Nodal Analysis

       A method of determining voltage between the nodes in electric circuit analysis is what it called a Node-Voltage or Nodal Analysis. 

These are the two cases in solving the circuit problem:

1st Case:  When the voltage source is connected to non-reference node and a reference node or ground.

2nd Case: The Supernode. It is when the voltage source is not connected to the ground or reference node.    
    
Let us analyze such cases by these examples:

#1: Consider this circuit. 

Here are the steps in solving a circuit:

1. Analyze the problem.

2. Assign nodes. Assume that I am going to assign V1 as my first node, and V2 as my second node.

3. Since the 12 volts source is given and is connected to the non-reference and a reference node, then V1 is 12V. We have now the exact value of the first node.

4. Solve V2 by adding the quantities of the following:
    
    Since V1 = 12 /_0V; then

    V2(1/3 - 1/j1 + 1/j2) - 12V/3 = 0
    0.33 + j0.5V2 - 6 = 0
    0.33 + j0.5V2 = 6
    V2 = 5.538 - j8.307 or

    V2 = 9.98 /_ -146.309 Volts

And now we get Io value:
    
      Io = V/R
       = 9.98/_ -146.309 / j2

    Io = 4.99/_ -146.309 Amps

#2: Consider this Circuit:

Solve for V1 and V2

Here are the steps in solving a circuit:

1. Analyze the problem.
2. Look at the figure again. As what we saw, the 12 volts source is not connected to the reference node. Then I conclude that it is a supernode. 

3. Assign nodes again. Assume that I am going to assign V1 as my first node, and V2 as my second node. 

4. Join the two nodes as a supernode.

5. Remove the voltage source by changing it in a short circuit. (Take note as well: that if the current source is given and has to be removed, change it in an open circuit.)

6. In getting the value of a supernode, get the equation value by using Kirchoff's Current Law(KCL):

@KCL:

I1 + I2 + I3 = 0

• I1 = V1/R1 = V1/3
• I2 = V2/-j1
• I3 = V2/j2

V1/3 + V2/-j1 + V2/j2 = 0

@KVL:

V1+12/_0 - V2 = 0
V1-V2 = -12/_0
V1 = V2 - 12/_0

Now get the values:

V2 - 12/3 + (V2/-j1) + V2/j2 = 0
V2/3 - 6 +(V2/-j1)+V2/j2 = 0
V2(1/3 + (1/-j1) + 1/j2) = 6
0.33V2 + j0.5 = 6

V2 = 9.98/_ -56.309 V

V1 = 9.98/_ -56.309 - 12
V1 = 10.52/_ -127.89 V