# Node analysis

Discussion in 'Homework Help' started by kidi3, Oct 16, 2012.

1. ### kidi3 Thread Starter New Member

Oct 16, 2012
4
0
I currently trying to train my skills in analyzing a Electric circuit using the node analysis and mesh method.

But the answers I get isn't the same as the one giving.

http://personal.georgiasouthern.edu...43-lec/exercises-quiz3-mesh-nodal-w-ans.pdf

I tried calculating Exercise 1 and 5 using Node analysis.

these are my calculations for exercise 1
http://snag.gy/Rfcp7.jpg

And these are my calculations for excersice 5
Har også lige prøvet opgave 5

For V1 = 12v

V2 = -2V

I1 =6+(-3) = 3A

V1 = 12V

I2 = -1A

-1 * 2 = -2V...

And here is another one http://snag.gy/Dnuqf.jpg

http://snag.gy/Rm2fs.jpg

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,513
1,272
You have here two unknown nodal voltages Vo and where we have 1Ω resistor

Are you sure that this is a solution for exercise 5
For node V1

3A + V1/4Ω = 6A + (V2-V1)/1Ω

And for V2

5A = V2/2Ω + 6A + (V2-V1)/1Ω

And solution
V1 = 4V l V2 = 2V

http://www.wolframalpha.com/input/?i=3+%2B+V1%2F4+%3D+6+%2B+%28V2-V1%29%2F1%2C+5+%3D+V2%2F2+%2B+6+%2B+%28V2-V1%29%2F1+

You made a error in calculations X = - 12/11
http://www.wolframalpha.com/input/?i=(12+-+X)/6+++(0+-+X)/4+-+(X+-+(-6))/2+==+0

Last edited: Oct 16, 2012
3. ### kidi3 Thread Starter New Member

Oct 16, 2012
4
0

I see there is 2 resistors for the voltage source at 40V.
Are those 2 in series or parallel.

The equation I wrote gave me V_0 to become 25,0714V. But thats incorrect?..

So should i change the resistors in parallel?

I didn't know how i should use the node equation here, because i seemed to be simple.
I shoudn't have thought so..

But using my knowlegde i get these equations.
Are they correctly written?
http://snag.gy/LWdjY.jpg

And about the last one..
Yeah I see my mistake.. - to +..

Thanks for your reply.. I hope you help me with my other questions.

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,513
1,272
You can not do that because you have a 5A current source

So we need to nodal equations for two unknown voltages V1 and Vo.

No, you forget about current sources.

For V1 node we can write

I1 + I2 = I3 + I4

So we can write

3A + V1/4Ω = (V2-V1)/1Ω + 6A

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5. ### kidi3 Thread Starter New Member

Oct 16, 2012
4
0
Ok.. I think i understand it now..

is it the direction of I0?

Oct 16, 2012
4
0
7. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,513
1,272
Hmm, This current source has 1mA or 1A current ??

First I assume that all current flow into the Va node.
And all current which entering the node are positive ( give them "+").
So I write

1mA + (-Va)/R3 + (Vo - Va)/R2 = 0

1mA - Va/510Ω + (0.5V - Va)/330Ω = 0

And the solution

Va = 503.929mV

As for your equations, you made slight error.

(0.5 - V)/330 - (-V)/510 - 1 = 0

You also assume that all current entering into a Va node.
but you assume that all entering currant are negative.
So the correct nodal equations look like this:

-(0.5 - V)/330 - (-V)/510 - 1 = 0

-(0.5 - V)/330 +(V)/510 - 1 = 0

Va = 200.661V

Next time you should use this convention:
Assume that all current flow out from the node and give them "+" if they flow out from the node.

So we can write

Va/R3 + (Va - Vo)/R2 + (-1A) = 0

Va/510 + (Va - 0.5)/330 -1A = 0

and solution

Va = 200.661V

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Last edited: Oct 17, 2012