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Circuit Analysis Help

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

upload_2015-10-4_11-32-12.png


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.
 
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
 
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...
 
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
 
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.
 
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



 
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...
 
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
 
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 ."
 
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