# Homework Help: Node Voltage Method

1. Feb 3, 2016

### Marcin H

1. The problem statement, all variables and given/known data
Find the voltage, V, across the 6A current source

2. Relevant equations
V=IR
Node Voltage Method

3. The attempt at a solution
Did I set this up correctly to find my voltage?

2. Feb 3, 2016

### Staff: Mentor

Almost. Be careful of the signs of the terms.

Judging by the sign you gave to the 6A supply term you're summing currents leaving the node. So why are your last two terms negative?

3. Feb 3, 2016

### Marcin H

Oh, woops. Read the polarities incorrectly. So if I change the last 2 to positive, it's good?

4. Feb 3, 2016

### Staff: Mentor

Yes, that would be a correct expression. Make it an equation by setting it equal to zero

5. Feb 3, 2016

Thanks!!!

6. Feb 3, 2016

### Marcin H

Got stuck again! The next question asks "What is the power associated with the 10V source?" I can find it using P=IV=I^2R=V^2/r, but how do I find the current through the 10V source?

7. Feb 3, 2016

### Staff: Mentor

Each term in the node equation represents a branch current. One of the terms corresponds to that current. Pick it out and use it now that you've found the node voltage.

8. Feb 3, 2016

### Marcin H

Ohh. So I use (Vx-10)/20= (36-10)/10 = 2.6A And then using P=IV I can do (2.6A)(10V)= 26W***?

Last edited: Feb 3, 2016
9. Feb 3, 2016

### Staff: Mentor

That would be the idea, yes.

10. Feb 3, 2016

### Marcin H

Ok thanks!! 13W***

11. Feb 3, 2016

### Staff: Mentor

Check your calculation for the current. (36 - 10)/10 = ?

12. Feb 3, 2016

### Marcin H

Woops! 26 watts.

13. Feb 3, 2016

### Staff: Mentor

14. Feb 3, 2016

### Marcin H

What? Where did I go wrong? I = (36-10)/10 = 26/10 = 2.6A. Using that current and P=IV I can find the power in the 10V source. P = (2.6A)(10V) = 26 Watts

15. Feb 3, 2016

### Marcin H

ORIGINAL QUESTION: what is the power associated with the 10V source?

16. Feb 3, 2016

### Staff: Mentor

Ah. My mistake. Sorry. I was thinking of the power associated with the branch as a whole. You have the correct result.