12+4)
=48/16
=3k

15. So by voltage division rule, the voltage across the equivalent 3k resistor is
V= (3/6)x12
=6 V

16. So the voltage across 3k resistor is V = (3/12) x 6 =1.5 volts.

17. The same voltage will be drop across 4k resistor.

18. KIRCHHOF’S LAW KIRCHHOF’S CURRENT LAW
Sum of all the currents entering in the node is equal to sum of currents leaving the node.
It can also be defined as
sum of entering currents +
sum of leaving currents = 0

19. ASSUMPTIONS All the entering currents are taken as negative.
All the leaving currents are taken as positive.

20. NODE It is the junction of two or more than two elements OR
It is simply a point of connection between circuit elements.

21. BRANCH
It is the distance or link between two nodes.

22. LOOP
It is the closed path for the flow of current in which no node is encountered more than once.
Lets take some examples of node analysis or Kirchhof’s current law.

23. Formula for Writing Equation
Number of equations in node analysis is one minus than the number of total nodes.
Number of equations = N – 1
Where N is the number of nodes.

24. Ground This is a common or reference point
among all the nodes without insertion
of any component between.

25. 60mA
40mA I6 I5
20mA I1 I4
30mA 3 1 2 5 4
Assuming the currents leaving the node are positive, the KCL equations for
node 1 through 4 are

26. For node 1 -I1 +0.06 + 0.02 =0 - I1+0.08= 0 I1 = 0.08 A 3 1 2 5 4 I1
60mA
40mA I6 I5
20mA I1 I4
30mA 3 1 2 5 4
For node 1
-I =0
- I1+0.08= 0
I1 = 0.08 A

27. For node 2 I1 - I4 + I6 =0 0.08 - I4 + I6=0 - I4 + I6 = -0.08 A 3 1 2
60mA
40mA I6 I5
20mA I1 I4
30mA 3 1 2 5 4
For node 2
I1 - I4 + I6 =0
I4 + I6=0
- I4 + I6 = A

28. For node 3 -0.06+ I4 –I5 + 0.04=0 I4 – I5 = 0.02 A 3 1 2 5 4 I1 60mA

29. For node 4 -0.02 +I5 -0.03 = 0 I5 = 0.05 A 3 1 2 5 4 I1 60mA 20mA I5

30. Now putting the value of I5 in equation of node 3 I4 -0.05 = 0.02
60mA
40mA I6 I5
20mA I1 I4
30mA 3 1 2 5 4
Now putting the value of I5 in equation of
node 3
I = 0.02
I4 = 0.07 A

31. Putting the value of I6 in the equation for node 2 - 0.07 +I6 =-0.08
60mA
40mA I6 I5
20mA I1 I4
30mA 3 1 2 5 4
Putting the value of I6 in the equation for
node 2
I6 =-0.08
I6= A