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# How inefficient are resistors? Answered

I know how to calculate resistive losses, but do different types of resistors have different amounts of losses (power in vs power-out, in watts) From my own observations, connecting a 10 ohm resistor across a barrey will cause significant heating in both the resistor and battery, and cause the wires to get warm. All of this is resistive loss, but when I exchange the resistor for a 10k resistor, there is virtually no heating at all. What if I use a 0 ohm resistor (direct short). The only thing getting hot would be the power supply, due to internal resistance.

Does this mean higher resistance is less lossy and by definition, more efficient, or is this simply due to the fact that there is less current flow, and less power loss, and efficiency (% of power loss)

With an ideal constant current source, will the losses though any resistive load be equal? ( X amount of watts lost/dissipated @ 1A)

Is it possible to limit current like a resistor without losses? (I know PWM techniques are more efficient, but I want actual resistance rather than chopping current flow and filtering with an inductor/capacitor RC filter)

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## 6 Replies

steveastroukBest Answer (author)2013-06-17

Power = V^2/R or I^2 x R That sums it all up, with a simple formula.

There is no such thing as a "0 ohm" resistor, except as a thought experiment, and yes, in that case the dissipation would be entirely in the internal resistance of the source.

No, if you discount active control, there is no method of limiting current without losses. If you permit it, then yes, you can control current loss-lessly.

-max- (author)2013-07-05

OK, thank you everyone, I think my mistake is the definition of efficiency, if the resistor is shorted out, there is an 100% loss as heat/EM radiation, And the loss depends on the voltage drop across it and current flowing through it.

David97 (author)2013-06-18

The reason that shorting the battery doesn't get hot is that when enough current is drawn from the battery, it starts to drop voltage. The voltage drop would be enough that there would not be enough voltage through the wire to heat up.

Resistors are very inefficient, they are designed to be. The power that they limit is got rid of in heat. The wattage rating of the resistor shows how much power can be dissipated.

As far as I no there is no way of limiting current without with %100 efficienty.

steveastrouk (author)2013-06-18

Sorry, that's wrong. The battery would get VERY hot, because all the power would be dissipated in the internal resistance of the source.

There ARE ways of limiting current with very high efficiency.

David97 (author)2013-06-21

Sorry my bad. Looks like I need to do a bit more reading :p

steveastrouk (author)2013-06-21

Its possible to cause a lead acid battery to explode by shorting its terminals.....