Power/Amps/Voltage

[MedPR]

Having trouble conceptually with these concepts.

Light bulbs that you use around the house are designed to have 120V across their terminal. For instance, a 120W bulb uses 12 watts of power when placed in a socket with a 120V potential difference.

Ok, I understand that P=IV=I^2R=V^2/R

I don't understand how this relates to the light bulbs though. A 120W bulb requires 120W of power to generate 120V within the light bulb filament? So it only has 1Amp flowing through it at any given time? By the same rules, a 30W bulb requires less power (so it saves you money on your electric bill...?) to generate the same 120V in the bulb?

If you connect the 120W and 30W bulb in series, and plugged it into a wall outlet, which bulb would be brighter?

NOVA says the 30W because it has a higher resistance. What makes one bulb brighter than another?

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[milski]

The light bulb does not create the potential, it's is already there created by the electricity source- battery, generator, whatever. When you connect the bulb, the current starts flowing through it. You can determine that current from Ohms law, I = U/R. The light bulb with lower resistance will have more current going through it for the same potential, so it will be the brighter one.

When you have the two light bulbs in series, the current through them will be the same but the potential drop will be different. You can calculate from Kirchoff's law or you can use P=I^2/R. For the same I, the light bulb with lower resistance will have higher power/will be brighter.

You can try to plug the numbers for the two bulbs that you describe or if you want me, I can do it for you, if the general argument is not clear enough.

 

[MedPR]

The light bulb does not create the potential, it's is already there created by the electricity source- battery, generator, whatever. When you connect the bulb, the current starts flowing through it. You can determine that current from Ohms law, I = U/R. The light bulb with lower resistance will have more current going through it for the same potential, so it will be the brighter one.

When you have the two light bulbs in series, the current through them will be the same but the potential drop will be different. You can calculate from Kirchoff's law or you can use P=I^2/R. For the same I, the light bulb with lower resistance will have higher power/will be brighter.

You can try to plug the numbers for the two bulbs that you describe or if you want me, I can do it for you, if the general argument is not clear enough.

I understand that the bulb with lower resistance will be brighter, but NOVA says the opposite. It says that if you connect the 30W and 120W bulb in series, then plug in to the outlet, the 30W bulb is brighter.

Since P=I^2/R, the power is proportional to the resistance. So the 480ohm bulb (that is, the 30W bulb) is brighter because the bulbs are in series. (Usually the 120W bulb is brighter because the bulbs are connected in parallel).

I'm not understanding any of that, especially the part about bulbs being connected in parallel?

 

[milski]

I understand that the bulb with lower resistance will be brighter, but NOVA says the opposite. It says that if you connect the 30W and 120W bulb in series, then plug in to the outlet, the 30W bulb is brighter.

There are two separate cases here but let's set thing up first.
U=120 V
Bulb A, 30 W, from P=V^2/R, R=V^2/P=120^2/30=480 Ohm
Bulb B, 120 W, same reason P=120^2/120=120 Ohm

- if you plug bulb A on its own, the power will be P=V^2/R=30 W
- if you plug bulb B on its own, the power will be P=V^2/R=120 W
When connected on its own, the bulb with lower resistance will be brighter.

Now let's consider the bulbs connected in series. That means that their total resistance is 120 Ohm + 480 Ohm = 600 Ohm. From I=U/R, the current through the circuit will be 120/600=0.2 A. From P=I^2R we have that bulb A has P=0.2^2*480 =19.2 W and bulb B has P=0.2^2*120=4.8 W. In that case the bulb with higher resistance will be brighter.

What we can say in summary is that when you make the bulbs work with the same potential, the one with lower resistance will be brighter. But when you make the current through them the same and the potential different (when connected in series), the one with higher resistance will be brighter.

I'm not understanding any of that, especially the part about bulbs being connected in parallel?

I would expect the case when they are connected in parallel to be more or less the same as connecting them separately - you have the same potential across each bulb but different current -> the one with lower resistance will be brighter.

 

[MedPR]

There are two separate cases here but let's set thing up first.
U=120 V
Bulb A, 30 W, from P=V^2/R, R=V^2/P=120^2/30=480 Ohm
Bulb B, 120 W, same reason P=120^2/120=120 Ohm

- if you plug bulb A on its own, the power will be P=V^2/R=30 W
- if you plug bulb B on its own, the power will be P=V^2/R=120 W
When connected on its own, the bulb with lower resistance will be brighter.

Now let's consider the bulbs connected in series. That means that their total resistance is 120 Ohm + 480 Ohm = 600 Ohm. From I=U/R, the current through the circuit will be 120/600=0.2 A. From P=I^2R we have that bulb A has P=0.2^2*480 =19.2 W and bulb B has P=0.2^2*120=4.8 W. In that case the bulb with higher resistance will be brighter.

What we can say in summary is that when you make the bulbs work with the same potential, the one with lower resistance will be brighter. But when you make the current through them the same and the potential different (when connected in series), the one with higher resistance will be brighter.



I would expect the case when they are connected in parallel to be more or less the same as connecting them separately - you have the same potential across each bulb but different current -> the one with lower resistance will be brighter.

Perfect, thank you. I was thinking that a 30W bulb always has a Power of 30W. But, as you showed in your calculations for the bulbs in series, that is not the case.

I still don't quite understand why more power means brighter though.

 

[milski]

Yes, the 30 W designation is just a convention - it shows the power at the nominal potential for the light bulb.

Power = energy/sec. Most of the power in a light bulb becomes heat. More power -> more heat -> the higher the temperature of the filament -> the brighter it is.