# Light Brightness Relationship with Potential and Current

1. Jan 12, 2014

I understand that generally, power and light bulb brightness are directly proportional. However, because there are so many ways to write the power formula (P=IV = I^2 * R = V^2 / R), I was a bit confused about what factors actually determine power and wanted to confirm whether the following is correct:

If Potential = Constant*, bulb with LOWER RESISTANCE = Brighter (since I is higher from V=IR)

If Current = Constant and potential different#, bulb with HIGHER RESISTANCE = Brighter (since V is higher from I=V/R)

*Have a single bulb connected to battery; or have multiple bulbs connected in parallel

#Have multiple bulbs connected in series

Are there any other scenarios I should be aware of when determining brightness?

Can I make any sweeping statements like "brightness and current are directly proportional" that are applicable to all scenarios?

Thanks!

2. Jan 12, 2014

### Simon Bridge

The electrical energy is transformed, in the bulb, into heat and light.

The amount of heat and light depends on the rate that the energy is transformed.

The rate of change of energy is power.

The current and the voltage are both needed to work out the power. P=VI

However, if you happen to know that the voltage and the current have a particular relation to each other - say the filament obeys Ohm's Law (seldom true - resistance normally depends on temperature), then you can take a shortcut in the calculation knowing the resistance and either the voltage or the current. This is where the other equations come from.

If P=VI and V=IR then P=I2R = V2/R
So you can still find the power dissipated in a resistor even if you don't know the voltage (1st one) or the current (2nd one).

This is useful because it is sometimes not convenient to determine both voltage and current.

You cannot make sweeping statements like "brightness is proportional to current" - because the voltage is also important. From your investigations you can say that, at the same voltage, the brightness is proportional to the current.

The situation can get as complicated as you like - the general energy principle in the first sentence will always be true.

3. Jan 13, 2014

Okay I understand now that I cannot make broad generalizations.

Would you (or someone else) be able to confirm whether my conceptual understanding of the two bolded scenarios (holding potential constant, holding current constant) is correct?

4. Jan 13, 2014

### sophiecentaur

Those (bolded) statements are not particularly well put together but I think they are broadly true. I suggest you look at the way the text books express things rather than trying for your own words. The 'official' language was developed after a very long development time so it's probably best to use it. Needless to say, using formulae is always a better idea than just verbal descriptions and all the basic relationships can be expressed very elegantly and unambiguously with a very few algebraic symbols.

5. Jan 13, 2014

### Simon Bridge

Good. Did you also get this bit:
i.e. you can check your statements by finding out the power dissipated.

If potential is a constant, and bulb resistance changes, then the power dissipated is P=v^2/R ... so a higher resistance means a lower power dissipation, so brightness is inversely proportional to resistance.

i.e. you got the right relationship out the end.
I'm a bit iffy about saying that this is because the current has increased - Ohms law describes a quantitative relationship not cause and effect. That must be inferred from other factors.

Certainly if the voltage remains the same, and the current increased, the resistance must have decreased.
I don't think you get into a lot of trouble with this sort of thinking at this level though.
As Sophie suggests - you should compare what you have written with how the text books phrase it... see if you can spot the difference between what they say and what you said.

If the current is held constant, how is the power dissipated related to the resistance.

6. Jan 13, 2014

### alva

I would say that electrical power is transformed into heat. Heat increases temperature. And, if the temperature is high enough, then the filament will emit visible light.

7. Jan 13, 2014

### alva

If you are talking about changing current/voltage for the same light bulb "directly" is wrong.

If you connect a 12 V light bulb to a 6 V source you will get, say, a 1/20 of the brightness.

The relationship between the power supplied to a light bulb and the (visible) brightness is very complex.

8. Jan 13, 2014

Well P = I^2 * R so if current is constant, greater R would reflect more power (since R is in the numerator in the equation).

I was confused because P = V^2/ R as well (although in this case, current is not in the equation and not being held constant).

9. Jan 13, 2014

### sophiecentaur

We took a few thousand years to develop a very good language for describing this sort of situation. That language is called Mathematics. It is spot on for showing relationships that apply to simple models like this one. If you ever use a calculator or computer to work out your finances or your credit card statement then consider the 'verbal' alternative - pretty well unthinkable, I would say. So why do you want to discuss basic electricity problems by verbally waving your arms about instead of going with the few formulae and getting sense out of them?
A little investment in getting to grips with basic algebra, solving and combining very simple equations and perhaps plotting the occasional graph would remove all the aggro you seem to be encountering with this elementary problem.

10. Jan 13, 2014

### CWatters

Correct.

It's not brighter because I is higher it's (as Simon said) because more power is dissipated. It doesn't matter which way you calculate the power..

Power = I*V (I is larger)
or
Power = V2/R (R is smaller)
or
Power = I2*R (I2 increases faster than R reduces)

Correct
Again it's not because V is higher but because more power is dissipated. It doesn't matter which way you calculate the power...

Power = I*V (V is larger)
or
Power = V2/R (V2 increases faster than R increases)
or
Power = I2*R (R is larger)

11. Jan 13, 2014

### Simon Bridge

Energy may undergo intermediate transformation - yes.
Indeed do many things come to pass :)

We've probably beaten this topic to death by now.

We tend to push people like this when we feel they are ready for the next step ;)

12. Jan 14, 2014

### CWatters

Only thing I can think of to add is that light bulbs don't behave like ideal resistors. Their resistance can vary with temperature. So using a multi meter to measure the resistance with the bulb cold and using that to predict the current at the rated voltage can give confusing results. Likewise putting two bulbs in series may not half the current exactly.

13. Jan 14, 2014

### sophiecentaur

Between cold and normal operating temperature, there can be a ratio of 1:10 in Resistance of a tungsten filament. Even over the mains AC cycle, the resistance will vary. A normal 230V bulb flickers noticeably - showing a variation in temperature (and resistance) even over the 20ms period.

You have to use the multimeter to measure I and V whilst the bulb is in actual operation if you want any sense from your measurements.

14. Jan 14, 2014

### alva

I apologize.
I thought we could have several "layers" of discussion: the simplified OP problem, the more explained problem, and the issue of how to teach the problem.
It won't happen again.

15. Jan 14, 2014