I'm currently in the process of working through some circuit analysis books on my own. Though it's easy enough to compare my answers to those given in the appendices, I figured that it would be worthwhile to learn how to simulate the problems in MultiSim as well.

However, I ran into trouble when trying to simulate the circuit shown below.

Original Schematic

Utilizing the node-voltage method, I determined that the voltages at the nodes will be the following, with respect to the ground node:

Node A = 40 [V]
Node B = 172 [V]
Node C = -64 [V]

I'm sure that these answers are correct. The problem is that if I run the simulation as shown in the schematic, I don't get the voltages I expect with respect to ground at the three nodes. I figured out that V1 is the reason for this.

I can fix this problem by changing the voltage of V1 from 40 [V] to -40 [V], but I don't understand why this works.

In short, why do I need to change the value of V1 in order to get the expected results from the simulation (rather than just simulating the circuit as it is given in the problem)?

I'm sure that I'm overlooking something important here. Help would be greatly appreciated!

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    \$\begingroup\$ Try a simpler problem where you remove the two current sources (you might have already solved this problem if you used the superposition method). Does your hand analysis now agree with the simulation? If not, I think you are just misunderstanding the voltage source symbol: The longer bar is on the side with the "positive" terminal. Meaning that if its value is V, then V(long bar side) - V(short bar side) = V \$\endgroup\$ – The Photon Jun 30 '14 at 0:15
  • \$\begingroup\$ Thanks for the comment! I explained what I did wrong above if you're interested. I really hate careless errors. xD \$\endgroup\$ – user32297 Jun 30 '14 at 5:35
  • \$\begingroup\$ Please write your solution up as an answer, and then accept it (you might have to wait 24 hrs or something). This will keep your question from reappearing on the front page as "unanswered". \$\endgroup\$ – The Photon Jun 30 '14 at 16:37
  • \$\begingroup\$ Yes, it looks like I'll have to wait until tomorrow to accept my own answer. \$\endgroup\$ – user32297 Jun 30 '14 at 18:13

You've installed V1 "upside down" in the simulation. The long bar is the positive terminal, the short bar is the negative.

One simple mnemonic is that the long bar is "more" than the short bar, so it's +. Also the short bar looks like the - symbol.

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  • \$\begingroup\$ Thanks for the comment! The problem was that I solved for the node voltages with the positive terminal of V1 on top, rather than on the bottom like the problem called for. I can finally move on now since I know what I did wrong! Live and learn, I guess. :) \$\endgroup\$ – user32297 Jun 30 '14 at 5:37

It seems that ignorance was the true issue at hand in this case. The voltage values listed above are correct IF the voltage source, V1, is connected with the positive terminal above the negative terminal at a value of 40 [V]. However, this isn't the case in the diagram. As a result, the node voltages with the negative terminal above the positive terminal at 40 [V] will be:

A = -10 [V]
B = 132 [V]
C = -84 [V]

After accounting for this error, MultiSim gave me precisely the results I was looking for. Unfortunately, it turns out that this was little more than a dumb question. Problem solved. :P

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