# question about an odd nodal analysis problem

Discussion in 'Homework Help' started by hunterage2000, Nov 16, 2010.

1. ### hunterage2000 Thread Starter Active Member

May 2, 2010
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Can someone tell me how the equation for the first node is:

(e1-1/10) + (e1/10) + (e1-e2/5.1) = 0

From the analysis the current over R3 = 0

so what I want to know is (i) why (e1-e2/5.1) is included in the equation if the current = 0 and (ii) why is (e1-e2/5.1) + not - as it is leaving the node.

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May 2, 2010
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3. ### Jony130 AAC Fanatic!

Feb 17, 2009
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I think that we have some freedom in choosing a current direction in nodal analysis.
For example for this diagram I choose

So I can write

(V1 - E1)/R1 = E1/R2 + (E1 - E2)/R3 (1)

I1 = (E1 - E2)/R3 - (E2 - V2)/R4 (2)

And after solving I get
E1=0.5V and E2=0.5V

But I can choose that all current flow out from the node
Then

-(E1 - V1)/R1 - E1/R2 - (E1-E2)/R3 = 0

-(E2 -E1)/R3 - (E2 - V2)/R4 - I1 = 0

And again after solve this I get the same result.

And finally I can choose that all current flow out from the node and give them "+" if they flow out from the node.

(E1 - V1)/R1 + E1/R2 + (E1-E2)/R3 = 0

(E2 -E1)/R3 + (E2 - V2)/R4 + I1 = 0

4. ### hunterage2000 Thread Starter Active Member

May 2, 2010
400
0
can you tell me why the current direction from V2 isnt the other way or is it because the current source is in parallel with it. Does the current source have an influence over the whole circuit or only those in parallel with it?

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
Becaues when we start nodal analysis we don't now in which direction current will flow.
So we can assume any direction we want.

Last edited: Nov 17, 2010