Do resistors consume power?

Question

Say I had a 9 volt battery, 9 volt lightbulb, and a resister that would reduce the voltage by 3 volts

If i were to wire all 3 of these in series, would the battery be discharged at the same rate if I did not use a resistor (compared to the circuit with the resistor?)

Answer 1

Heb, I think you overcomplicated it.

If you add external resistance to a lamp, the current will go down. No question.

Why? if the current went up, then the lamp would have more voltage across it than the battery voltage, which leads to a contradiction. The lamp VI curve is not linear, but it is monotonic.

So the answers are:

yes

Battery will be discharged at slower rate.

Answer 2

Resistors do consume power. The formula is:

W = (I**2)*R

Where I is the current in Amperes, which is squared and multiplied by the resistance in Ohms. W is in Watts.

However, the second question is a little trickier as the light bulb will be running at a different resistance (due to electrical heating of the filament) at 6 volts then it would at 9 volts. If you were to measure the light bulb's resistance, it would read almost zero. But when lit up, the current heats the filament which raises its resistance. This can be seen by the fact that you know that your little 9 volt battery isn't supplying an infinite amount of current! Thus, the light bulb has more resistance when lit than otherwise.

Now, that increase in resistance depends on the current. Which depends on the voltage applied. With a resistor in series, there is less voltage. Is there enough to significantly heat up the filament? I don't know; you didn't give me all the specs. If it isn't, you could actually draw more current than a fully lit bulb.

But wait (as they say in late night TV ads), it gets more complicated. The battery never puts out an infinite amount of current. (How could it?) But the filament reads 0 Ohms, so why not? Well, it turns out that batteries can be modeled as a pure voltage generator in series with an internal resistor. That internal resistance will limit the maximum amount of current that can be drawn, even in a dead short. But that amount of current is sufficient to heat up the filament of a 9V light bulb. If you knew the size of that internal resistance, you would know all there is to know about that circuit.

Whew!

So now you need to lookup how temperature is related to resistance  (it is pretty linear over a large portion of the curve and non-linear otherwise).

If the range for 6 volts to 9 volts is linear (probably), then exactly the same amount of power would be drawn with or without the resistor. But it is a pretty inefficient dimmer since it draws the same power for less light!

Sure hope that wasn't a homework question.