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.

Ok, that makes sense now - thanks.

 

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 will set aside this issue 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?

As to supernodes, I defer to ratch since he seems to have greater comfort with that concept. I don't use supernode in my analysis. My analysis approach of choice is Millman's Theorem. The two voltages in your circuit are in series.

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.

I have no reason to refute the truth of this statement from your book.

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.

Once more, I defer to ratch on the interpretation of the supernode identification in this circuit.

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?

Actually what I said was the resistor is connected across (not in series with) the two series batteries. I suppose you could say that value of the current changes. Right or wrong, I have always considered that the current flows through the load. This concept has served me well over my 35 year career.

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.

A short circuit is not an infinite resistance. An open circuit is an infinite resistance. A short circuit is a zero resistance. If you plug zero ohms into the Ohm's Law expression in an attempt to calculate the current, you end up with division by zero which is not defined. You can say that as the resistance across an "ideal" voltage source approaches zero, the current approaches infinity in the limit.

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?

You statement above seems reasonable.

hgmjr

 

ihaveaquestion,

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.

I showed you how to combine 3 nodes connected by voltage sources into one supernode. Only one equation was needed to solve for one of the node voltages (V1 in this case). The other two voltages were easily found by knowing the values of the voltages between each of the nodes. You should review that calculation again. If you want to desigate only two nodes like V1,V2 or V2,V3 as a supernode, fine, but you will then need two equations to solve for the voltages. If you do not want to implement any supernodes, then you will need three equations to solve the node voltages. Any number of nodes can be defined as a supernode as long as they are adjacent and connected by a voltage source. Supernodes help to reduce the number of equations to be solved. Is there anything about this you do not understand?

Ratch

 

Actually what I said was the resistor is connected across (not in series with) the two series batteries. I suppose you could say that value of the current changes. Right or wrong, I have always considered that the current flows through the load. This concept has served me well over my 35 year career.

A short circuit is not an infinite resistance. An open circuit is an infinite resistance. A short circuit is a zero resistance. If you plug zero ohms into the Ohm's Law expression in an attempt to calculate the current, you end up with division by zero which is not defined. You can say that as the resistance across an "ideal" voltage source approaches zero, the current approaches infinity in the limit.

hgmjr

So when you said across did you not mean in series?

This is what I interpretted in your first situation:
http://img32.imageshack.us/img32/4121/21044467.jpg

(Sorry that was a typo what I said in the second part... I meant what you explained. A short circuit has infinite CURRENT not resistance due to I = V / R(zero). So if I put a battery with a wire connected to the +/- terminals in front of me on a table... theoretically there's infinite current being drawn throughout the short circuit... but in reality it's just a really high value that we can measure with a ammeter?)

Anyway back to the picture above... when I add that resistor the current slows down...

What you said in your other message also confused me and I forgot to mention it. You said add a resistor in parallel and the current will not change. Like this?

http://img530.imageshack.us/img530/6553/59544676.jpg

If we did that the equivalent resistance value would become smaller and again according to Ohm's law we would get a different value, correct?

Also, according to the definition of elements being in series that I know, I believe I am in disagreement with your statement that the two voltage sources in the circuit of the original problem in this thread are in series. For two elements to be considered in series, they have to share a single node that is shared only between those two elements and no others. In this circuit the node that the two voltage sources are sharing is also being shared by the 4 ohm resistor.


Ratch:

I looked over your equation again. I'm surprised I didn't notice earlier that all you did was express everything in terms of V1 so you could solve for it... I would have written V1 - 10 as simply V2 for example then go through more math to arrive at your conclusion. HOWEVER, I need help on a small issue with your equation:

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

Here's my circuit picture linked for convenience:
http://img33.imageshack.us/img33/1389/scan0081w.jpg

The last term being added on in your equation:
I understand the 5v1/2 part.... and all over 3 because of the resistance... just not the v1-10 part... perhaps you could help me understand that part of the calculation - the current being drawn over the 3 ohm resistor from v3 to ground in terms of v1...

thanks

 

ihaveaquestion,

I looked over your equation again. I'm surprised I didn't notice earlier that all you did was express everything in terms of V1 so you could solve for it... I would have written V1 - 10 as simply V2 for example then go through more math to arrive at your conclusion. HOWEVER, I need help on a small issue with your equation:

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

Here's my circuit picture linked for convenience:
http://img33.imageshack.us/img33/1389/scan0081w.jpg

The last term being added on in your equation:
I understand the 5v1/2 part.... and all over 3 because of the resistance... just not the v1-10 part... perhaps you could help me understand that part of the calculation - the current being drawn over the 3 ohm resistor from v3 to ground in terms of v1...

Sure, V3 = V2 + 5*V1/2 . V2 = V1 - 10 . Substitute the second equations into the first and you get the last term of my solution equation you are inquiring about. What don't you understand?

Ratch

 

Don't know why, but I only understood it after you broke it down a step further for me like you just did... I see now - thanks.