Circuit Analysis Help

[Chris Bargman]

Hello all. I am taking my first class on circuit analysis and I have a problem that has got me pretty well confused.




Using Nodal Analysis:
The equations I come up with are:
A = 21
B-C = 9
(1/4+1/3)B-(1/4)A+(1/2+1/6)C=0

But plugging it into my calculator I get: A=21 B=9 and C=0

(1/3)*9 is clearly not -0.6

But I have no idea what I am doing wrong.

Using loop analysis it works out fine but I am trying to get a better grasp of nodal analysis.

 

[Arouse1973]

Hi Chris. So you know the value of current through the 2R. What voltage does this develop across the resistor? I bet it's less than 9 Volts. To make sure the voltage across the 9 Volt source is 9 Volts what do you think has to happen to the other side of the voltage source?
Adam

 

[Chris Bargman]

Am I missing something obvious? I do not see how I would know what the current through 2R is... I mean I can figure it out by doing loop analysis but I am trying to avoid doing that.

I would also bet its less than 9 volts as the 9 volt source is being split between a R6 and an R2.

I guess I could say that the total resistance is 3/2 Ohms and then using ohms law say that V/R= I but the I that I would be solving for is for both resistors not just the one...

The voltage source has to be grounded... Which it is not. I see your point there...

 

[Arouse1973]

H Chris, yeah sorry I misread your post, of course you wouldn't know the current in the 2R. I would have to dig out my books to help you work it out properly, it's been a long time. I do my own thing for problems like this, I guess it could be a combination of loop and node, I don't get too hung up about what the methods are called as long as I get the right answer.
Thanks
Adam

 

[Chris Bargman]

That is fine and I would generally agree. But when a test asks you to do one specific method over another...

But I'm really not to worried about it at this point.

Thanks anyway.

 

[Ratch]

Hello all. I am taking my first class on circuit analysis and I have a problem that has got me pretty well confused.

View attachment 22352


Using Nodal Analysis:
The equations I come up with are:
A = 21
B-C = 9
(1/4+1/3)B-(1/4)A+(1/2+1/6)C=0

But plugging it into my calculator I get: A=21 B=9 and C=0

(1/3)*9 is clearly not -0.6

But I have no idea what I am doing wrong.

Using loop analysis it works out fine but I am trying to get a better grasp of nodal analysis.

Correct, you don't know what you are doing. What is A, B, and C?

You have a supernode (two nodes separated by a voltage) in that circuit that makes evaluation easy and simple. Assume "v" is on the negative of the 9 v battery. Then the node equation is (v-21)/4+v/3+(v+9)/2+(v+9)/6=0 . Solving for v below gives -0.6 v. Then it is easy to calculate (9-0.6)/2 = 4.2 .

Ratch

 

[Chris Bargman]

Ahh of course I was forgetting to add in the voltage. But why are you adding in the 9 volts to the 2 and 6 ohm resistor?

When using a super node I thought that power source was left out as it is accounted for in the super node...

 

[Ratch]

Ahh of course I was forgetting to add in the voltage. But why are you adding in the 9 volts to the 2 and 6 ohm resistor?

When using a super node I thought that power source was left out as it is accounted for in the super node...

Voltage is never left out in node analysis. The voltage across the 2 ohm and 6 ohm resistor is 9 volts higher than the voltage "v" across the 3 ohm resistor.

Ratch

 

[Chris Bargman]

Okay thank you. You have cleared a lot of things up for me

 

[Arouse1973]

Well done @Ratch.

 

[MadlinZax]

If you want the solution from us then please specify your whole calculation here, so that it will be easy to find out your mistake.
If you have a supernode in that circuit that makes evaluation easy and simple, then let assume V is on the negative of the 9V battery. I am agree withe the @Ratch for this "Then the node equation is (v-21)/4+v/3+(v+9)/2+(v+9)/6=0 . Solving for v below gives -0.6 v. Then it is easy to calculate (9-0.6)/2 = 4.2 ."