# Find "v". Nodal Analysis

Discussion in 'Homework Help' started by truettct, Feb 27, 2011.

1. ### truettct Thread Starter New Member

Feb 27, 2011
10
0
1. The problem statement, all variables and given/known data

For I[SUB]s1[/SUB]=3.5A
V[SUB]s2[/SUB]=25V
R[SUB]1[/SUB]=2 ohm
R[SUB]2[/SUB]=7 ohm
R[SUB]3[/SUB]=3 ohm
R[SUB]4[/SUB]=13 ohm
R[SUB]5[/SUB]= 4 ohm

R[SUB]23[/SUB]= 2.1 ohm

2. Relevant equations

For the equations I put R[SUB]2[/SUB] and R[SUB]3[/SUB] in series.

Then i used nodal analysis to come up with the equations for v[SUB]1[/SUB] and v[SUB]2[/SUB] which are the two potentials on either side of R[SUB]23[/SUB].

3. The attempt at a solution

I attached the circuit and a copy of my attempt at the solution but my answers are wrong. I feel like my equations are right but I went wrong with the algebra. I have no one to check my work so if someone can show me where my algebra or equations are flawed so I can stop making the same mistake.

Thanks

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2. ### Heavydoody Active Member

Jul 31, 2009
140
11
I have not gone all the way through your work yet, but look closely at your simplification of the equation for $U_m$ and note that:
$(\frac {1}{R_1}+ \frac {1}{R_{23}}) \neq (\frac {1}{R_1+R_{23}})$

3. ### truettct Thread Starter New Member

Feb 27, 2011
10
0
Ok I understand that I didnt do that part right, but how would you do it. I am just kind of confused and cant grasp the algebra part to solve that part of the equation.

4. ### Heavydoody Active Member

Jul 31, 2009
140
11
Well, first of all, since you know all of your resistor values, why not substitute them in initially. It will make your algebra manipulation a lot easier. Also, I probably would not have combined $R_2$ and $R_3$ as you have done since the voltage you are solving for is in between the two and you will eventually end up having to separate them somehow anyway. You could do a source transformation on the left by replacing the current source with a voltage source equaling $I_{s1} \times R_1$ in series with $R_1$. Then you can add $R_1$ to $R_2$ while still maintaining the node you are solving for. Have you done source transformations yet?

5. ### Heavydoody Active Member

Jul 31, 2009
140
11
BTW, just in case you didn't understand my first post:
$(\frac {1}{R_1}+ \frac {1}{R_{23}}) = (\frac {R_1+R_{23}}{R_1 R_{23}})$

6. ### truettct Thread Starter New Member

Feb 27, 2011
10
0
No. Right now we are going over nodal analysis and mesh analysis.

we were supposed to use nodal analysis on this specific problem, but I'm having a lot of trouble with this specific one.

7. ### Heavydoody Active Member

Jul 31, 2009
140
11
Also, I may be misreading your handwriting, but it looks as though you have added 7+3 and come up with 2.1 when combining the two resistors.

8. ### truettct Thread Starter New Member

Feb 27, 2011
10
0
I tried to rework the problem, but to no avail. I am stumped and have no clue where to go with this problem.

9. ### Heavydoody Active Member

Jul 31, 2009
140
11
When combining two resistors in series you should simply add their resistances. You combined two resitors in parallel to come up with the 2.1. Be that as it may, it does not change the fact that you have been solving for the wrong voltage by combining those two resistors. So, without source transformation, I would start out like this:

10. ### truettct Thread Starter New Member

Feb 27, 2011
10
0
I see where you are getting all of those equations and where mine were wrong but i keep trying to solve the equations, but keep coming up with totally wrong values.

11. ### Heavydoody Active Member

Jul 31, 2009
140
11
Unfortunately, I have a midterm in the morning and must retire. Perhaps someone else can pick up here. But a couple of points before I go. First, post what you have done to solve those three equations. It seems as though you have the analysis technique pretty well down, because your previous equations, while different from mine, were correct, even if your technique for simplifying the circuit was not. Your difficulties seem to lie with algebra, and a second set of eyes always helps there. Also, source transformation would simplify this problem greatly. Your transformed circuit would look like this:

Well, good luck, and hang in there. You will get it.

12. ### Heavydoody Active Member

Jul 31, 2009
140
11
Actually, you could even get it down to here:

13. ### truettct Thread Starter New Member

Feb 27, 2011
10
0
using nodal analysis and solving for the unknown potential between r4 and r5 i was able to calculate that unknown to be 16.65V. Then I plugged this into the equation of KCL of potential v since i had the unknown and solved for v to get 14.23V and it was right. The transformation really helped. to solve fo the first unknown i put R12 and R3 in series. then i split them back up when solving for v.

File size:
2.3 MB
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