# Electrical Engineering - Branch, Loop and Node Analyses - Discussion

Discussion Forum : Branch, Loop and Node Analyses - General Questions (Q.No. 1)

1.

Find the node voltage

*V*_{A}.Discussion:

39 comments Page 1 of 4.
Thirumal said:
1 decade ago

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)).

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)).

Charishma said:
1 decade ago

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.

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.

Naveen Pal said:
9 years ago

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.

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.

Abdullah U of R said:
7 years ago

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.

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.

Prosper said:
6 years ago

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.

Where I49 = (12-VA)/49,

I80 =(VA-6)/80,

I24 = VA/24,

This will give an answer of 2.2945.

Nami-Swan Jyv said:
1 year ago

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.

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.

(1)

Surajit said:
1 decade ago

((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.

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.

Sharathkumar said:
1 year ago

((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.

((24Va-288 + 49Va)/1176) + (Va-6)/80))=0

1920Va-2340 + 3920Va + 1176Va-7056=0

7016Va-30096 =0,

7016Va = 30096,

Va = 30096/7016,

= 4.28.

(2)

Ganesh said:
8 years ago

@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.

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.

Neeraj said:
6 years ago

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.

i1 + i2 - i3 = 0.

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

Va = 4.2V.

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