Electrical Engineering - Branch, Loop and Node Analyses - Discussion

Steps for Nodal Analysis :.

1. Find the number of Nodes = n.

2. Represent the Nodal voltages = V1, V2, V3, . , Vn.

3. Number of Linear Equations required = Number of nodes = n.

4. Choose any one node as reference node and ground it, so the potential of Reference node = 0.

So one equation is obtained now.

5. To fine the remaining (n-1) equ, Find the voltage sources in the circuit and write equations. If total number of equ are obtained, stop the analysis, otherwise go to the step 6.

6. Find the nodes which have no voltage source and apply KCL. If total number of equ are obtained, stop the analysis otherwise go to step 7.

((NOTE : KCL SHOULD APPLY ONLY AT NON REFERENCE NODES)).

7. Super Node Analysis :.

In this analysis short a voltage source and check step 6. Repeat this step till all the equations are obtained.

((Note : although it's looks like lengthy, but using these steps you can solve any circuit)).

STEPS FOR NODAL ANALYSIS...
STEP1: Choose the nodes and node voltages to be obtained.
STEP2: Choose the currents preferably leaving the node at each
branch connected to each node.
STEP3: Apply Kirchoff's current law at each node with proper sign
convention.
now apply the procedure suggested by our friends.

STEPS FOR NODAL ANALYSIS.

STEP 1: Choose the nodes and node voltages to be obtained.

STEP 2: Choose the currents preferably leaving the node at each
branch connected to each node.

STEP 3: Apply Kirchhoff's current law at each node with proper sign
convention.

Now apply the procedure suggested by our friends.

Easy way to calculate node analysis:

i1+i2+i3 = 0 it is equal to:

(12-VA)/49 + -VA/24 + (6-VA)/80 = 0.

12/49 - VA/49 - VA/24 + 6/80 - VA/80 = 0.

12/49 + 6/80 = VA (1/49 + 1/24 + 1/80).

So, VA = (12/49 + 6/80)/(1/49 + 1/24 + 1/80) = 4.28 V.

Current flowing through the 49ohms resistor is equal to the sum of current flowing through the 24ohms and 80ohms resistor. eg. I49 = I80 + I24.

Where I49 = (12-VA)/49,
I80 =(VA-6)/80,
I24 = VA/24,
This will give an answer of 2.2945.

KCL: Iin = Iout.
I1 + I3 = I2.
I1 - I2 + I3 = 0.

Nodal Analysis:

(Start node minus End node)/Resistor.

((V1- Va)/R1) - ((Va-0)/R2) +((V2-Va)/R3) = 0.

Shift Solve;

Va= 4.28v.
or
Va = 4.25v.

((V-12)/49)+(V/24)+((V-6)/80)=0
or,V=(3762/877)
=4.28 volt
Basic formula of nodal analysis,
For any closed circuit, Summation of the all voltage sources is equals to zero.

((Va-12)/49) + (Va/24) + (Va-6)/80))=0
((24Va-288 + 49Va)/1176) + (Va-6)/80))=0
1920Va-2340 + 3920Va + 1176Va-7056=0
7016Va-30096 =0,
7016Va = 30096,
Va = 30096/7016,
= 4.28.

@Krishna.

Is right according to nodal analysis. The algebraic sum of currents in a network of conductors meeting at a point is zero.

Here we take i=v/r.

At node vA we can see that Tge current i3 is leaving so the eq must be like,
i1 + i2 - i3 = 0.
(12 - vA)/49 + (6-vA)/80-VA/24 = 0,
Va = 4.2V.