# Nodal analysis and Supernodes

Discussion in 'Homework Help' started by SilverKing, May 29, 2014.

1. ### SilverKing Thread Starter Member

Feb 2, 2014
72
0
Hi everyone,

I've the following problem:
http://s9.postimg.org/6wgvejisv/Untitled.png

I've assumed that that the node between 4 ohms resistor, 3 ohms resistor and 9 Volts source, and the node between 9 Volts source, 2 ohms resistors and 6 ohms resistor, are forming a Supernode.

So, Applying KCL on the Supernode:

$I_{4 \Omega}=I_{3 \Omega}+I_{2 \Omega}+I_{6 \Omega}$

$\frac{V_{1} - 21}{4}=\frac{V_{1}}{3}+\frac{V_{2}}{2}+\frac{V_{2}}{6}$

$V_{1}+8V_{2}=-63 \rightarrow (1)$

Applying KVL on the Supernode:

$-V_{1}-9+V_{2}=0$

$V_{1}-V_{2}=-9 \rightarrow (2)$

Solving (1) and (2):

V1= -63, V2=-9

Which - as you can see above - not the right answer.

Any help?

2. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
Stop being lazy and sloppy in your work. Your error is a direct result of this.

Your terms such as $I_{4\Omega}$ is meaningless because you don't designate a polarity for it. Similarly, you use the designators $V_1$ and $V_2$ but never define them. As a result, we have to reverse engineer your work and/or guess at what you mean. But engineering is NOT about guessing.

Your first equation is only correct if $I_{4\Omega}$ is the current flowing left-to-right through the resistor and all of the other three are the currents flowing downward through their respective resistors. But then, on the next line, you replace $I_{4\Omega}$ with an expression that corresponds to the current flowing right-to-left. You couldn't catch it because you were too lazy and sloppy in your work to take the time to properly define your terms.

Along with this is that you need to start properly tracking your units throughout your work. Heck, you aren't even tacking the units onto the final answer!

Please note that I am not making a personal remark against you, but rather how you are approaching your work. Doing work properly is a learned skill and we can only correct our weaknesses if they are pointed out to us.

3. ### SilverKing Thread Starter Member

Feb 2, 2014
72
0
Okaaaaaaay. Hold it on.

First of all, If I was "lazy" I wouldn't bother myself to ask here and there while my - supposed to be - professor is having some nice nap. Anyway, that's belongs to me.

I also want to say that if you knew the education I got and the way I get to the internet, you wouldn't say that.

Anyway, I've solved the problem and knew my mistake, it was in identifying the ordinary and the extraordinary nodes.

Thanks anyway.

4. ### freemindbmx Member

Mar 5, 2014
72
0
I know Wbahn posted before but I figured I post what I got and how I got there.

I chose my ground to be directly under the 21V independent source.And by quick inspection I can tell that there is going to be a supernode problem in my work. The 9V independent source has no connection to my ground,thus its a supernode.And current leaving is positive as my convention.

V1+V2:

V1(1/4Ω+1/3Ω)-(21V/4Ω) + V2(1/2Ω+1/6Ω)=0

AUX Equation:
9V=V2-V1

Finding what V1 is equal to in our AUX equation: V1=V2-9V

Plugging this into V1+V2:
(V2-9V)(1/4Ω+1/3Ω)+V2(1/2Ω+1/6Ω)=(21V/4Ω)
Multiplying everything through:
V2(1/4Ω+1/3Ω+1/2Ω+1/6Ω)=(21V/4Ω)+(9V/4Ω)+(9V/3Ω)
V2= 8.4V
Plugging this value into V1=V2-9V
V1=-0.6V
to solve for "v": "v" is the voltage across the 3ohm resistor and due to my convention and since its already labeled as + top,-bottom.The voltage of "v" is the voltage of the node V1.
To solve for I: I is through the 2ohm resistor,and can be expressed as V2/2Ω-> 4.2A

I think I covered everything,let me know. Also Wbahn if you read this, how do you answer these posts so fast?Do you have forum posts hooked up to you email and phone?

5. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
As I stated, I did not NOT say that YOU were lazy. I said that you are being lazy and sloppy IN YOUR WORK. What you presented (and didn't present) are adequate to establish that. And it says nothing pro or con about you in any other aspect of your life.

And your mistake was in not defining your current polarities and, as a consequence, using two different polarities for the same current. It's that simple.

So let me show you by example what it takes to be properly diligent in your work.

First, annotate your diagram with the definitions you will use.

The entire process of capturing the image, annotating it in MS Paint, and uploading it took less than three minutes. Not too much to expect.

Then, express your governing equations in a manner such that their correctness can be readily determined by inspection of your diagram. Remember, these equations represent all of the science and engineering in the solution. Everything else is 'just' math.

KCL at the supernode (the blue box):

$I_4 \; = \; I_3 \; + \; I_2 \; + \; I_6$

$\frac{$$21V-V_1$$}{4\Omega} \; = \; \frac{V_1}{3\Omega} \; + \; \frac{V_2}{2\Omega} \; + \; \frac{V_2}{6\Omega}$

Supernode constraints:

$V_2 \; - \; V_1\; = \; 9V$

There. The EE work is done. Compare these to your equations. Do you see the error and how it is that you used one polarity for $I_4$ in the KCL equation but then used the opposite polarity when you expressed that same current in terms of the voltages and resistances in the circuit? Do you see how having prepared a properly annotated diagram significantly reduces the chance of making that mistake or, even if you do make it, of catching it?

Now it's time to crank the handle on the math.

$7V_1 \; + \; 8V_2 \; = \; 63V$

$-V_1 \; + \; V_2 \; = \; 9V$

Multiplying the second equation by 7 and adding to the first yields:

$15V_2 \; = \; 126V$

$V_2 \; = \; \frac{126V}{15} \; = \; 8.4V$

Multiplying the second equation by -8 and adding to the first yields:

$15V_1 \; = \; -9V$

$V_1 \; = \; \frac{-9V}{15} \; = \; -0.6V$

$v \; = \; V_1 \; = \; -0.6V$

$i \; = \; \frac{V_2}{2\Omega} \; = \; \frac{8.4V}{2\Omega} \; = \; 4.2A$

Then, don't stop there just because the supposedly correct answers are provided for you. First, they are sometimes wrong. Second, you aren't going to get them in the real world and will have to accept responsibility for verifying your own work. So start with your answers and use the problem to verify their validity.

Check #1:

$V_2 \; - \; V_1 \; = \; 9V \ ?$

$8.4V \; - \; -0.6V \; = \; 9V \ check$

Check #2:

$V_1 \; + \; I_4 $$4\Omega$$ \; = \; 21V \ ?$

$V_1 \; + \; $$I_2 \; + \; I_6 \; + \; I_3$$ $$4\Omega$$ \; = \; 21V \ ?$

$-0.6V \; + \; $$\frac{8.4V}{2 \Omega} \; + \; \frac{8.4V}{6 \Omega} \; + \; \frac{-0.6V}{3 \Omega}$$ $$4\Omega$$ \; = \; 21V \ ?$

$-0.6V \; + \; $$4.2A \; + \; 1.4A \; - \; 0.2A$$ $$4\Omega$$ \; = \; 21V \ ?$

$-0.6V \; + \; $$5.4A$$ $$4\Omega$$ \; = \; 21V \ ?$

$-0.6V \; + \; 21.6V \; = \; 21V \ check$

Another very good check for most problems of this nature is to compute the power delivered by the sources and the power absorbed by the loads and confirm that they match.

If you will start presenting your work in a manner similar to this, you will be surprised how much your grades will improve. More importantly, you will learn to do quality work that is verified as being correct once you get out into the real world. That will greatly increase your value to your employer and decrease the likelihood that you will make a mistake that goes uncaught and gets someone killed.

Finally, consider that if it is not unreasonably for me to present this problem this way for your benefit (I'm sure not getting any points for it!), then it probably is not too unreasonable to expect a similar level of effort on your part.

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6. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
I get e-mailed whenever there is a new response to a thread I have responded to, but that's it. Whenever I check the forum I always see if there are any new threads in Homework Help and, if there are, I see if I can respond. This is particularly true for threads that haven't received any responses yet.