Simple Q Ohms law

sci.electronics.basic is the correct group for this sort of question.

The power supply is probably current limited, with 'foldback', which means that if you put a way-too-small resistance on it the current goes lower than the maximum to keep the power supply internal parts from toasting.

The power relation is P = I*E. Combine this with ohms law and you get P = V^2/R. 25V * 25V / 2000ohm = 0.31W, so that's not bad at all. Strangely, 200V * 200V / 200kohm = 0.2W, which should be better.

Four watts into a 5 watt resistor would get things hot, but you won't get four watts out of that power supply unless you load it with a 10k ohm resistor. A 20k ohm would give you 2W, which will generally make things pretty hot unless you're using a 10W resistor or something similarly big -- are you sure you're not using a 20k resistor?

--

Tim Wescott
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200V across 2Kohms would be 0.1A and 20 watts. Your supply is rated at 4W, apparently, and likely has some current limiting built in to prevent you from damaging the supply to too high a load current. 200V across 200kohms is 1mA, and 200 milliwatts dissipation. That is well within the rating of your supply, so the current limit is not activated. I'm surprised you say it gets "very hot"; it must be physically small, or perhaps not really 200k ohms. 200V at 4W is 20mA. 200V at 20mA represents a 10k ohm load. 4 watts dissipated in a small volume should be expected to make it quite warm; the heat must go somewhere or the temperature will continue to increase. (That's a whole 'nuther resistance question--thermal resistance. Temperature rise equals thermal resistance times dissipated power. If the part gets rid of heat just by conduction, it's a pretty linear relationship, but if by radiation, it's not linear at all.)

I trust that wasn't a homework problem...

Cheers, Tom

With a 4 W supply, the lowest value resistance load you can have is

10K. The supply is probably going into current limiting with the 2K load. Regarding the smell; you are only dissipating 0.2W with the 200K load. Are you sure it's 200K?

The OP said the burning smell is with a 2k resistor - and the burning smell is most likely the PSU!

You need to recognize that any real world source has an internal resistance:

.----Internal resistance-------. | | - - |S| |L| |o| |o| |u| |a| |r| |d| |c| - |e| | - | | | . . ------------------------------

The "internal resistance" could be an active current limiter or simple resistance. If your numbers are right, then it has to include an active limiter. A useful exercise for you is to solve the circuit for I and P with first the 2K load resistance, then the 200K load resistance, then solve for Vload. You will see why the limiting must be active, not merely a resistance.

The power dissipated in the load resistor is E^2/R, or .2 watts. A 1/4 200K resistor should get hot, but not cause a burning smell. What size resistor did you use?

As to the load resistor always having 200 volts across it, look at the impact of the source internal resistance. Ed