|Apr4-09, 09:34 PM||#1|
Need guidance on this circuit question using nodal analysis. Confused.
it asks to solve for V1 and V2 using nodal analysis however I'm not sure how to solve this particular one because there is a voltage source and a resistor between the reference node and V1and I'm not sure what to do. Here is what I did so far because I'm not sure which way to go. I'm hoping someone can tell me which way is correct and why.
What I did so far.
equation 1: ((v2-v1)/1) = 2 + (v1/1) or ((v2-v1)/1) = 2 + ((v1+6)/1) " because of the 6v voltage source?"
equation 2: ((v3-v2)/2) = (v2/1) + ((V2-V1)/1)
equation 3: ((V3-V2)/2) + 3/0? or just + 3i?
equation 1 and equation 3 is what i'm confused about.
Also I just need V1 and V2 for the problem.
|Apr4-09, 09:41 PM||#2|
Also if someone wouldn't mind verifying my solution to this problem by superposition?
It took me a while but I think I got it.
I used KVL for the voltage source and got -12 -3i -3i = 0 so i' = 2Amps
for the Current source I used current division and got both currents to equal 1.5Amps so i'' = 1.5Amps going the other direction
When adding them both together since i'' was going the other way against i' yielded i'-i'' = 0.5 Amps flowing to the right. so i = 0.5 A. Is that right? I can't see why not but I spent so much time and confusion I can never be sure.
|Apr5-09, 12:58 AM||#3|
please someone, the first problem I have finally given up on. I just am confused and cannot find any information on google to answer my question. I don't know what to do for V1 since I am not sure how to set up the equation with a resistor and a voltage source between V1 and the reference node, no example has ever shown that. I assume that V1 would still equal 6 regardless and it would be 6/1 but I'm not certain.
As for the node on the far right that isn't labeled. I don't know what to do with that, is it just equal to 3 volts?
I'm not asking for anyone to solve it just point me to the right equations and I know how to take it from there. I really need to get past this problem.
|Apr5-09, 02:07 AM||#4|
Need guidance on this circuit question using nodal analysis. Confused.
Yep you got the superposition problem right. Voltage source causes 2 amps, but current source causes 1.5 amps in opposite direction. So current in 3 ohm resistor is 0.5 amps.
Word of advice though. Don't use KVL for the voltage source. Its much easier to do this.
When doing superposition the current source is equivalent to an open circuit. So you erase it. Then the series combination of 3 ohm resistors is in parallel with the 6 ohm resistor. And parallel elements have the same voltage.
So the 6 ohm has 12 volts across it and so does the series combo of 3 ohm resistors. So the current in the series combo of 3 ohm resistors is 12 volts / 6 ohms equals 2 amps. But the 3 ohm resistors are in series so they both have 2 amps through them.
This is much much faster than doing KVL with matrices and etc.
About your nodal analysis problem, I am still trying to do that one. Does your book teach of Supernodes and supermeshes? Thats the way they taught us. When I figure it out I will tell you how to do it.
Wait a sec you dont need supernodes here. Hold on a sec. This is actually not hard.
|Apr5-09, 02:31 AM||#5|
Okay I understand why you dont get it. I'm finishing up circuits 2 right now, and I had the same problems as you a year ago. But eventually I got it. It just takes practice.
You dont need a 3rd equation!!!!
KCL simply says. The sum of all currents entering or leaving a node is zero.
So if you only want V1 and V2 you do this. Here are the 2 equations. I am picking current leaving as being positive here.
(v1-6)/1 +2 + (V1-V2)/1 = 0 (Node V1)
(V2-V1)/1 + V2/1 + (V2-3)/2 = 0 (Node V2)
The third node above the 3 volt source is just 3 volts. Remember the reference ground node on the bottom of the circuit. All the voltages are between the nodes in question and it.
But since you know the voltage of the third node you dont need an equation for it.
Since you are only interested in V2 thats all you need.
Oh and for the 6 volt source, that node also has 6 volts because its 6 volts of the source minus 0 volts of the reference node (ground) = 6 volts.
The first branch (6 volt source) isn't really a node, its a branch. A node is more of a junction. Like a fork in the road. Electrons get to that fork and they have a choice which branch to go into. They always choose the path of least resistance.
Anyway you don't need to know that to solve the problem, but it helps to really understand what is going on in terms of electrons traveling in the circuit. You figure stuff out without math that way.
I hope I helped.
|Apr5-09, 01:36 PM||#6|
You did help a lot, thank you so much. I'm a computer engineering major so I don't think I'll have to take circuits II if that is AC analysis and stuff. So I think I'll luck out but it looks hard, is it? I'm use to the software side of things and all this circuit stuff is new. The algebra part isn't hard it's just setting up the problem and getting the right equations for me. Anyways..
So I assume that you used (v2-6)/1 and (v2-3)/2 because the current is going through the positive terminal of the voltage sources correct?
The confusing thing is (I have consulted several textbooks) and they all use different orientations in terms of +'s and -'s and it's been confusing me to death. This is what I've stuck with is that the current flows through the - terminal and out the + terminal of any dependent source like a resistor (---> -|||+ --->) and it flows through the + terminal and out the - terminal of any voltage source (--->(+-)--->). Is that correct? Books seem to be different but I think I'll stick with that orientation for now.
So I originally thought because of that it would be (v2+6)/1 and (v2+3)/2 because of consistent orientation be but I guess not since it's going into the + terminal of the voltage sources. Thanks for clearing that up, I trust your correct more than me.
So using (v2-6)/1 + 2 + (v1-v2)/1 = 0 reduces to 2v1-v2 = 4
and (v2-v1)/1 + v2/1 + (v2-3)/2 = 0 reduces to -v1 + 5v2/2 = 3/2
so using both algebra and equation solver on my calculator I came out with v1 = 23/8 volts and v2 = 7/4 volts which I'm confident is correct. Thanks so much!
If I'm wrong in any of this feel free to correct me.
Now on to Thevenin and Norton Theorems.
PS: If anyone else wouldn't mind taking a look at the superposition problem. I just need to know if I got the setup and right methods. I'm not sure if canceling a voltage source in parallel with a resistor cancels out the current through that resistor too. If that is correct I believe I followed the right method.
|Apr5-09, 01:42 PM||#7|
Ooops I didn't see the reply about the superposition problem, thanks!
|Apr5-09, 01:59 PM||#8|
Hey, I used (v2-3)/2 because I chose current leaving the node is positive. So I move away from the node in the equation and don't need any minus signs at all.
Just look at it. V2-3 you start at V2 (the node) and move towards 3 volts source. So you are kinda going in the direction of the current. If you made current entering the node as positive then you would write 3-V2.
Also the current source arrow says that the current is leaving node V1 and so you write +2. If it was pointing into the node you would write -2.
It doesnt matter what you choose current leaving node is pos or current entering nod is pos, as long as you are consistent.
Just FYI, real current goes from negative side of battery to positive. But we use passive convention, from positive side of battery to negative. We can get away with that because it really doesnt matter, you just have to be consistent. The math doesn't care.
Oh and I know there are many computer engineers in circuits 2. I don't know about your college but some require it. Don't get the wrong impression that computer engineers don't need to know circuits. After all you guys aren't programmers, you are above programmers.
Circuits 2 isn't hard. Its all complex numbers, and more nodal and mesh analysis like you doing now.
You say youre good at algebra. Can you maybe help me. I have a math question.
How the heck do you solve for R L C in this equation. I am trying to compare the right equation with the left. On the right are my unknowns. On left is what I do know.
Math is evil
|Apr5-09, 02:58 PM||#9|
yeah I know what you mean.
I find it handy to have a TI 89 cuz it can do all that stuff for you however on the flip side you loose your math skills if your too reliant on it. Anyway
what I did:
I set it up like (s^2 + sqrt(2)*s +1)^-1 = (LC*s^2+RC*s+1)^-1 since fractions even if it's just a 1 tend to always confuse me for some reason. So basically the above is the same thing they are just inverses.
just algebraically solve for each unknown you want in terms of the other
for example with RC: the 1's cancel and you subtract LC*s from the right side and place it in the left side like (s^2 + sqrt(2)*s - LC*s^2)^-1 = ( RC*s)-1 then just divide both sides by s and you should get
RC = s(1-LC) + sqrt(2) in it's simplest form
using similar methods for the other unknowns I got
LC = (s-RC-sqrt(2))/(s)
and s = -(RC-sqrt(2))/(LC-1)
and my calculator showed the same.
So yea algebra has always made sense to me. Differential Equations which I'm in now is the hardest math I've struggled with yet and if that's not evil I don't know what is.
|Apr5-09, 10:07 PM||#10|
Thanks, but I needed actual values for R L and C. Not an abstract expression. Anyway I will ask my professor tomorrow and post in my thread how to do it for other people having the same question.
|Apr5-09, 11:51 PM||#11|
Oh how I understood your question I thought you needed expressions, as for the actual values I have no clue. I haven't even reached RC and RLC circuits yet in my course and I think I'd need more information.
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