Determine node voltages in circuit

[Color_of_Cyan]

Homework Statement

http://imageshack.us/a/img713/9389/homeworkprobsg23.jpg Determine node voltages for the circuit

Homework Equations

V = IR,

KCL

Supernode equations

The Attempt at a Solution

I am sorry to say that There is a LOT of stuff in this problem I still need help with, like how/ when to apply supernodes.

Is it true that where you see a ground node, then they do NOT want you to put supernode there? Or can you really put it anywhere with a voltage / current source? The only equation for this problem I know for sure to make out at this point is

Va - Vc = 10VPutting the supernodes like this:

http://imageshack.us/a/img94/5332/homeworkprobsg23edit.jpg I think I would get (with both currents in the middle going towards B, if I can set it that way):

5A - 2A = (Va - Vb)/10Ω

and then

5A - 2A = (Vc - Vb)/40ΩI do not know if this is right at all though.Sorry if I should know this well already too, but can a current from another source (ie from the 5A and 2A) pass freely through a wire with a voltage source (ie 10V and 12V)? Thank you.

 

[CWatters]

The voltage on one node is obvious. Look carefully at node b and what's it's connected to.

Sorry if I should know this well already too, but can a current from another source (ie from the 5A and 2A) pass freely through a wire with a voltage source (ie 10V and 12V)?

yes. A voltage source only regulates it's voltage.

 

[NascentOxygen]

Determine node voltages for the circuit

Denote the current through the 10Ω as i₁
and current through the 40Ω as i₂

That gives you 2 unknowns, so you must find two equations involving i₁ and i₂.

Does the textbook provide the answers?

 

[Color_of_Cyan]

My professor gave me this problem, but yeah he gave the correct answers too and that's it.

But I have little clue on how to get the correct answer. He says they're based on textbook problems though. Can I say that I#1 and I#2 both flow TO node b?

I'm thinking if they do, thenI#1 = (Va - Vb)/10Ω and

I#2 = (Vc - Va)/40Ω

then if both currents flow towards node b then they would have to go out through the 12V node path so then

I#1 + I#2 = 3 ampsWould this be right / the right idea so far?

 

[NascentOxygen]

Can I say that I#1 and I#2 both flow TO node b?

You are free to mark each in which ever direction you like. But once chosen, mark the direction with an arrow and stick to that direction rigidly.

I'm thinking if they do, then


I#1 = (Va - Vb)/10Ω ✔ and
I#2 = (Vc - Va)/40Ω ✗

Using the notation of the diagram, what is the voltage across the 40Ω?

then if both currents flow towards node b then they would have to go out through the 12V node path

On what basis exactly do you claim this?

I#1 + I#2 = 3 amps

You don't explain your thinking, so I can't say whether you are on the right track. Maybe you are.

 

[Color_of_Cyan]

Sorry, I made a dumb typo for the second equation:I meant:

I#2 = (Vc - Vb)/40Ω

along with

I#1 = (Va - Vb)/10Ω

now would it be right that they would both be going out through the 12V wire?

ie:

I#1 + I#2 = 3 amps, because then it's the last current for the ground and so 2A + 3A = 5A

Would that be along correct now?

 

[NascentOxygen]

Yes, 3A flows to ground via the 12V source. So can you now determine the node voltages?

 

[Color_of_Cyan]

Okay so the starting equations (simplified to my needs) are:

Va = 10V + Vc

i1 = (Va - Vb)/10Ω

i2 = (Vc - Vb)/40Ω

i1 + i2 = 3ASo then:

Plugging Va into I1:

i1 = (10V + Vc - Vb) / 10ΩNow plugging this i1 and i2 into i1 + i2 = 3A, then:

(10V + Vc - Vb)/10Ω + (Vc - Vb)/40Ω = 3AMultiplying 10Ω and everything above it by 4 to get 40Ω as the denominator:

(40V + 4Vc - 4Vb)/40Ω + (Vc - Vb)/40Ω = 3A

(40V + 5Vc - 5Vb)/40Ω = 3A

40V + 5Vc - 5Vb = 120V

5Vc - 5Vb = 80V

0Va - 5Vb + 5Vc = 80V

And from here it looks like I need to use Cramer's rule. Looks like I get stuck again here, if I was going correct.

Would I need another equation to do Cramer's rule? If so, how would I get it? Or was there another way?

Thanks for all your help and patience so far, by the way.

 

[NascentOxygen]

Okay so the starting equations (simplified to my needs) are:

Va = 10V + Vc

i1 = (Va - Vb)/10Ω

i2 = (Vc - Vb)/40Ω

i1 + i2 = 3A


So then:

Plugging Va into I1:

i1 = (10V + Vc - Vb) / 10Ω


Now plugging this i1 and i2 into i1 + i2 = 3A, then:

(10V + Vc - Vb)/10Ω + (Vc - Vb)/40Ω = 3A

Doesn't Vb have a numerical value?

And from here it looks like I need to use Cramer's rule.

Cramer's Rule?! I don't think that will be necessary. http://img41.imageshack.us/img41/7325/winkingw.gif

 

[Color_of_Cyan]

Wait... so Vb was just -12V all along?

That's because it's the only source of potential touching it right, and the other node it touches is the ground?

Okay so then.

5Vc + 60V = 80V

5Vc = 20V

Vc = 4V

then

Va - Vc = 10V

Va = 10V + 4V

Va = 14VThanks. Can't believe I didn't see Vb = -12V like that...