# To supernode or not to supernode...

Discussion in 'Homework Help' started by badCircuitAnalyst, Feb 13, 2015.

1. ### badCircuitAnalyst Thread Starter New Member

Feb 12, 2015
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0
When I first saw this problem my initial reaction was to solve using a supernode since we have a voltage source between two non-reference nodes. Then I figured that you're probably not allowed to have another component in series within a supernode.

So I defined a new node between the 2Ω resistor and voltage source and approached the problem this way:
(I've included ground as Vg and try to show all my work and make it as clear as possible)

I'm slightly worried about my statement of equality for i1. I know current across a resistor in series is the same but I'm not sure where else I could've gone wrong. I know the solution can be obtained through regular nodal analysis like as I've show below, but I don't understand where I went wrong in the method above and why it doesn't work.

Maybe I'm over-thinking things, but I want to try to understand these concepts as thoroughly as possible. Any input is appreciated!!

2. ### Dodgydave Distinguished Member

Jun 22, 2012
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I would like to know where in the "REAL WORLD" will you find a circuit like this?

3. ### Jony130 AAC Fanatic!

Feb 17, 2009
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If you have a supernode in your circuit you treat it as if you had a one big node (supernode) in your circuit.
So to write a KCL we "short" V2 node with Vx node.
-15A + V2/4Ω + (Vx - V1)/2Ω = 0,
And KCL for V1 node
V1/8Ω + (V1 - Vx)/2Ω = 0
And one additional equation for Vx or V2.
Vx = 10V+V2 or V2 = Vx - 10V;

4. ### WBahn Moderator

Mar 31, 2012
17,454
4,701
First, it doesn't matter since the point is to develop analysis skills.

Second, lots of transistor problems reduce to circuits along these lines in both small signal and large signal.

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5. ### WBahn Moderator

Mar 31, 2012
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4,701
It's hard to follow your work because you have bits and pieces located at random locations on your paper. Organize your work so that it flows from beginning to end.

You need to track your units much better. Look at your equation #1: V1 + 2·V2 = 120A. Do you really believe that if I add two voltages together I'm going to get a current?

Now look at your fourth and fifth lines from the bottom. Do you really believe that 5·2=8 ?

Once you found that your answer didn't work, how much effort did you put into reviewing your work to see if you did everything right?

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6. ### badCircuitAnalyst Thread Starter New Member

Feb 12, 2015
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I apologize for the oversight on my part. I did look over my work several times but was thinking I most likely made a conceptual error. I understand why you would assume the worst of people on here; regardless thank you for your response.

7. ### WBahn Moderator

Mar 31, 2012
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4,701
It's not a matter of assuming the worst of you at all. The fact that you showed your work at all, and in your initial post no less, puts you head and shoulders above most of the newbies here -- I meant to specifically thank you for that and forgot to do so. Then you obviously didn't just accept your answer but did it a different way and got a different result and proceeded to try to find out why. Again I applaud you for that.

My intent was to point out several errors that would have been a lot easier for you to catch if you had been more organized in your work (and your organization was a LOT better than many folks, so you just need to clean it up and pay attention to detail a bit better, as opposed to starting over from scratch in your approach.

We all make mistakes, including stupid mistakes like thinking that 5·2 is 8 (and actually it is pretty easy to see where you goofed since you have 4/5 in there and so you just latched onto the 4 instead of the 5 -- we all do it). You are more likely to make those kinds of mistakes when dealing with new concepts and techniques, but you will still make them with some regularity no matter how experienced you get. So by developing a few simple habits, you will find that you will make a LOT fewer of these kinds of mistakes to begin with AND that you will be able to detect most of the ones you do make almost immediately, AND you will be able to track down the ones you don't catch right away much more easily. Those things are (among others):

1) Always, always, ALWAYS properly track your units throughout your work -- and tracking them means that you don't just pay them lip service, but you treat them for what they are, part of the work that must be manipulated according to the rules that govern them.

2) Always, always, ALWAYS ask if the answer makes sense.

3) Keep your work neat and organized. Even if you are doing stuff on scratch paper, do it as though someone else will need to follow your work later and do it the way that you would want someone else to do it if you were to have to follow their work afterwards.

4) Don't skip too many steps or do too much on a calculator between steps. The person reviewing your work -- which includes you when trying to track down an error -- only has what is written down to work from.

5) When you review your work, particularly looking for the source of an error, make no assumptions about what the error might be. Go from line to line making sure that each line is correct based on what has led up to it -- and that means doing all of the math involved, including making sure that the 8 is correct (at which point you would have slapped yourself on the forehead and been quite relieved at having found the mistake.

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8. ### atferrari AAC Fanatic!

Jan 6, 2004
2,611
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1) The son of a...mother teaching us Thermodynamics at the Academy always required:
Statement of the problem at the top, data WITH units on the right and sequence of calculations on the left.
I should have to thank him the equivalent of several lifes, for teaching me that, what I followed religiously for the rest of my life, when doing manual calculations of vessel's trim using moments.

3) The first "someone else" will be YOU. No doubt.

4) Helped me to revise my own calculations.

5) Ditto, as in 4)

Last edited: Feb 16, 2015
9. ### badCircuitAnalyst Thread Starter New Member

Feb 12, 2015
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You bring up great points. I would definitely benefit from taking a more standardized approach to calculations (even in my math classes). You've been nothing but helpful since I joined this forum so I do sincerely appreciate that! Good day sir!

10. ### Rayaarito New Member

Jan 8, 2015
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I know this is a very old post but I am having trouble with this problem. I can solve it when I create a node inbetween the 10V source and the 2 Ohm resistor but I can not understand how you get (V2+10)/(8+2). Can somebody please explain that to me? Every problem that i have with a resistor next to a super node, I create that extra node inbetween the two and it makes the solving the problem longer than it has to be

11. ### WBahn Moderator

Mar 31, 2012
17,454
4,701
Let's see if this helps.

Normal Node:

The node equation is simply KCL, summing the currents leaving the node and setting that sum to zero, applied at Node X:

$
\frac{$$V_X-V_A$$}{R_A} \; + \; \frac{$$V_X-V_B$$}{R_B} \; + \; \frac{$$V_X-V_C$$}{R_C} \; = \; 0
$

Super Node:

The node equation is simple KCL, summing the currents leaving the supernode and setting the sum equal to zero, applied at the boundaries of the supernode consisting of X and Y and Z.

$
\frac{$$V_X-V_A$$}{R_A} \; + \; \frac{$$V_Y-V_B$$}{R_B} \; + \; \frac{$$V_Z-V_C$$}{R_C} \; = \; 0
$

You are merely treating the supernode as a black box and recognizing that KCL applies to the black box as a whole every bit as much as it applies to a normal node.

But in this case you have three unknowns associated with the supernode, Vx, Vy, and Vz, and so you need three equations. The supernode equation gives you one. That's looking at the circuit outside the supernode. To get the other two you have to look at the circuit inside the supernode and come up with two equations relating the voltages at the boundary to each other (if you have N boundary nodes you need (N-1) such "auxiliary" constraint equations since, as always, you get to pick one boundary node as a reference and write the others relative to that node).

12. ### Rayaarito New Member

Jan 8, 2015
2
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Thank you so much for the quick response

Where i wrote "Where the correct answer is" I meant "BUT the correct answer is:"
I see what you're saying with that formula but in this case [referring to (Vx-Va)/Ra] Vx is equal to V1. Va is equal to 0 since it is the ground node. And Ra is 8 ohms. I don't understand why you would add the 2 ohms to the 8 ohms if the 2 ohms is before the node. If that's the case then wouldn't you add the 2 ohms to the 4 ohms in this scenario? (the i2 case)

13. ### WBahn Moderator

Mar 31, 2012
17,454
4,701
The quickest way to see that the correct answer is, indeed, correct is to ask what the voltage is at the positive terminal of the 10V source. It is simply (v2+10V). From that node, you have 10Ω (the 2Ω and 8Ω resistors in series) to ground, so the current through those resistors is

i1 = (v2+10V)/(2Ω+8Ω) = (v2/10Ω) + 1A

The claim that v1 = 10V + v2 is wrong except in the case in which there is no current flowing in the 2Ω resistor.

Your constraint equation between v1 and v2 has to take the voltage drop across the 2Ω resistor into account, so it should be

v1 = v2 + 10V - i1(2Ω)

which becomes

v1 = v2 + 10V - [(v1-0V)/8Ω](2Ω)
v1 = v2 + 10V - [v1/8Ω](2Ω)
v1 = v2 + 10V - (v1/4)
v1 + (v1/4) = v1(5/4) = v2 + 10V
v1 = (4/5)(v2 + 10V)
v1 = (4/5)v2 + 8V

Then you have that

i1 = v1/8Ω
i1 = [(4/5)v2 + 8V]/8Ω
i1 = (v2/10Ω) + 1A

In order for a supernode to be most useful, it should partition the circuit into two pieces such that the relationships between the boundary nodes depend only on things inside the node and not on things that are influences by things outside of it, such as i1. It's still valid if you violate this rule, just not nearly as useful.