Node Voltage: Where to choose ground?

[rugerts]

Homework Statement

Find the branch currents.

Relevant Equations

KCL and Ohm's -> Node Voltage

Hello,
Here's the example I'm looking at now.

I am wondering if I swap ground and Node C... will this still be correct? I can see how it's more useful to have ground where it is above, but I'd still like to know if it would work for a swap of C and GND.

Here's my work trying this:

Does anyone see where I went wrong?

 

[gneill]

Check the signs you've used for the voltage sources in your node equations. Also, I see that node B involves currents i2, i4, and i5, but not i3.

 

[rugerts]

Check the signs you've used for the voltage sources in your node equations. Also, I see that node B involves currents i2, i4, and i5, but not i3.

Why should it involve i3?

And, for the signs, it feels like there's something wrong with the Vc - 125 - Va part since, if GND were in C's spot, i1 would = 125- Va. Could you provide any clues as to what's going on here?

 

[gneill]

Why should it involve i3?

It should not. In your work you wrote for node B:

And, for the signs, it feels like there's something wrong with the Vc - 125 - Va part since, if GND were in C's spot, i1 would = 125- Va. Could you provide any clues as to what's going on here?

Let's examine the expression for the $i_1$ branch. If you do a "KVL walk" from node C to node node A then your starting potential is $V_C$, then you "walk" through the 125 V supply increasing your potential to $V_C + 125$. Next you pass through the resistor, dropping your potential to reach $V_A$. Hence the expression for $i_1$ is:

$i_1 = \frac{V_C + 125 - V_A}{1}$

 

[rugerts]

It should not. In your work you wrote for node B:
View attachment 252668Let's examine the expression for the $i_1$ branch. If you do a "KVL walk" from node C to node node A then your starting potential is $V_C$, then you "walk" through the 125 V supply increasing your potential to $V_C + 125$. Next you pass through the resistor, dropping your potential to reach $V_A$. Hence the expression for $i_1$ is:

$i_1 = \frac{V_C + 125 - V_A}{1}$

Interesting. What you say for the KVL walk makes sense, but I had thought that for KVL walks you simply write down the first sign you encounter. Here, for a clockwise loop, I hit the minus terminal first, which results in -125 instead of +125. What is the flaw in this "first sign you encounter" approach? I've definitely heard my prof. mention this method for KVL walks along with another (which is most likely the one you mention).

 

[gneill]

I've never used your "first sign encountered" rule before. I've always just done the KVL walk in the direction of assumed loop (or branch) current when those current directions are specified in the problem statement.

Otherwise, if they are not given, I just assume that all currents at a given node are flowing out of the node in question and that they sum to zero. The math takes automatically takes care of sorting out the signs (hence directions) of the actual currents.

To me it makes sense that the potential at the left end of the resistor in that branch must be $V_C + 125$

 

[rugerts]

I've never used your "first sign encountered" rule before. I've always just done the KVL walk in the direction of assumed loop (or branch) current when those current directions are specified in the problem statement.

Otherwise, if they are not given, I just assume that all currents at a given node are flowing out of the node in question and that they sum to zero. The math takes automatically takes care of sorting out the signs (hence directions) of the actual currents.

To me it makes sense that the potential at the left end of the resistor in that branch must be $V_C + 125$

View attachment 252672

Here's a screenshot of the relevant slide.
What you say of course works and makes sense. It's just strange the other alternative doesn't work here. Do you think that alternative way may not apply here because we're working with node voltages, and the node voltage doesn't have a sign component to it? That is, there aren't any terminals (and hence no signs to go off of) to work with for Vc. Vc is just the sum of the voltages on its branch.

 

[gneill]

Thanks for the screenshot.

I suspect that the reason the method works for KVL full loops is that the summation of potential changes is set equal to zero regardless of whether you are summing potential rises or potential drops (multiply through by -1 and the equation remains valid).

When you're doing nodal analysis you're not doing the summation over a closed loop but just along a branch, so using the "correct" signs for potential changes along the walk matters.