# B Which makes a bulb glows brighter?

#### Angela Liang

My textbook says a bulb glows brighter due to its high resistance. But it also says a bulb glows brighter when a larger current flows through it. Which is correct?

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#### Khashishi

Both. P=IV, so you need both resistance and current.

#### Delta Kilo

100W bulb has less resistance than 40W bulb.
But P=IV = I^2R = V^2/R.
If you connect them in parallel (as you would normally do), more resistance means less current at the same voltage and less power.
However if you connect them in series then more resistance means more voltage at the same current and consequently more power.

#### jim hardy

Gold Member
2018 Award
My textbook says a bulb glows brighter due to its high resistance. But it also says a bulb glows brighter when a larger current flows through it. Which is correct?
"My textbook says a bulb glows brighter due to its high resistance." That's a strange and (misleading) thing for an author to write.
Can you provide the actual text ? Perhaps he said 'high temperature' ?
Larger current will heat an incandescent filament to higher temperature increasing its light output, ie making it brighter ..

#### Angela Liang

Both. P=IV, so you need both resistance and current.
But how to achieve the condition where both current and resistance are high?

#### davenn

Gold Member
But how to achieve the condition where both current and resistance are high?
that is a contradiction ... higher resistance = lower current

eg for a given voltage, as the resistance increases, the current will decrease

but before we continue, answer Jim Hardy's question

#### Merlin3189

Homework Helper
Gold Member
I agree, the comment is nonsensical as written. I have many perfectly good bulbs which don't glow at all - because they are in their boxes in the cupboard. Whatever their resistance bulbs don't glow until you connect them to the electric power source. Therein lies the key to this issue.
Mains bulbs are connected to what is essentially a constant voltage source. Therefore their power is determined by P = V2/R and (for the same physical size, which we haven't yet mentioned) brightness is inversely proportional to resistance.

If you want power proportional to resistance, with P = I2R then you need a constant current source, which you come across in some LED drivers, but not for incandescent bulbs.

A battery or dynamo with low internal series resistance is a good approximation to a constant voltage source. Constant current sources are more difficult to make and usually involve wasting power.

Power inversely proportional to resistance is inherently more stable (for metallic filaments) because, if the filament temperature rises, the resistance increases, thence the power decreases and the temperature falls. This is better than rising temp -> rise in resistance -> increase in power -> rise in temp that would happen with a constant current source and metallic filament.

Carbon filament bulbs have a falling resistance with temperature, so are happier with P ∝ R as with a current source. Carbon filament bulbs powered from mains are often run in series with a metallic resistor ballast (eg. an ordinary bulb!) This is usually thought of as a combined resistance having a neutral or slightly positive temperature coefficient of resistance, but could be thought of as moving a low resistance constant voltage mains supply towards a constant current supply by adding series resistance.

Since real power supplies are neither constant voltage nor constant current, there is an optimum value of resistance where the power dissipated in the bulb is maximum. If the bulb has a resistance less than that, then it will get brighter if you increase its resistance. If the bulb has higher resistance than optimum, then you get more power by reducing the resistance. But you don't normally want maximum possible power: when connected to a mains supply that would be 10's of kW at least; with a battery not so much, but an equal amount of power is dissipated in the battery, so not a good idea.

You may have noticed that I have vacillated between talking about power and brightness. This is because I am making the assumption that brightness is proportional to the power dissipated by the bulb, which may not be exactly the case.
If you want to change the resistance of your filament, you may do this by changing the length or thickness of the filament. Much easier than making it out of a different alloy with different resistivity (I'd have thought.) Once you change the shape or size of your filament, the rate at which it radiates energy can change, thence the temperature at which it is in equilibrium for a given power dissipation. Different temperature gives different spectrum and different visual brightness. All a bit messy and probably not essential to your question.

#### Angela Liang

"My textbook says a bulb glows brighter due to its high resistance." That's a strange and (misleading) thing for an author to write.
Can you provide the actual text ? Perhaps he said 'high temperature' ?
Larger current will heat an incandescent filament to higher temperature increasing its light output, ie making it brighter ..
Ok. "The actual text is that Tungsten is used in light bulbs. It converts electrical energy to light and thermal energy due to its high resistance". So I interpreted it wrongly as it is suitable because it glows lighter. So the essential point is that it converts electrical energy to thermal energy? But you said 'larger current' will heat...' Do you mean that given its high resistance, more thermal energy is produced when a large current passes through? So that's why both high resistance and large current are needed?

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#### Angela Liang

I agree, the comment is nonsensical as written. I have many perfectly good bulbs which don't glow at all - because they are in their boxes in the cupboard. Whatever their resistance bulbs don't glow until you connect them to the electric power source. Therein lies the key to this issue.
Mains bulbs are connected to what is essentially a constant voltage source. Therefore their power is determined by P = V2/R and (for the same physical size, which we haven't yet mentioned) brightness is inversely proportional to resistance.

If you want power proportional to resistance, with P = I2R then you need a constant current source, which you come across in some LED drivers, but not for incandescent bulbs.

A battery or dynamo with low internal series resistance is a good approximation to a constant voltage source. Constant current sources are more difficult to make and usually involve wasting power.

Power inversely proportional to resistance is inherently more stable (for metallic filaments) because, if the filament temperature rises, the resistance increases, thence the power decreases and the temperature falls. This is better than rising temp -> rise in resistance -> increase in power -> rise in temp that would happen with a constant current source and metallic filament.

Carbon filament bulbs have a falling resistance with temperature, so are happier with P ∝ R as with a current source. Carbon filament bulbs powered from mains are often run in series with a metallic resistor ballast (eg. an ordinary bulb!) This is usually thought of as a combined resistance having a neutral or slightly positive temperature coefficient of resistance, but could be thought of as moving a low resistance constant voltage mains supply towards a constant current supply by adding series resistance.

Since real power supplies are neither constant voltage nor constant current, there is an optimum value of resistance where the power dissipated in the bulb is maximum. If the bulb has a resistance less than that, then it will get brighter if you increase its resistance. If the bulb has higher resistance than optimum, then you get more power by reducing the resistance. But you don't normally want maximum possible power: when connected to a mains supply that would be 10's of kW at least; with a battery not so much, but an equal amount of power is dissipated in the battery, so not a good idea.

You may have noticed that I have vacillated between talking about power and brightness. This is because I am making the assumption that brightness is proportional to the power dissipated by the bulb, which may not be exactly the case.
If you want to change the resistance of your filament, you may do this by changing the length or thickness of the filament. Much easier than making it out of a different alloy with different resistivity (I'd have thought.) Once you change the shape or size of your filament, the rate at which it radiates energy can change, thence the temperature at which it is in equilibrium for a given power dissipation. Different temperature gives different spectrum and different visual brightness. All a bit messy and probably not essential to your question.
Thanks:)

#### jim hardy

Gold Member
2018 Award
So the essential point is that it converts electrical energy to thermal energy?

Yes ! That's the essence of it.
Resistance is analogous to friction , rub your hands together and feel the heat.
The harder you press the more resistance to motion you'll feel and the more heat you'll make.

But you said 'larger current' will heat...' Do you mean that given its high resistance, more thermal energy is produced when a large current passes through?
Yes ! That's the essence of it. Power is the product of resistance and the square of current
power equals resistance X current2
and that power leaves the filament as radiated heat and light ,
in SI units
power is watts
current is amps
resistance is ohms

So that's why both high resistance and large current are needed?
Yes, with this caveat..
In the grand scheme of things lightbulbs are fairly puny handlers of power
To give you some perspective.......
one amp through 100 ohms would be 100 X one 2 = 100 watts, once a very typical bulb for a household room ceiling fixture
an automobile headlamp is around 35 watts low beam, nominal 12 volts & 3 amps; 50 watts high beam

the element in your water heater is probably 2500 watts
and the motor in your vacuum sweeper is perhaps 700 watts.
hope above helps

old jim

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#### Angela Liang

hope above helps

old jim
Thanks a lot! Now I can understand it better:)

#### russ_watters

Mentor
Ok. "The actual text is that Tungsten is used in light bulbs. It converts electrical energy to light and thermal energy due to its high resistance". So I interpreted it wrongly as it is suitable because it glows lighter. So the essential point is that it converts electrical energy to thermal energy? But you said 'larger current' will heat...' Do you mean that given its high resistance, more thermal energy is produced when a large current passes through? So that's why both high resistance and large current are needed?
I think their use of the word "high" was unnecessary and led you to the wrong interpretation. *Any* resistance results in heat dissipation and as it happens, lower resistance, not higher resistance, results in more heat dissipation(and light) at a fixed voltage.

#### Khashishi

A lower resistance means a higher current will flow, but if you consider the whole circuit for a light bulb, the bulb filament should have significantly higher resistance than the wires in the lamp and wall. The light bulb is in series with the wires in the lamp and wall. Since the current through all elements in a series is the same (neglecting capacitance), the elements with higher resistance will dissipate more heat. You want the light bulb to heat up, not your wires in your walls (which can cause fires).

It all goes back to the basics: $P = IV$. For Ohmic resistors, you have $V=IR$, so you can write it as $P=I^2 R$ or $P=V^2 / R$. Now these equations work for either the single element or for the whole circuit. If you look at just the filament as a single element, you have $P=V^2/R$, but the $V$ is the voltage across the filament, which is a little less than the voltage from the source, since some voltage is being dropped in the wires in the walls and lamp. Since $I$ is constant across the circuit, you see that the voltage drop is proportional to resistance, and the filament needs to have high resistance compared to the wires.

#### Angela Liang

I think their use of the word "high" was unnecessary and led you to the wrong interpretation. *Any* resistance results in heat dissipation and as it happens, lower resistance, not higher resistance, results in more heat dissipation(and light) at a fixed voltage.
Why? Because of the formula: P=I^2R? So under a fixed voltage, I affects more due to the power of 2?

#### russ_watters

Mentor
Why? Because of the formula: P=I^2R? So under a fixed voltage, I affects more due to the power of 2?
Right. And V=IR (I=V/R) so decreasing R increases I by the same amount and thus increases P.

#### Angela Liang

Right. And V=IR (I=V/R) so decreasing R increases I by the same amount and thus increases P.
Thanks :)

#### sophiecentaur

Gold Member
But how to achieve the condition where both current and resistance are high?
You feed it from a high voltage supply. Because it's made of Tungsten, its resistance will increase by ten times from cold to operating temperature (white hot). Despite that, given enough Volts you can achieve a high resistance and a lot of current - compared with the situation at room temperature.
This is a strange way of looking at the situation, though.

#### jim hardy

Gold Member
2018 Award
Angela,
have they covered Ohm's Law yet?
have they introduced what are the basic units yet? Charge, Current, Voltage ?

We who are decades into the subject tend to forget how awkward it was when we were just starting out.

Current Voltage and Resistance are related by simple ratios and that's Ohm's law.
You need them to be intuitive so you can work simple circuits in your head.

If you need a hand with the basic concepts don't hesitate to ask.
It's a helpful bunch here but sometimes we're not quite sure what to offer up.

old jim

#### sophiecentaur

Gold Member
Current Voltage and Resistance are related by simple ratios and that's Ohm's law.
There is always a caveat to be born in mind. Ohm's Law is not really a 'LAW'. It's more of a 'law', which just describes behaviour under certain circumstances. A light bulb filament does not follow Ohm's Law because the conditions of Ohm's Law are not followed in a glass envelope. Hence the reason for this ever lengthening thread. R is always the ratio of V/I but R is not always constant so it can't always be used on its own to jiggery pokery with V to get an I that you wanted.
"R=V/I" is not Ohm's law; it just gives the Resistance of a component when it was measured.
Otoh, Newton's gravitational LAW (under non-relativistic circumstances, at least): g=Gm1m2/r2 is a law because it applies to all combinations of m and r for all (as far as we know) values.
I think this is important to point out but, of course, I can never hope to change the almost universal inaccurate usage of the term Ohm's Law.

#### jim hardy

Gold Member
2018 Award
R=V/I" is not Ohm's law; it just gives the Resistance of a component when it was measured.
Otoh, Newton's gravitational LAW (under non-relativistic circumstances, at least): g=Gm1m2/r2 is a law because it applies to all combinations of m and r for all (as far as we know) values.
I think this is important to point out but, of course, I can never hope to change the almost universal inaccurate usage of the term Ohm's Law.

I'd say it's "inaccurate" only in that we don't finish the explanation by pointing out when R becomes a function of current f(I) , as in a lightbulb or semiconductor, it's no longer just a simple linear ratio . That's algebra not physics. well, maybe some physics and differential equations if we calculate resistance of the element as f(temperature), temperature in turn by heat balance with heat input f(I) .

Yet an ohm is a volt per amp , simple ratio at the time of measurement
as you said
"...it just gives the Resistance of a component when it was measured..". which takes time out of the formula , but R can be a function of I .

I was under impression OP might be struggling with basics so didn't go there yet. I might be wrong.

old jim

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#### Angela Liang

Angela,
have they covered Ohm's Law yet?
have they introduced what are the basic units yet? Charge, Current, Voltage ?

We who are decades into the subject tend to forget how awkward it was when we were just starting out.

Current Voltage and Resistance are related by simple ratios and that's Ohm's law.
You need them to be intuitive so you can work simple circuits in your head.

If you need a hand with the basic concepts don't hesitate to ask.
It's a helpful bunch here but sometimes we're not quite sure what to offer up.

old jim
OK. Thanks very much:)

#### Angela Liang

OK. Thanks very much:)
We have covered Ohm's Law, also only ohmic conductors obey Ohm's Law.

#### Angela Liang

OK. Thanks very much:)
However I have a question- it is said that no matter ohmic or non-ohmic conductors, the resistance generally increases with increasing temperature, but I find that according to Ohm's Law, if V is fixed, the larger the current, the lower the resistance( R=V/I). As I believe that larger current means higher temperature, the whole thing becomes contradictory. How is it that possible?

#### sophiecentaur

Gold Member
Jim. I know you know and you know I know you know about this. My only point is that a simple ratio does not make a LAW. It's just the Definition of Resistance.

On a parallel line, we have Newton's Second LAW of motion: F=ma. That really does come under the heading of a LAW because (unless you stray into Relativistic conditions) it describes a relationship that always applies. Otoh, the 'Equations' of motion (SUVAT) under linear acceleration are mere Equations and people just use them to work things out. R=V/I is one of those Equations and it also comes in handy in electrical problems.

Am I really being too picky here? Go on, you can tell me. I shall not take umbrage (haha).
I would be much happier if people said "When the component follows Ohms Law" and not just "That's Ohm's Law"

#### jim hardy

Gold Member
2018 Award
My only point is that a simple ratio does not make a LAW. It's just the Definition of Resistance.

My initial schooling was practical in nature
Mr Davis (high school electronics teacher) taught us boys
'all resistors have some temperature coefficient. Carbon's happens to be negative so you will find that low-ohm carbon resistors actually decrease as they heat up. We know Ohm's law is the thinking tool describes ideal components, always remain aware that real ones are not quite ideal.'

So i never thought about definition of "law", ,, Ohm's is perhaps a ' thought tool' that goes by the name "Law" ?
Perhaps as teaching evolves it'll become "Ohm's Ratio"

My only point is that a simple ratio does not make a LAW. It's just the Definition of Resistance.
You know i have a soapbox about the term "Ground"

Am I really being too picky here? Go on, you can tell me.
no, it's a fine point that should be made just as Mr Davis did for us boys back in 1962.
Ms Angela seems aware of the distinction, she says ( i presume gender from the name) "only ohmic conductors obey Ohm's Law."
back to Lavoisier:
Abbé de Condillac adds: "But, after all, the sciences have made progress, because philosophers have applied themselves with more attention to observe, and have communicated to their language that precision and accuracy which they have employed in their observations:
In correcting their language they reason better."

One term at a time. I'll try "Ohm's Ratio" out on next beginner i help , see how it works.
How does "only ohmic conductors obey Ohm's Ratio" sound to you ?.

old jim

old jim

"Which makes a bulb glows brighter?"

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