How does the expression 2I_x I_2 = 16W relate to power in a circuit?

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I understand everything in the solution of this problem, except the expresson
[tex]2I_{x}I_{2}=16W[/tex] supplied. can some one please tell me what's 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?http://img511.imageshack.us/img511/3175/screenshot003zo4.th.jpg
 

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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 [tex]P = I^2R[/tex].
 
No, sorry there are not resistors in the diagram.
 
Mesmer said:
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?
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.
 
Mesmer said:
I understand everything in the solution of this problem, except the expresson
[tex]2I_{x}I_{2}=16W[/tex] supplied. can some one please tell me what's 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?

Are you doing mesh power(?) analysis?

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

Any other questions?
 
Mindscrape said:
Are you doing mesh power(?) analysis?

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

Any other questions?

I see now. Thank you!