# Passive sign convention question

Discussion in 'Homework Help' started by katarina, Apr 16, 2013.

1. ### katarina Thread Starter New Member

Apr 15, 2013
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0
Hi everyone!

I am currently studying for my exam and thought this would be a nice place to ask about some stuff I wondered about.

The question is:

Find the power associated with each source in the circuit (Circuit 2 is attached as a picture) and state weather the source is delivering or absorbing power. I used the node-voltage method to find out what i(a) was. (i(a) = 2A). Than, to find the power associated with the 50 V source is (p = vi) 50 x 2 = 100 W. However, the book says that it is not supposed to be 100 W but -100 W. So, my sign is wrong.

I wonder, why is my sign wrong? Because I have attached a second picture (Circuit 3) which I believe models the situation in this case, and according to that model power should be positive.

Is the power perhaps negative because a voltage source always delivers power?

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2. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,993
1,116
Yes, you right. But voltage source not always delivers power.
We can treat voltage source as a battery. So voltage source can delivers power to the load. We discharge a battery. Or we can charging a battery. So our voltage source absorbing power from external circuit ( from the charger).

3. ### WBahn Moderator

Mar 31, 2012
18,093
4,920
The passive sign convention basically says that we will match the polarity of the current through a device to the polarity of the voltage across the device. If the device is a load (or if we declare it to be a load by assumption), then the polarities are chosen so that positive current enters the device at the positive voltage terminal and exits the device at the negative voltage terminal. Thus, if we compute a positive power the device is absorbing power and if it is negative the device is producing power. We could just call everything, including sources, a "load" and be done with it. But humans make sign errors very easily, so we flip this around and say that if the device is a source (or if we declare it to be a source by assumption), then the polarities are assigned such that positive current leaves the device at the positive voltage terminal and returns to it at the negative voltage terminal. That way, sources with positive power are delivering power. If we choose everything correctly, all of our powers will be positive and the sum of the powers of the sources will equal the sum of the powers of the loads. If any of them turn out to be negative, then we know we just made a wrong assumption and can flip things quickly and efficiently.

For this circuit, I agree with your answer of 2A and +100W for the 50V source. For their to be 2A flowing the other way and delivering power to the voltage source would require the current source to be delivering 9.5A.

The book may be using the "everything's a load" convention in which case the power associated with the source would be negative because it is a "load" that is delivering power to the circuit.