Nodal analysis

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  • #1
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hello!

I have some difficulty in nodal analysis

I want to know basically how to apply KCL and Ohm's Law to find the equations

few branches meet in a node
I know that KCL means that sum of currents that enter are equal to the sum of currents are leaving the node
the problem is that I may not know the directions of the currents
1) do I choose them arbitrarily?
2) after choosing them arbitrarily, how do I find the voltage difference across a branch? V1-V2 or V2-V1 ?

thanks!
 

Answers and Replies

  • #2
berkeman
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hello!

I have some difficulty in nodal analysis

I want to know basically how to apply KCL and Ohm's Law to find the equations

few branches meet in a node
I know that KCL means that sum of currents that enter are equal to the sum of currents are leaving the node
the problem is that I may not know the directions of the currents
1) do I choose them arbitrarily?
2) after choosing them arbitrarily, how do I find the voltage difference across a branch? V1-V2 or V2-V1 ?

thanks!
For KCL, you just label the node voltages that you don't know (like, V1, V2, V3, etc.), and then for each node, write the equation that is the sum of all currents *leaving* the node is equal to zero. (You can write the sum of all currents going into the node instead, but I personally like to write the equation with all currents leaving the node is equal to zero).

Does that make sense? Do you have a specific example you would like help working through?
 
  • #3
182
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my question is how to choose from V1-V2/R1 or V2-V1/R1
if you don't understand it, I can find an example
 
  • #4
berkeman
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my question is how to choose from V1-V2/R1 or V2-V1/R1
if you don't understand it, I can find an example
When you are writing the KCL equation for the node V1, to get the current out of node V1, you write (V1-V2)/R1.

When you are writing the KCL equation for the node V2, to get the current out of node V2, you write (V2-V1)/R1.

Makes sense?
 
  • #6
berkeman
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it doesn't really matter? but we have to be consistent somehow so that we won't end up with the wrong equations!
Correct, it doesn't matter whether you sum all currents out of a node or into a node. Just be consistent. :smile:
 
  • #7
182
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well at that website, he is not actually consistent

1) basically first, we need to define how to apply Ohm's Law in a branch

if we have a branch from node V1 to node V2 with a resistor in that branch
the current is V1-V2/R or V2-V1/R?
and is the answer different if we consider arbitrarily that the current has direction from V1 to V2 or V2 to V1?

2) because in his webpage, he takes the current at the branch between V1 and V2
as (V1-V2)/5 when applying KCL for node V1
and as (V2-V1)/5 when applying KCL for node V2 !!!!

is that consistency?
 
  • #8
818
67
1) basically first, we need to define how to apply Ohm's Law in a branch

if we have a branch from node V1 to node V2 with a resistor in that branch
the current is V1-V2/R or V2-V1/R?
and is the answer different if we consider arbitrarily that the current has direction from V1 to V2 or V2 to V1?
V = R*I is only true when you assign the reference polarity and direction correctly for V and I, respectively. See the figure here:

120px-OhmsLaw.svg.png


If you have a current I with a reference direction from V1 to V2, then V must also have a reference polarity from V1 to V2 as shown in the image. From KVL you have then -V1 + V + V2 = 0 or V = V1 - V2.

If the reference direction for I was from V2 to V1, then you'd have -V1 - V + V2 = 0 or V = V2 - V1 instead.

With practice you won't really need this amount of detail in your analysis.

2) because in his webpage, he takes the current at the branch between V1 and V2
as (V1-V2)/5 when applying KCL for node V1
and as (V2-V1)/5 when applying KCL for node V2 !!!!

is that consistency?
In both cases, he/she considers currents with reference directions outwards from the node. That's consistency.
 
Last edited:
  • #9
182
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why it's consistency?
basically what he does is, for a specific branch, let's say between V1 and V2, he considers the current going from V1 to V2 when applying KCL for V1 and from V2 to V1 when for V2!
that's not consistent, because he uses different current direction for the SAME current, when applying KCL !!!
 
  • #10
818
67
why it's consistency?
basically what he does is, for a specific branch, let's say between V1 and V2, he considers the current going from V1 to V2 when applying KCL for V1 and from V2 to V1 when for V2!
that's not consistent, because he uses different current direction for the SAME current, when applying KCL !!!
It's not the same current. He has one current, call it I1, with a reference direction from V1 to V2 and another, I2, with a reference direction from V2 to V1, so you have I1 = -I2.
 
  • #11
182
1
it IS the same current! it runs in the same branch between V1 and V2 !!!
how can it NOT be the same current?
can two different currents run in the same branch???
 
  • #12
818
67
it IS the same current! it runs in the same branch between V1 and V2 !!!
how can it NOT be the same current?
can two different currents run in the same branch???
Let's say you actually built this circuit and measured the current in the branch. With, for instance, V1 = 100 V and V2 = 50 V, you'd probably agree that you'd measure 10 A with a direction from V1 to V2?

Now consider the two currents I1 and I2 that I defined in #10:

I1 = (V1 - V2)/5 A = (100 - 50)/5 A = 10 A
I2 = (V2 - V1)/5 A = (50 - 100)/5 A = -10 A

So you see, both these currents give the same result. I1 is 10 A with a direction from V1 to V2, which is correct. I2 is -10 A with a direction from V2 to V1, which is the same as 10 A with a direction from V1 to V2.

You can define all the currents you want. All that matters is that you apply Ohm's law, KVL/KCL etc. correctly, so you get true equations.
 
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  • #13
182
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mmm I see
that was a bit tricky though, isn't it?

and if I applied KCL for V2 as this, would it be a mistake?
V1-V2/5 + V2/3 + (V2-V3)/2 = 0
 
  • #14
818
67
and if I applied KCL for V2 as this, would it be a mistake?
V1-V2/5 + V2/3 + (V2-V3)/2 = 0
You probably mean (V1-V2)/5 + V2/3 + (V2-V3)/2 = 0, but yes. V2/3 and (V2-V3)/2 are for currents with reference directions outwards from the node, but (V1-V2)/5 is for a current that's flowing into the node, which means they have to have opposite signs, e.g.:
-(V1-V2)/5 + V2/3 + (V2-V3)/2 = 0
 

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