Power dissipated without a load?

[jsaxton]

Greetings,

I recently took an introductory circuits test. The answer key was released today. I am convinced it's wrong, and if my argument is flawed, I'd like to know before bringing it up with the instructor (no, not the professor, the BME guy who appears to apply equations without actually knowing what they mean).

The http://www.ece.umn.edu/class/ee2001/quiz_1a_solution.pdf basically had three idealized voltage sources and a current source. There are no resistors (no load, as far as I can tell). However, power is dissipated in the circuit (according to the answer key). Now, as far as I can tell, this is not possible as P=R*I^2. R=0 (as far as I can tell). Hence no power is dissipated.

I have heard one counterargument: the current source implies a resistance in the voltage source, despite the fact it is ideal. Still how can one dissipate power without a resistor? Am I in the wrong? Can one even have current without a load?

From a physics perspective, what would happen is that the potential differences would simply equal out (we'd get a massive equipotential). Again, as this is an ideal circuit, no voltage would be dropped on the "wires" and no power would be dissipated. Is this reasoning sound?

Thanks,
John

 

[SGT]

In order to dissipate power you really need a resistor, but the voltage and current sources will be delivering or absorbing power according to p = VI. If the current I enters the higher potential, the power is positive, so the source is absorbing power. If the current enters the lower potential and leaves by the higher one, the source is delivering power.

 

[thtadthtshldntbe]

I have heard one counterargument: the current source implies a resistance in the voltage source, despite the fact it is ideal. Still how can one dissipate power without a resistor? Am I in the wrong? Can one even have current without a load?

I am an electronics student as well. The way that I learned it was that a current source implied a resistance that was present even if the source happened to be turned off, i.e. the equivalent circuit to represent the internals is an idealized current source in parallel with a resistor representing internal source resistance (for a voltage source, the internal resistance is in series, so open means no resistance effect).

If the current is turned off, the internal resistance of the current source is still in the circuit, whereas if a voltage source is turned off, you have an open circuit.

 

[SGT]

I am an electronics student as well. The way that I learned it was that a current source implied a resistance that was present even if the source happened to be turned off, i.e. the equivalent circuit to represent the internals is an idealized current source in parallel with a resistor representing internal source resistance (for a voltage source, the internal resistance is in series, so open means no resistance effect).

If the current is turned off, the internal resistance of the current source is still in the circuit, whereas if a voltage source is turned off, you have an open circuit.

In the examples given all sources are ideal, so the current sources have infinite parallel resistances and the voltage sources have zero series resistances.
There is no dissipated power, but there is delivered and absorbed power in each source.