Understanding Voltage and Current: How Reference Points Affect Calculated Values

[csmith23]

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

 

[Ackbach]

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.

 

[csmith23]

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.

 

[Dethrone]

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.

 

[csmith23]

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!