Physics past papers question

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Rosie1999
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Explain the effect on the resistance of the bulbs if they are operated at a much smaller current so that neither bulb lights up.
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cymbly schweig
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(Original post by Rosie1999)
Explain the effect on the resistance of the bulbs if they are operated at a much smaller current so that neither bulb lights up.
ooh hi
I think according to Ohm`s law R∝I then if we operarate at a low(much smaller current ) then the resistance will also decrease
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TheTechnoGuy
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Ohms Law tells us that if the current is lower, then the resistance will also be lower. You know the resistance will go down because less current means less ion collisions meaning a lower temperature - Less heat = Less resistance. So because the temperature has gone down, the effective resistance of the bulb will also go down. Less temp. also means less energy is lost to heat than before.
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Rosie1999
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but how does resistance go down? isn't due to the equation R=V/I Resistance is inversely propotional to current ? so as current decrease resistance increases? This is why am confused know :s
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Rosie1999
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but how does resistance go down? isn't due to the equation R=V/I Resistance is inversely propotional to current ? so as current decrease resistance increases? This is why am confused know :s
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TheTechnoGuy
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(Original post by Rosie1999)
but how does resistance go down? isn't due to the equation R=V/I Resistance is inversely propotional to current ? so as current decrease resistance increases? This is why am confused know :s
I know it sounds confusing, I may have misquoted ohms law, but a light bulb DOES not follow Ohms law, it has a curve-y I-V graph. You have to think of it practically. A light bulb (filament) is practically a very inefficient resistor that doesn't follow ohms law.

The hotter the filament gets, the resistance will go up because more heat will result in more ion collisions (More KE) - This is the result of more power going into the bulb (ie More voltage or more current)

So this works the opposite way too, if the current goes down, then less power will go into the bulb meaning the bulb will be less hot than it was before. Which means Less collisions (due to less KE) and less collisions means electrons can flow more easily hence resistance goes down.

Sorry if my first was a tad bit confusing, electricity is regarded as one of the hardest modules in Physics so don't worry if you still don't get it.
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Rosie1999
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(Original post by TheTechnoGuy)
I know it sounds confusing, I may have misquoted ohms law, but a light bulb DOES not follow Ohms law, it has a curve-y I-V graph. You have to think of it practically. A light bulb (filament) is practically a very inefficient resistor that doesn't follow ohms law.

The hotter the filament gets, the resistance will go up because more heat will result in more ion collisions (More KE) - This is the result of more power going into the bulb (ie More voltage or more current)

So this works the opposite way too, if the current goes down, then less power will go into the bulb meaning the bulb will be less hot than it was before. Which means Less collisions (due to less KE) and less collisions means electrons can flow more easily hence resistance goes down.

Sorry if my first was a tad bit confusing, electricity is regarded as one of the hardest modules in Physics so don't worry if you still don't get it.

SO it has nothing to do with ohms law basically with the power that goes into the bulb p=iv as current decreases, power decreases and then will be less hot which mean less flow of electrons so that's why resistance goes down

How can I know when resiatance increases or decreases? because there are loads of questions like these and I find like that one that resistance decreases which was pretty confusing at the beginning but know i guess it's better.

Thank you so much , you're so smart
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Rosie1999
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which book or resources do you study from? do you have any revision sheets you use? I really need them i'll be really grateful. and Thank you for the reply
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TheTechnoGuy
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(Original post by Rosie1999)
SO it has nothing to do with ohms law basically with the power that goes into the bulb p=iv as current decreases, power decreases and then will be less hot which mean less flow of electrons so that's why resistance goes downHow can I know when resiatance increases or decreases? because there are loads of questions like these and I find like that one that resistance decreases which was pretty confusing at the beginning but know i guess it's better.Thank you so much , you're so smart
Usually questions based on bulbs, they ask questions for you to answer with the theory rather than the maths.

A filament bulb doesn't follow Ohm's Law to it's working voltage. You only use the Ohm's Law equation (V=IR) when they give you you a single value such as the current or voltage at one point and then ask you to calculate the resistance.

For example, lets say a questions says "A bulb is powered at an constant voltage of 5v, a ammeter is connected in series and a value of 0.3 A is shown on the display" - THIS IS WHEN WE USE THE V=IR EQUATION, because it is at one point, so the answer to this would be 5/0.3 which would give 16.3 Ohms, this means the resistance is 16.3 Ohms when the bulb is operated at 5 volts.

However if it ask's to describe the trend through a range of voltages such as a question like "Describe what happens in terms of resistance as the voltage is increased from 0 to the bulbs working voltage", Now we use general knowledge, no calculation needed, as the voltage goes up, the power intake does too. As that power goes up, so does the temp. Hence as the voltage goes up, the bulbs resistance also increases up until the point of the working voltage.

I hope this clears it up, it's always a good idea to look at the IV graph. They look for key words such as ions colliding, more KE, more heat = more collisions = more resistance etc.
However you need to remember that the Current - Voltage relationship in a filament bulb is non linear hence it does not follow ohm's law. They asked that in one of the AS Physics exams last year and it threw everyone off because most people assumed that if V=IR works at points of the curve then it must be ohmic, well the answer was that it was non-ohmic (not following ohms law)

No problem, and thank you ( making me blush...lol)
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TheTechnoGuy
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(Original post by Rosie1999)
which book or resources do you study from? do you have any revision sheets you use? I really need them i'll be really grateful. and Thank you for the reply
Well, I'm on the edexcel board for AS Physics and well I use the old spec book for question - they're very useful. Do loads of past papers, try other exam board papers as well, AQA had a dedicated electricity back in the day, I used to do questions on that too.

I'm assuming you're on the linear AS spec?
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Rosie1999
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(Original post by TheTechnoGuy)
Well, I'm on the edexcel board for AS Physics and well I use the old spec book for question - they're very useful. Do loads of past papers, try other exam board papers as well, AQA had a dedicated electricity back in the day, I used to do questions on that too.

I'm assuming you're on the linear AS spec?
I'm using edexcel board AS physics as well , I'm using the new one and yeah I guess I'll have to try other exam board papers as well. Thanks
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Rosie1999
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(Original post by TheTechnoGuy)
Well, I'm on the edexcel board for AS Physics and well I use the old spec book for question - they're very useful. Do loads of past papers, try other exam board papers as well, AQA had a dedicated electricity back in the day, I used to do questions on that too.

I'm assuming you're on the linear AS spec?
I have another question
Two coherent sources emit waves of wavelengths lamda n phase . At a point where the two waves meet they have a phase difference of 90 degrees which of the following could be the path difference at this point?
A)2 lamda
B) lamda
c) lamda /2
D) lamda / 4
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TheTechnoGuy
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Name:  DSC_0057.jpg
Views: 148
Size:  473.8 KBThe answer would be D, the solution is enclosed but it's just inputting the values into the phase diff. equation and finding the path diff in terms of lamda.
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Rosie1999
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(Original post by TheTechnoGuy)
Name:  DSC_0057.jpg
Views: 148
Size:  473.8 KBThe answer would be D, the solution is enclosed but it's just inputting the values into the phase diff. equation and finding the path diff in terms of lamda.
Thank you so much, that was really Helpful
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cymbly schweig
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(Original post by TheTechnoGuy)
Name:  DSC_0057.jpg
Views: 148
Size:  473.8 KBThe answer would be D, the solution is enclosed but it's just inputting the values into the phase diff. equation and finding the path diff in terms of lamda.
wooow that`s splendid
thanks a lot
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