MHB Understanding Voltage and Current: How Reference Points Affect Calculated Values

  • Thread starter Thread starter csmith23
  • Start date Start date
  • Tags Tags
    Current Voltage
csmith23
Messages
16
Reaction score
0
Hello all,

I am working on a homework problem from class and I seem to be in another bind. I have found both voltages and the Current i1. However, both my voltages differ from what she has as answers, but my current for i1 is the same. She only gives us the final numerical value with no work in between. Just to make sure I am not crazy I got v1 = 4.7[V], v2 = 6.45[V](which is my super node), i1 = .645[A]. Her two voltages are v1 = 18.24[V], and v2 = 13.53[V] with the same matching current. I would assume she made a mistake because just to find i1 using both her voltages and ohms law neither one of them give the proper current for i1.

thanks for the help!View attachment 4012
 

Attachments

  • Picture.png
    Picture.png
    9.5 KB · Views: 96
Mathematics news on Phys.org
I don't know what your voltages are, but here's a link to the solution, using mesh current analysis, of a very similar problem (current and voltage sources):

http://www.calvin.edu/~svleest/circuitExamples/NodeVoltageMeshCurrent/soln2.mc.htm

Hope that helps! If you need additional help, please post your working, including any definitions of voltages, currents, mesh currents, etc.
 
I spoke with my professor through email. I asked her the same question on here. Which to my surprise she responded with "Take the reference point to the 20V power supply V2 will be 20V." I never responded to that email because it just confused me more and she never answer my original question as to why the voltages don't match and how she found that current. Because you won't get that current she listed if you use any other voltage. Here is my work. Maybe I'm crazy.https://onedrive.live.com/redir?res...authkey=!AKitp_nw8Ke9Gg4&v=3&ithint=photo,png

Sorry tried everything to embed the picture. I kept going over file size.
 
Hi csmith23,

I would like to point out that you might simply be getting different voltages because you chose a different reference point. Normally, a super node can be avoided (which is ideal) if you set that to be the reference point, which is what your professor did. If you do that, then $V_2$ would simply be $20V$ since it $20-0=20V$.

This is what your she did:

Nodal analysis about node 1:

$$-2+\frac{V_1-V_3}{2}+\frac{V_1-20}{5}=0$$
$$\left(\frac{1}{2}+\frac{1}{5}\right)V_1-\left(\frac{1}{2}\right)V_3=0$$

Nodal analysis about node 2 (the node that you used as your reference point (i.e bottom of the circuit)):
$$2+\frac{V_3-V_1}{2}+\frac{V_3-20}{10}+1=0$$
$$\left(\frac{1}{2}+\frac{1}{5}\right)V_3-\left(\frac{1}{2}\right)V_1=-1$$

Solving the two equations will yield $V_1=18.24V$ and $V_2=13.53V$, as desired.

In fact, what I mentioned earlier is precisely what happened, so your answer is correct as well. If we set node 2 as the reference node, then as you computed $V_1-V_{ref}=4.7$. Notice though, that the difference between the voltages I calculated is $V_1-V_2=4.7$ as well! So I believe the difference is only due to your choice of reference.
 
Last edited:
Rido12 said:
Hi csmith23,

I would like to point out that you might simply be getting different voltages because you chose a different reference point. Normally, a super node can be avoided (which is ideal) if you set that to be the reference point, which is what your professor did. If you do that, then $V_2$ would simply be $20V$ since it $20-0=20V$.

This is what your she did:

Nodal analysis about node 1:

$$-2+\frac{V_1-V_3}{2}+\frac{V_1-20}{5}=0$$
$$\left(\frac{1}{2}+\frac{1}{5}\right)V_1-\left(\frac{1}{2}\right)V_3=0$$

Nodal analysis about node 2 (the node that you used as your reference point (i.e bottom of the circuit)):
$$2+\frac{V_3-V_1}{2}+\frac{V_3-20}{10}+1=0$$
$$\left(\frac{1}{2}+\frac{1}{5}\right)V_3-\left(\frac{1}{2}\right)V_1=-1$$

Solving the two equations will yield $V_1=18.24V$ and $V_2=13.53V$, as desired.

In fact, what I mentioned earlier is precisely what happened, so your answer is correct as well. If we set node 2 as the reference node, then as you computed $V_1-V_{ref}=4.7$. Notice though, that the difference between the voltages I calculated is $V_1-V_2=4.7$ as well! So I believe the difference is only due to your choice of reference.
Wow. I am impressed. Thank you so much for your help. Now I have a better idea on how reference nodes change voltages. THANKS!
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.

Similar threads

Replies
12
Views
2K
Replies
1
Views
3K
Replies
26
Views
3K
Replies
6
Views
3K
Replies
13
Views
2K
Back
Top