What quantity affects the brightness in a bulb? Watch

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th3 hampst3r
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Is it current, voltage, power? Found a few 3 mark questions relating to this, and answer + explanation would be much appreciated. Ty
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OllieGCSEs
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(Original post by th3 hampst3r)
Is it current, voltage, power? Found a few 3 mark questions relating to this, and answer + explanation would be much appreciated. Ty
I also would like an answer to this question - I know that more power will make a bulb brighter, but I'm unsure as to whether the voltage increases the brightness or the current does... I have a feeling it's both, but I don't know for sure I'm afraid
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Stonebridge
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(Original post by th3 hampst3r)
Is it current, voltage, power? Found a few 3 mark questions relating to this, and answer + explanation would be much appreciated. Ty
All 3 can and do affect brightness.

It's better if you post the actual exam question. Then we can give you a more focussed answer.
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uberteknik
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(Original post by th3 hampst3r)
Is it current, voltage, power? Found a few 3 mark questions relating to this, and answer + explanation would be much appreciated. Ty
You need to assume the resistance of the light bulb is constant.

Then for d.c. Power dissipated = V*I

In the case of a light bulb, the current through the load resistance is determined by the supply voltage. The higher the voltage the more current is pulled and the therefore more power is consumed.
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Namige
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Can anyone explain how the thickness of the wire(resistance) will affect the brightness?

From what I can see higher resistance will reduce the power through the wire, as P=VI. This is what confuses me. Won't the increased resistance in itself, increase the power dissipated, as in the rate of energy lost as heat? For example, why doesn't a think wire glow, but a very thin wire glow under the same voltage? Even though it has a higher resistance.
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(Original post by th3 hampst3r)
Is it current, voltage, power? Found a few 3 mark questions relating to this, and answer + explanation would be much appreciated. Ty
Directly, it is the power that determines the brightness of a bulb. That is why lightbulbs have a power rating, i.e. a 60W bulb is enough to light a typical bedroom.

But power is determined by the current and voltage which, in turn, is determined by the current.
Let's look at a few equations
P = IV
Since
V = IR

P = I * IR
hence
P = I^2R
Spoiler:
Show
where:
P = Power
V = Voltage
I = Current
R = Resistance


I hope this helps you to understand it a little better.
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uberteknik
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There is some confusion here. As Stonebridge asks, it is better to see the context of the original question.

A current is a measure of the flow of electrons and electrons cannot flow unless there is an emf driving them. ergo no emf, no current.
So current is determined by voltage and resistance in a real world application.

The following is not required to answer a GCSE/A-level question. It does however show why there is difficulty in interpreting the OP's question:

Chapmouse is correct in identifying that standard light bulbs are marketed as having a brightness as a function of power consumed at the standard mains supply voltage of 230V a.c. This is so that consumers can easily choose a bulbs relative brightness based on the specified power rating.

The reason power is used to identify brightness is that there are a multitude of variables that affect the light output: dimensions and shape of the filament, conductor material (normally tungsten), gas filling (xenon, halogen etc), transparent enclosure (glass, quartz etc) etc.

But there is no industry standard that says any given light bulb at a certain power rating, must produce a given output of light. Moreover there is no specification for the colour temperature of that light either and hence the perceived brightness of the bulb.

So a manufacturer will state a power rating and will have designed the light bulb to consume a standardised power at the standardised mains supply voltage. i.e. 30W, 60W, 100W, 150W etc @230V a.c.

If the supply voltage is varied but resistance is kept constant, then the current drawn by the bulb will also vary and hence the brightness will alter.
If the filament resistance is varied but the supply voltage is held constant, then the current drawn by the bulb will vary and hence the brightness will alter.
However if the current supplied to the bulb is held constant (constant current source) the only way of then varying the brightness is by controlling both the resistance and supply voltage as a combined function. i.e. the ratio V/R must also be held constant. But crucially P = V2​/R. So the higher the current, the power dissiapated increases exponentially.

Stonebridge is correct that power, voltage, current, resistance are all linked and that varying any one will result in a change of brightness - on paper.
This is however purely mathematical in that the only physical parameters in practice that can be changed are the ones stated earlier.

And the only electrical physical parameter that is within a manufacturers control is the resistance of the filament.

Hence we get back to the OP's question: Is the context purely academic or is it real world application?
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uberteknik
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(Original post by Namige)
Can anyone explain how the thickness of the wire(resistance) will affect the brightness?

From what I can see higher resistance will reduce the power through the wire, as P=VI. This is what confuses me. Won't the increased resistance in itself, increase the power dissipated, as in the rate of energy lost as heat? For example, why doesn't a think wire glow, but a very thin wire glow under the same voltage? Even though it has a higher resistance.

P = V*I

but I = V/R

therefore

P = V*V/R

P = V2/R

Hence reducing the resistance whilst keeping the voltage constant will increase the power dissipated.
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Namige
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(Original post by uberteknik)
P = V*I

but I = V/R

therefore

P = V*V/R

P = V2/R

Hence reducing the resistance whilst keeping the voltage constant will increase the power dissipated.
Then why does a thick wire not glow, whereas a thin wire with higher resistance, does?
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hecandothatfromran
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'Brightness' = intensity = power of the light/area swept by light. As the area swept out by the light remains the same (I.E. the light always radiates in the same direction/distribution from the filament ). So its power....

Edit: I think the confusion is probably that Power is the sole variable in determining brightness. Increasing either voltage or current won't necessarily increase the brightness as there's another independent variable that must remain constant (raising voltage only increases brightness if current is constant as P=VI, and vice-versa)!
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hecandothatfromran
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(Original post by Namige)
Then why does a thick wire not glow, whereas a thin wire with higher resistance, does?

Edit: sorry, misread :P.
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uberteknik
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(Original post by Namige)
Then why does a thick wire not glow, whereas a thin wire with higher resistance, does?
Current density and hence power density.

A thick wire has a lower resistance than a thin wire made from the same material.

However, it's the current density through either wire which produces the heat resulting in light emission.

The thicker wire even though it has lower resistance, must pass a substantially higher current to achieve the same current density and hence energy dissipation per unit area to reach the same temperature as the thin wire.
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Namige
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(Original post by uberteknik)
Current density and hence power density.

A thick wire has a lower resistance than a thin wire made from the same material.

However, it's the current density through either wire which produces the heat resulting in light emission.

The thicker wire even though it has lower resistance, must pass a substantially higher current to achieve the same current density and hence energy dissipation per unit area to reach the same temperature as the thin wire.
So since power dissipated is v^2/r, lower resistance means more power across the wire but less is wasted though heat and light emission? Normally, brightness is proportional to power, but in this case it isn't true? Still confused.
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uberteknik
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(Original post by Namige)
So since power dissipated is v^2/r, lower resistance means more power across the wire but less is wasted though heat and light emission? Normally, brightness is proportional to power, but in this case it isn't true? Still confused.
Brightness is still proportional to power.

The key is understanding that power dissipation results in a temperature increase but light will not be emitted until the power density (power dissipated per unit volume) reaches a threshold (governed by the dimensions of the filament) for that to happen.

A small filament takes a lot less energy to heat up (and glow) than a large chunky filament.

The large filament will eventually glow, but the power needed for it to do so is much greater than compared to that needed to make the small filament glow.

And as the power dissipated increases further, the glow will get brighter.
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th3 hampst3r
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Hi again, thankyou all very much for your time.

Here is the kind of question I'm struggling with;

http://filestore.aqa.org.uk/subjects...2-QP-JAN12.PDF

Page 9, question 8c

I guess what I was expecting was a specific quantity I could deduce to be increasing/decreasing. It seems now its not as simple as that. The question above I have answered on 2 seperate occasions, each time arriving at a different answer following alternate reasoning.

Take the faulty headlamp as lamp A, and the lamp which is affected by this as lamp B.

On the one hand, treating the system as similar to a potentiometer (I know its not, its in parallel), lamp B 'takes' a greater portion of the voltage in order to overcome its greater resistance. Now taking a greater share of the voltage, it is brighter (P= V^2 / R).

On the other hand, now having the greater resistance, less current flows through B. This means it now has less power (P= I^2 *R).

Each method relies on either voltage or current staying constant, which you seem to say neither would in practice. At a bit of a loss as to what quantity I can use to determine the brightness :L

Thankyou again for your time.

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th3 hampst3r
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For some reason my replies aren't view-able when I refresh... not sure whats going on :L
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cuckoo99
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Power and the things that are directly proportional to it. e.g current and voltage as intensity of light = power incident per cross sectional area eg P/A
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th3 hampst3r
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Hi again, thankyou very much for your time.

Here is the kind of question I'm struggling with;

Jan 2012, AQA AS Physics B, Unit 2, Page 9, question 8c. The rest of the question I am fine with.

Take the fault headlamp as lamp A, and the lamp in question (the one for which a change in brightness is to be deduced) as lamp B.

I have answered this question twice, each arriving at separate answers with alternate reasoning. I guess what I was hoping for was a specific quantity which I could deduce to be increasing/decreasing, perhaps even by introducing figures into the equation.

On the one hand, treating the system as something close to a potentiometer (I know its not, its in parallel), lamp B takes a greater share of the voltage in order to overcome its greater resistance; resulting in increased power and work done (P = V^2 / R).

On the other hand, now having a greater resistance, current flows less through lamp B, meaning it has less power (P = I^2 *R).

Both methods rely on either voltage or current remaining constant, which you seem to be telling me would not happen in practice.

Thankyou again for your time
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uberteknik
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(Original post by th3 hampst3r)
Hi again, thankyou very much for your time.

Here is the kind of question I'm struggling with;

Jan 2012, AQA AS Physics B, Unit 2, Page 9, question 8c. The rest of the question I am fine with.
Hi, please could you scan the question and post or give a location address, as I am having trouble locating it on the most common sites.

Thanks.
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th3 hampst3r
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When I include links in a post, the post doesn't seem to appear when it is submitted. If you google 'aqa as physics b past papers', the first option will take you to a site. Ty
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