0
$\begingroup$

I'm confused about resistance's relationship to current and power.

Scenario:

CIRCUIT A

  • 10 V battery
  • 100 W incandescent light bulb

Calculate resistance:

I = P/E
I = 100/10
I = 10 amps

R = E/I
R = 10/10
R = 1 ohm

CIRCUIT B

  • 100 V battery
  • 100 W incandescent light bulb

Calculate resistance:

I = P/E
I = 100/100
I = 1 amp

R = E/I
R = 10/10
R = 100 ohms

Has the resistance of the bulb's filament increased 100x because of a characteristic of the material? Assuming the filament is made from tungsten, according to the above math, it seems like if there are 10x more electrons moving through tungsten per unit of time, tungsten's resistance decreases by 100x. Is this true, or am I profoundly confused about the math here?

$\endgroup$
2
  • $\begingroup$ A "100 Watt" bulb only consumes 100 watts when plugged into a circuit of some defined voltage. If you put it in a different circuit, it will not consume 100W (you should consider it constant resistance, not constant power). It is difficult to tell from your question if the bulb is supposed to be a standard US bulb (100W @ 120V), or something else is going on. $\endgroup$ – BowlOfRed Dec 3 '18 at 18:54
  • $\begingroup$ The same bulb would not consume the same power if it's connected between different voltages. (A bulb connected between 10V will consume less power than the one connected between 100V if they are the same bulb) $\endgroup$ – SmarthBansal Dec 3 '18 at 19:32
3
$\begingroup$

The trick to this is that a "100W" bulb is a bulb that will consume 100W when plugged into a standard wall socket voltage (120V in the US). For most calculations people want to do (such as "how much power are might lights consuming when I leave them on?" or "how bright is this bulb anyways?") this is convenient. However, for electrical engineering purposes it is less than convenient, as you can see.

A 100W bulb plugged into a 10V circuit will actually consume quite a lot less than 100W of power, despite it's name.

The actual answer to your question is that filaments do the exact opposite: their resistance goes up as the current increases. The reason for this is temperature. As the current goes through the filament, it heats up. Many materials increase resistance as they heat up (though there are some fun ones whose resistance goes down!).

This, unfortunately, makes it very difficult to determine how much power a 100W bulb will consume if 10V is applied to it. 10V will not put as much current through the bulb as 120V would have. This means the bulb won't heat up as much, so the resistance will be lower.

How much lower? Hard to say. You'd need to know more about the particular bulb. But I will note that if you apply a very low voltage, like 1V, you won't heat up the filament much at all. At this low voltage, the filament will act like... well.. a wire. It's resistance will be very low indeed, approaching 0 ohms.

$\endgroup$
1
0
$\begingroup$

Mathematically you are correct, but the question seems strange when considering common usage.

Usually when you buy a lightbulb rated for 100 W, the manufacturer assumes that you are going to plug it into a standard set voltage. If you put it at a much lower voltage it will not make full use of its capabilities and will certainly not output 100 W.

Intuitively you know that a given light bulb will not be as bright (here we make the assumption that brightness = Watts) if you connect it to a lower voltage (this is how dimmers work basically). To get the same brightness when connecting to a lower voltage battery you will need a different bulb, with a different resistance for sure.

$\endgroup$
0
$\begingroup$

Has the resistance of the bulb's filament increased 100x because of a characteristic of the material?

Yes. I don't remember the actual numbers, but the bulk resistance of white-hot tungsten is much higher than the bulk resistance of room-temperature tungsten.

$\endgroup$
1
  • $\begingroup$ Fig. 62 here, for example, shows one set of measurements for W resistivity vs. temperature. $\endgroup$ – Chemomechanics Dec 3 '18 at 21:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.