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.

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

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

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

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

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.

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

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.

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

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

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

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

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

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

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

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

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

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

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

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

BRANCH It is the distance or link between two nodes.

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.

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.

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

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

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 -I1 +0.06 + 0.02 =0 - I1+0.08= 0 I1 = 0.08 A

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 0.08 - I4 + I6=0 - I4 + I6 = -0.08 A

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

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

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 I4 -0.05 = 0.02 I4 = 0.07 A

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 - 0.07 +I6 =-0.08 I6= -0.01 A