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Power through current sources

Constant

Feb 15, 2015
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So the voltage across the two upper branches is 8V.

Working through say the top branch, the voltage difference across the current source must be 8V-6V=2V.

Hence P=IV = 2 X 4mA = 8mW.

Why is my answer incorrect?
 

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Harald Kapp

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Why is my answer incorrect?
I say your answer isn't incorrect. I arrive at the same value both by a back of the envelope calculation and by simulation. In my view the answer provided in the task description is wrong.
 

Ratch

Mar 10, 2013
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I say your answer isn't incorrect. I arrive at the same value both by a back of the envelope calculation and by simulation. In my view the answer provided in the task description is wrong.

Yes, the answer key is wrong. The 4 ma source absorbs 8 mw and the 2 ma source supplies 20 mw.

Ratch
 

KrisBlueNZ

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Nov 28, 2011
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Just trying to clarify...

I guess the original circuit can be simplified down to this:

epoint 272717.png

Do you agree with that diagram?

Then the power dissipated in each current source can be calculated using the Power law, taking into account the direction of current flow.

So you guys are assuming that the arrow in the current source represents the direction of conventional current flow? AFAIK that's true; I've never seen a current source where the arrow direction represented electron flow.
 

Ratch

Mar 10, 2013
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Just trying to clarify...

I guess the original circuit can be simplified down to this:

View attachment 18665

Do you agree with that diagram?

Then the power dissipated in each current source can be calculated using the Power law, taking into account the direction of current flow.

So you guys are assuming that the arrow in the current source represents the direction of conventional current flow? AFAIK that's true; I've never seen a current source where the arrow direction represented electron flow.

I don't know how the rest of this group interprets the diagram, but I go by what a voltmeter and ammeter would indicate.

Ratch
 
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