## Brightness of a bulb

Today in a lecture, prof asked that for 2 bulbs connected in series(100W and 40W), which was brighter? And the answer was the 40W bulb.

His explanation was that we use P = I2 R. Higher power rating means higher R, and thus less bright. I argued that since P = VI, and I is constant, P is directly proportional to V and thus the 100W bulb should be brighter. I could not really comprehend his explanations later on.

Also, for a parallel circuit, it was mentioned that a higher power rating bulb would be bighter instead.

Can someone here please enlighten me?
 If these are usual light bulbs, the power rating refers to the standard voltage (120 or 220V). So the bulb rated at higher power has lower resistance, but this is true for nominal conditions. The resistance depends on temperature. When you put them in series they don't work at they nominal parameters. Assuming that they keep the same relationship between their resistances, the voltage across the bulb rated at higher power (and so with lower resistance) will be lower and (the current being the same) this bulb will use less power than the low power bulb. However the brightness may depend on how close is the bulb to nominal power. A bulb rated at 40 W and functioning at 35 W may be brighter than one rated at 100 W and functioning at 60 W. It depends on the properties of the filament.
 Mentor Higher power rating does not mean higher R, but it does mean less bright when in series. The analysis by your method, assuming they don't change their resistance with temperature: The nominal currents are: 100/120=.83 A 40/120=.33 A And resistances: V=IR 120/.83= 144 Ohm 120/.33= 363 Ohm In series, the amperage is: 120/(144+363)=.236A The voltage drops are: 120*144/(363+144)=34V 120*363/(144+363)=86V And the powers are: 34*.236=5.7W 86*.236=20.3W

Recognitions:

## Brightness of a bulb

 Quote by russ_watters The analysis by your method, assuming they don't change their resistance with temperature:
Thanks for this- I think you just gave me a good test question!
 Thank you all for the replies. Been reading a few sources here and there, and I think I get a bigger picture as a whole.

 Quote by Ambidext Thank you all for the replies. Been reading a few sources here and there, and I think I get a bigger picture as a whole.
Hi,
I found a similar question in a physics book and I also arrived at the 40W bulb being less bright. Here's my argument- Since the bulbs are in series, therefore the current will be same for both. Now, P=I2R, hence P is directly propotional to R as I2 is constant for both bulbs. Therefore, R is greater for the 100W bulb. And R is directly proportional to V. Therefore the 100 W bulb has greater voltage and should be brighter.

I know this argument is wrong. Can you please clarify?? Thanks in advance.

 Quote by shivamhzb Hi, I found a similar question in a physics book and I also arrived at the 40W bulb being less bright. Here's my argument- Since the bulbs are in series, therefore the current will be same for both. Now, P=I2R, hence P is directly propotional to V as I2 is constant for both bulbs. Therefore, R is greater for the 100W bulb. And R is directly proportional to V. Therefore the 100 W bulb has greater voltage and should be brighter. I know this argument is wrong. Can you please clarify?? Thanks in advance.

The affirmation in bold does not follow from your argument and is not true.
It was already said several times that the higher power bulb has the lowest resistance. This has nothing to do with the series connection.
Use P=v^2/R to see this. All bulbs (in this problem) are rated at the same voltage.
 Thanks. But why can't we use P=I2R? Also, both the bulbs will have different V's, hence I don't think we should use P=V^2/R. As for the affirmation in bold- P=I2R and I is constant because of which R will be greater for the 100 W bulb. Now, V is directly proportional to R. Therefore the 100W bulb will have greater V(making it glow brighter).

 Quote by shivamhzb Thanks. But why can't we use P=I2R? Also, both the bulbs will have different V's, hence I don't think we should use P=V^2/R. As for the affirmation in bold- P=I2R and I is constant because of which R will be greater for the 100 W bulb. Now, V is directly proportional to R. Therefore the 100W bulb will have greater V(making it glow brighter).
why is P constant? Since I is the same for both bulbs, and R is greater for the 40W bulb, this shows that P is biigger for the 40W bulb

Mentor
 Quote by shivamhzb Thanks. But why can't we use P=I2R?
You don't know P, I or R! So before you use it, you need to use another equation to find at least one of those (namely, R, since you know the I's are equal).

 Quote by willem2 why is P constant? Since I is the same for both bulbs, and R is greater for the 40W bulb, this shows that P is biigger for the 40W bulb
.

Do you mean to say that the 40W bulb will have a greater power than the 100W one?
 Mentor Yes.
 Thanks

 Quote by shivamhzb Thanks. But why can't we use P=I2R? Also, both the bulbs will have different V's, hence I don't think we should use P=V^2/R.
So you still don't completely understand the concept of "nominal power" or "power rating".
The power marked on the bulb is power that the bulb would draw when is connected to the nominal voltage, also given on the bulb usually. These values have nothing to do with the circuit in which the bulb will be used. It just tells you that IF the bulb is connected to the nominal voltage (for example, 120 V in North America) , the power will be the one indicated (40 W, for example). In most cases this is enough info as all the bulbs in a household are connected to the same voltage. Using P=V^2/R you can find the resistance of the bulb when it works at nominal parameters. If you neglect the change in resistance with temperature, this will be the resistance of the bulb in any situation, not just when it works at nominal parameters. So between two bulbs designed to work at same nominal voltage, the one with higher power rating will have lower resistance.

Now when you connect the same bulbs in "unusual" ways, like your series circuit, neither voltage nor power will have the nominal values. However the resistance of the bulb is still the same, as I said before.

 Quote by nasu So you still don't completely understand the concept of "nominal power" or "power rating". The power marked on the bulb is power that the bulb would draw when is connected to the nominal voltage, also given on the bulb usually. These values have nothing to do with the circuit in which the bulb will be used. It just tells you that IF the bulb is connected to the nominal voltage (for example, 120 V in North America) , the power will be the one indicated (40 W, for example). In most cases this is enough info as all the bulbs in a household are connected to the same voltage. Using P=V^2/R you can find the resistance of the bulb when it works at nominal parameters. If you neglect the change in resistance with temperature, this will be the resistance of the bulb in any situation, not just when it works at nominal parameters. So between two bulbs designed to work at same nominal voltage, the one with higher power rating will have lower resistance. Now when you connect the same bulbs in "unusual" ways, like your series circuit, neither voltage nor power will have the nominal values. However the resistance of the bulb is still the same, as I said before.
Thanks a lot. Though I arrived at the same conclusion, you post explains it perfectly. Thanks once again.
 You are welcome. I am happy it helped.