# Help with nodal analysis problem

Discussion in 'Homework Help' started by leoisleo, Jan 28, 2015.

1. ### leoisleo Thread Starter New Member

Jan 25, 2015
10
0
I'm stuck on a nodal analysis problem. I'll post a picture of the problem as well as my attempt on it.

The problem:
https://www.dropbox.com/s/zovi5e0knoxcy6w/Photo 2015-01-27 11 45 26.jpg?dl=0

My attempt:
https://www.dropbox.com/s/jttay1paw0ls6vo/Photo 2015-01-28 14 13 43.jpg?dl=0

In the picture on the link below I put vb = 0 and vc = vd in my equations and tried to solve vx but got the wrong answer. It should be -4 V.

https://www.dropbox.com/s/2lgm6aknrvq1aw6/Photo 2015-01-28 19 45 50.jpg?dl=0

So I don't know where I have gone wrong with this. Is there something wrong with my equations? Maybe vc isn't equal to vd? I really don't know, too confused. Would appreciate help with this.

2. ### WBahn Moderator

Mar 31, 2012
18,079
4,917
I'm tired of going out to drop box or this site or that site, almost all of which want to install cookies on my machine, to view images that are supposed to be posted HERE.

Please scale and save your images so that they are between 300 and 500 pixels wide and that the files (PNG for line art and JPG for photos) are under 200KB (50KB preferred) and attach them to your post (and embed them in the post if small enough) and you will get much more of a response.

3. ### leoisleo Thread Starter New Member

Jan 25, 2015
10
0
Alright here they are:

4. ### WBahn Moderator

Mar 31, 2012
18,079
4,917
Remember that Node Voltage Analysis (or just Nodal Analysis) is nothing but a formalized way of applying KCL. You need to be sure that you account for all of the current at each node.

In your first equation, you accounted for the red currents but not the blue one.

In your second equation, you accounted for the green currents but not the yellow one.

Do you see the one you missed in your third equation?

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
I think that the simplest and the fastest way to find Vx is to pick Vs- (Vx) as a reference point.
Also notice that R3 is short so Vc = Vd = Vs = 5V.

So all we have left with is Vb as a unknown.

Vb/R1 + (Vb - Vs)/R2 + (Vb - Vs)/R4 = 0

Vb/10kΩ + (Vb - 5V)/5kΩ+ (Vb - 5V)/5kΩ = 0