Discussion in 'Homework Help' started by ihaveaquestion, Aug 23, 2009.

May 1, 2009
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2. t_n_k AAC Fanatic!

Mar 6, 2009
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While it doesn't really help in checking your solution I have attached a novel source pushing solution which reduces the math somewhat - if you can accept the approach!

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3. ihaveaquestion Thread Starter Active Member

May 1, 2009
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I have no idea what source pushing is haha

4. hgmjr Retired Moderator

Jan 28, 2005
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As it turns out, the 6 ohm resistor in this particular circuit can be ignored. That is because the circuit contains two voltage sources and the current on which the current-controlled voltage-source depends is not dependent on the 6 ohm resistor. By eliminating the 6 ohm resistor, the problem simplifies to a two loop circuit.

hgmjr

5. ihaveaquestion Thread Starter Active Member

May 1, 2009
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Could you think of either another way to explain that or another reason why we don't count the 6 ohm resistor please? I can see what you're saying, but it still doesn't make sense for me to just omit it from the circuit... why would current not flow either clockwise or counterclockwise across the 6 ohm?

6. hgmjr Retired Moderator

Jan 28, 2005
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The 6 ohm does draw current from the two series voltage sources, I am only pointing out that the current it draws does not have any impact on the voltages V1, V2, or V3.

hgmjr

7. ihaveaquestion Thread Starter Active Member

May 1, 2009
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1) How do I know which way the current is drawn? Can I arbitrarily pick?

2) If it actually is being drawn and I'm including in my calculations, the shouldn't I get the right answer?

8. hgmjr Retired Moderator

Jan 28, 2005
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You can draw the current direction arbitrarily but you will need to take into account the direction you choose in order to interpret your results.

Not sure what you are referring to as "included" in this comment.

hgmjr

9. ihaveaquestion Thread Starter Active Member

May 1, 2009
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If you look at my answer provided in the link in the first thread you'll see what I mean.

10. hgmjr Retired Moderator

Jan 28, 2005
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Including the 6 ohm in an analysis does not mean that the solution will be wrong. All that will happen is the terms containing the 6 ohm resistor will cancel each other out.

hgmjr

11. ihaveaquestion Thread Starter Active Member

May 1, 2009
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Then I suppose we are back to my original question in asking where I went wrong with my solution.

12. Ratch New Member

Mar 20, 2007
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ihaveaquestion,

Your calculation does not use the supernode method. You said in your first post of this thread that the supernode method was to be used. Since three nodes are connected by two voltage sources, you only need one equation. The equation is:

V1/(2.0)+(V1-10)*(1/4)+(V1-10+5*V1*(1/2))*(1/3) = 0

Notice that the 6 ohm resistor does not enter into the supernode equation because it cancels out.

Solving for V1 we get 3.043478260 volts.

V2 = V1-10 = -6.956521740

V3 = V2+5*V1/2= 0.652173910

If you want the current through the 6 ohm resistor, do (V1-V3)/6 .

Ratch

13. ihaveaquestion Thread Starter Active Member

May 1, 2009
314
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It's hard for me to see where it 'cancels out,' especially mathematically since it appears no where in your equation.

Also I'm not sure how you're getting v1-10 at all...

From my understand of the way the book explains it:

If a voltage source is connected between the reference node anda nonreference node, we simply set the voltage at the nonreference node equal to the voltage of the voltage source.

If the votlage source (dependent or independent) is connected between two nonreference nodes, the two nonreference nodes form a generalized node or supernode; we apply oth KCL and KVL to determine the node voltages.

I agree that there's only one supernode, I see that now.

So I guess I don't see how to setup the equation?

I wrote everything out nicely in my attempt linked, so it would help me understand what's going wrong to use my answer a reference.

14. t_n_k AAC Fanatic!

Mar 6, 2009
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The attachment shows why current i4 plays no part in the analysis.

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15. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Looking at your analysis the primary supernode current equations are

i1+i2+i4=0

and

i4=i2+i3

After eliminating i4, one obtains the equation ...

i1+2*i2+i3=0

Which is inconsistent with the requirement that i1+i2+i3=0. So there is an error in your initial assumptions regarding the node currents.

16. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
BTW. The other far more obvious way to conclude that i1+i2+i3=0 is to note that these three currents all converge at the common (ground) node. As hgmjr points out the solution then reduces to a simpler analysis since you will note that i1,i2 & i3 can all be defined without any reference to the 6Ω resistor.

17. Ratch New Member

Mar 20, 2007
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ihaveaquestion,

The supernode is composed of three single nodes (V1,V2,V3) connected by two voltage sources. The currents leaving the supernode equal the currents entering according to KCL. The current leaving V1 is (V1-V3)/6 and the current leaving V3 is (V3-V1)/6 . Adding the two together is zero, which explains why we say they cancel out. So connecting a resistor to two nodes of the same supernode does not change the voltages of its member nodes.

The current leaving V2 is V2/4 . V2=V1-10 as shown in the schematic. Are you sure you understand how to set up node equations?

I agree. Back to the book.

Yes, but you did not use the supernode method. You indicated the problem was to be worked using supernodes. If you want a solution using loop or node analysis, then you should not ask for a supernode solution.

Ratch

18. hgmjr Retired Moderator

Jan 28, 2005
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218
Here is an intuitive way of looking at the lack of an impact by the 6 ohm "in this particular" problem.

Imagine a circuit composed of two batteries (ideal voltage sources) in series. Then place a resistor load across the two batteries. This resistor draws a current from the batteries that is easily calculated using Ohm's Law. Now place a resistor in parallel with the first resistor. Notice that the added resistor will have no effect on the current that was calculated for the first resistor.

Remember that the two voltage sources in this example as in the case of the problem with which you are dealing are "ideal" and therefore they have no internal source resistance.

The 6 ohm resistor in your particular example is connected across the two voltages in your example. It is also in parallel with the other resistors in your example. It therefore has no impact on the voltages calculated for V1, V2, V3.

I hasten to add that the 6 ohm resistor plays no role in your particular circuit, however it would become critical to your problem if for example the current on which the current-controlled voltage source were a function of the current through the 6 ohm resistor rather than the 2 ohm resistor.

hgmjr

19. ihaveaquestion Thread Starter Active Member

May 1, 2009
314
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That confused me hgmjr... first we're saying that ANY current flowing through the 6 ohm resistor is negligible in this circuit... but NOT if the ccvs depends on it? Let's not worry about my not understanding that for now.

I think I need to go back to the basics...

1) Why are the two voltage sources that are next to each other (they are not in series to my understanding) in the circuit considered ONE supernode?

This is from the book:
If the voltage source (dependent or independent) is connected between two nonreference nodes, the two nonreference nodes form a generalized node or supernode; we apply both KCL and KVL to determine the node voltages.

The way the book draws the circuit, there are three nodes - v1, v2, v3. So is this exact configuration an exception to the rule above? I originally thought that since I have two cases of the rule stated above, I have two super nodes, v1&v2 and v2&v3.

2) You state that in the batteries in series analogy ONCE the resistor is added in series with the two voltage sources, a current is drawn. Is it accurate to say that the current value CHANGES? The way I interpet a current being "drawn" is that it is created. With the two batteries in series with only a short circuit connecting everything, a theoretical infinite resistance flows correct? Then by adding the resistor the current slows down.

3) As another fundamental question... since according to Ohm's law, when dealing with DC... current is proportional to voltage... so if I increase the voltage of something without changing anything else, more current will flow... if I have a battery hooked up to a resistor and wire... a 9v battery and a 3 ohm resistor.. I have 3 amps flowing... if I change it to a 20v battery... more current will flow - correct?

Ratch: Is it possible that your equation can be expressed a different way mathematically? The way the book has been doing it is by doing KCL for the supernode... so all the currents into/out of the supernode. Then write the currents in terms of the voltages... I don't see how to get v2 = v1-10 by the diagram.

tnk: I believe I see your way mathematically now, but the fact that I can ignore the resistor didn't immediately catch my attention. I'd like to get to the point where I can notice these omissions by inspection.

20. Ratch New Member

Mar 20, 2007
1,068
4
ihaveaquestion,

Not if it is to be done by the supernode method.

Which is exactly the way I have been doing it.

There is a 10 volt source between V1 and V2. Therefore V1 and V2 will differ by 10 volts. I don't know how to explain it better than that.

Ratch