Question about power drawn from a current limited supply when it is short circuited

Consider the schematic below:

When the switch is open, the 10 ohm resistor will draw 0.5 amps of current and dissipate 2.5 watts of power. No issue there.

When the switch closes:

We know the power supply output current is limited to 1 amp. So when the switch closes, the resistor is bypassed and the entire 1 amp of current from the power supply will flow through the short to ground. What is the power being drawn from the power supply?

The short circuit current is 1 amp. Power is $P = {I_{short}}^2 \cdot R_{short}$.

$I_{short}$ will be limited to 1 amp, and $R_{short}$ (the resistance of the short) is essentially 0 ohms. So is the power drawn from the supply $P = 1^2 \cdot 0 = 0$?

• No, it doesn't work this way. The current always takes the path of least resistance, which in this case is the short. Shorting an ideal voltage source doesn't work either. – Mario Oct 9 '16 at 18:39

I think this is a sort of brain teaser. The trick is in definitions, what does it mean "drawn power", versus "power delivered to a user's load"?

The paradox resolution is yes, the power delivered to the user load is ZERO (assuming ideal short with zero resistance). However, the power drawn from this power source is 5W, which will be all dissipated inside, on internal current-limiting circuitry. In reality this power supply will be pretty hot. You can try this with any AA battery to see the effect.

CORRECTION: As Chris Stratton pedantically commented, the dissipated power can be different from 5W if the power supply, say, has switching topology, such that it can dynamically change internal effective EMF and effective internal output impedance.

• This is incorrect. The source drawn has no "inside". The actual consumption of any possibly realizable device would depend on its implementation detail, which is not knowable. So the claim of 5 watts is categorically wrong. – Chris Stratton Oct 9 '16 at 23:12
• This is correct under the standard model of voltage source within the standard Spice framework. You allude that the "Electromotive force" (5V) can depend (be reduced) as load increases. It is possible, but the question stems from the standard model environment, and all standard assumptions should apply. More, these are very reasonable assumptions, because EMF of a battery is determined by fundamental electro-chemistry of electrode materials, which does not change. But if you have any data that, say, alkaline batteries reduce their EMF under heavy load, please post a link. – Ale..chenski Oct 10 '16 at 3:15
• As the question says nothing about spice or batteries you are not addressing the question asked, and thus giving an answer which is incorrect. If you are only able to think in examples, anything with switching regulation would be an obvious counterexample to demonstrate the falsehood of your claims about internal dissipation. But that is also entirely beside the point, as the subject is not internal dissipation. – Chris Stratton Oct 10 '16 at 3:19
• @AliChen The internal resistance of chemical cells can change a lot under load due to polarisation and present as a dropping EMF. In the worst case the polarisation (this is the change in the local electrode conditions due to ion migration or gas build up) could alter cell chemistry if an ion species becomes predominant that is not otherwise present. If you look at vintage chemical EMF reference cells like the Weston Cell they are specified at zero current because the measured EMF is not reliable under load. – KalleMP Dec 26 '16 at 20:08

In engineering, we must apply real world solutions.

As has been said before, the actual load will not be zero. The contact resistance of your switch and any and all connections to ground will have a resistance around a few milliohms. Your power supply will also have a source resistance/impedance that must be taken into account.

Power supplies react to shorts/overcurrents differently depending on type and quality. If a power supply output is shorted (in my experience), it will either clamp back the voltage, so that the power delivered does not exceed the maximum rating or it will enter a hiccup mode in which the power supply will disable itself and periodically attempt to re-enable itself. In the second condition, if the shorted load is removed, the power supply will re-enter regulation.

In your ideal case, the load/switch will consume zero power, and all power will be burned in the output/source impedance of your supply. You should always draw source impedances into your diagrams to ensure accuracy.

If this is a physics problem. The ideal conductor/switch could not have a voltage drop >0, so your voltage supply would attempt to suppy an infinite amount of current while never exceeding a ∆V >0

• "your voltage supply would attempt to supply an infinite amount of current" - No, it wouldn't. By definition, it's a current-limited supply, and so a physics problem would need to replace modelling as a voltage source with modelling as a current source. – Reinderien Aug 26 '17 at 18:17
• Perhaps when I wrote this, the current limit wasn't there. ;-) – Optimizer Aug 27 '17 at 18:29

A practical voltage source will have a series resistance to it. So when the 2 terminals of the voltage source is shorted the power delivered by the voltage source would be V^2/R. Where R is the series resistance. So, technically speaking when u short the terminals of a practical voltage source there is no load at all. So there will not be any power dissipated from the short-circuited wire. But the power delivered by the voltage source will by dissipated obviously as heat from its own series resistance.

• Under current-limited conditions, you cannot model the source as an ideal voltage source, and should not model it as a voltage source with an inline resistance. Instead, it should be modelled as a current source. – Reinderien Aug 26 '17 at 18:15