# Very easy circuit question.

1. Oct 3, 2006

### Mesmer

I understand everything in the solution of this problem, except the expresson
$$2I_{x}I_{2}=16W$$ supplied. can some one please tell me whats going on here?

I thought that Power was the product of current and voltage. How can I get power out of the current source in the center of the mesh?

#### Attached Files:

• ###### Doc1.pdf
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Last edited: Oct 3, 2006
2. Oct 3, 2006

### Saketh

I cannot see your document (approval usually takes a while), but I have a question: is the 2 referring to 2 ohms of resistance? Based on your description, it seems like the 2 is 2 ohms of internal battery resistance.

Also, remember that:
V = IR
P = IV
Therefore $$P = I^2R$$.

3. Oct 3, 2006

### Mesmer

No, sorry there are not resistors in the diagram.

4. Oct 3, 2006

### Staff: Mentor

Attachment is still pending approval, but you get power from a current source based on the voltage that is across the current source as it supplies the current. If you supply current into a 10V load, that's 10x the power compared to if you source that current into a 1V load.

5. Oct 3, 2006

### Mesmer

Can anyone see the image that I have linked to?

6. Oct 3, 2006

### Stevedye56

Its approved now.

7. Oct 3, 2006

### Mesmer

any suggestins now that you can view the file?

8. Oct 3, 2006

### Mindscrape

Are you doing mesh power(???) analysis?

Anyway, to answer your question, the $$2I_x I_2 = 16W$$ is a power because it uses the dependent voltage source, which has a voltage of $$2I_x$$.

Any other questions?

9. Oct 3, 2006

### Mesmer

I see now. Thank you!!