Hereis the question:
I get how the resistance decreases in the LDR, and I would have gotten all the marks. I have a question though.
The mark scheme says When the LDR resistance is reduced, that decreases the pd across the LDR, OR increases the pd across the fixed resistor.
I get how a lower LDR resistance means a lower pd across the LDR, as V=IR so a smaller value of R would mean a smaller value of V.
However, I don't get how a lower LDR resistance would mean the pd would increase across the fixed resistor. I know the current would be the same everywhere in the circuit (series circuit)., Why would the pd increase across the fixed resistor? i don't get that, as R doesn't change, so V=IR, why would that increase the value of V?
Thanks
Simple circuit question involving an LDR and a fixed resistor  I have a question. Watch
Announcements

blobbybill
 Follow
 2 followers
 8 badges
 Send a private message to blobbybill
 Thread Starter
Offline8ReputationRep: Follow
 1
 26012017 22:06
Last edited by blobbybill; 26012017 at 22:10. 
Laurenx
 Follow
 34 followers
 11 badges
 Send a private message to Laurenx
Offline11ReputationRep: Follow
 2
 26012017 22:14
Kirchhoff's Second Law means that...
p.d. of LDR + p.d. of fixed resistor = 6V
Therefore, to keep the sum at 6V: if p.d. of LDR decreases, then the p.d. of the fixed resistor must increase 
blobbybill
 Follow
 2 followers
 8 badges
 Send a private message to blobbybill
 Thread Starter
Offline8ReputationRep: Follow
 3
 26012017 22:18
(Original post by Laurenx)
Kirchhoff's Second Law means that...
p.d. of LDR + p.d. of fixed resistor = 6V
Therefore, to keep the sum at 6V: if p.d. of LDR decreases, then the p.d. of the fixed resistor must increaseLast edited by blobbybill; 26012017 at 22:24. 
 Follow
 4
 26012017 22:24
(Original post by blobbybill)
Thanks, but how can you explain that in terms of V=IR? R is constant (75 ohms) and I would be the same everywhere in the circuit, so why does it change in terms of the equation? V=IR seems to disprove your answer (but I know it is correct because the mark scheme also says it), as I and R are constant, so it would seem V shouldn't change, but it does. 
blobbybill
 Follow
 2 followers
 8 badges
 Send a private message to blobbybill
 Thread Starter
Offline8ReputationRep: Follow
 5
 26012017 22:26
(Original post by h3rmit)
If the resistance of the LDR changes, R for the LDR isn't constant 
Laurenx
 Follow
 34 followers
 11 badges
 Send a private message to Laurenx
Offline11ReputationRep: Follow
 6
 26012017 22:31
(Original post by blobbybill)
Thanks, but how can you explain that in terms of V=IR? R is constant (75 ohms) and I would be the same everywhere in the circuit, so why does it change in terms of the equation?
For V=IR of the fixed resistor, you know that R cannot change (since it's fixed). That means V ∝ I for the fixed resistor. Since I has increased, it follows that the p.d. of the fixed resistor must also increase. 
blobbybill
 Follow
 2 followers
 8 badges
 Send a private message to blobbybill
 Thread Starter
Offline8ReputationRep: Follow
 7
 26012017 22:34
(Original post by Laurenx)
I would be the same everywhere in the circuit  but if there is a smaller resistance across the LDR, that means that more current can flow through the LDR; hence the current flowing through the entire circuit is increased (like you said, I is the same everywhere).
For V=IR of the fixed resistor, you know that R cannot change (since it's fixed). That means V ∝ I for the fixed resistor. Since I has increased, it follows that the p.d. of the fixed resistor must also increase.
To clarify, as you said, the current in the circuit would increase as a result of the lower resistance in the LDR. However, as V=IR, if the current had increased by quite a lot and the resistance of the LDR had only decreased a small amount, couldn't that cause the voltage to increase rather than decrease?
How do you know the voltage would decrease across the LDR? Even though V=IR, if the current increases (which it does when the resistance drops), and the resistance falls due to increasing light, how do you know that the voltage would decrease? Surely it could increase in some circumstances?
I am really confused right now. Please help.Last edited by blobbybill; 26012017 at 22:43. 
 Follow
 8
 26012017 22:45
(Original post by blobbybill)
Thanks.
To clarify, as you said, the current in the circuit would increase as a result of the lower resistance in the LDR. However, as V=IR, if the current had increased by quite a lot and the resistance of the LDR had only decreased a small amount, couldn't that cause the voltage to increase rather than decrease?
How do you know the voltage would decrease across the LDR? Even though V=IR, if the current increases (which it does when the resistance drops), and the resistance falls due to increasing light, how do you know that the voltage would decrease? Surely it could increase in some circumstances?
I am really confused right now. Please help. 
Laurenx
 Follow
 34 followers
 11 badges
 Send a private message to Laurenx
Offline11ReputationRep: Follow
 9
 26012017 22:54
(Original post by blobbybill)
Thanks.
To clarify, as you said, the current in the circuit would increase as a result of the lower resistance in the LDR. However, as V=IR, if the current had increased by quite a lot and the resistance of the LDR had only decreased a small amount, couldn't that cause the voltage to increase rather than decrease?
How do you know the voltage would decrease across the LDR? Even though V=IR, if the current increases (which it does when the resistance drops), and the resistance falls due to increasing light, how do you know that the voltage would decrease? Surely it could increase in some circumstances?
I am really confused right now. Please help.(Original post by h3rmit)
You'd get free force if the total voltage increased without the power supply emf increasing.
Since you know that voltage across the fixed resistor MUST increase when the current increases, the voltage across the LDR cannot also increase as V_{LDR }+ V_{resistor }= 6V.
Or you can think of it like 6V  V_{resistor }= V_{LDR }
Which shows that if p.d. across the resistor increases, then the p.d across the LDR cannot increase as 6 is a constant.
I'm so sorry if I'm not making any sense!!Last edited by Laurenx; 26012017 at 22:57. 
 Follow
 10
 26012017 22:59
(Original post by blobbybill)
Thanks.
To clarify, as you said, the current in the circuit would increase as a result of the lower resistance in the LDR. However, as V=IR, if the current had increased by quite a lot and the resistance of the LDR had only decreased a small amount, couldn't that cause the voltage to increase rather than decrease?
How do you know the voltage would decrease across the LDR? Even though V=IR, if the current increases (which it does when the resistance drops), and the resistance falls due to increasing light, how do you know that the voltage would decrease? Surely it could increase in some circumstances?
I am really confused right now. Please help.
Battery = 10V
fixed resistor = 100 ohms
LDR = 100 ohms
find the current in the circuit and the PD across the LDR
see what happens to the current and the PD across the LDR as the resistance of the LDR changes while the fixed resistor and battery remain fixed  see if there's any possible way you can make the PD across the LDR go up by making it's resistance smaller.
do this for 15 minutes and you'll understand more than posting on here for 1 hour 
blobbybill
 Follow
 2 followers
 8 badges
 Send a private message to blobbybill
 Thread Starter
Offline8ReputationRep: Follow
 11
 26012017 23:01
(Original post by Joinedup)
try plugging in some numbers and see what happens
Battery = 10V
fixed resistor = 100 ohms
LDR = 100 ohms
find the current in the circuit and the PD across the LDR
see what happens to the current and the PD across the LDR as the resistance of the LDR changes while the fixed resistor and battery remain fixed  see if there's any possible way you can make the PD across the LDR go up by making it's resistance smaller.
do this for 15 minutes and you'll understand more than posting on here for 1 hour
I messed around with numbers in it about 30 mins ago, but I made up the currents and didn't work them out. Even still only once I was able to make it so the PD across the LDR to increase. I guess if I did it using proper currents that I worked out, that would make it so that the LDR pd would decrease?Last edited by blobbybill; 26012017 at 23:06. 
Laurenx
 Follow
 34 followers
 11 badges
 Send a private message to Laurenx
Offline11ReputationRep: Follow
 12
 26012017 23:07
(Original post by Joinedup)
try plugging in some numbers and see what happens
Battery = 10V
fixed resistor = 100 ohms
LDR = 100 ohms
find the current in the circuit and the PD across the LDR
see what happens to the current and the PD across the LDR as the resistance of the LDR changes while the fixed resistor and battery remain fixed  see if there's any possible way you can make the PD across the LDR go up by making it's resistance smaller.
do this for 15 minutes and you'll understand more than posting on here for 1 hour
@blobbybill I hope it makes sense soon 
 Follow
 13
 26012017 23:08
(Original post by blobbybill)
Will do. I'd need to work out new values of I (the current) when I decrease the LDR resistance wouldn't I?
you'll do I=V/(R_{fixed}+R_{LDR}) each time (resistances are added because they're in series)
then
V_{LDR}=I x R_{LDR}
then on to the next value of LDR resistance and calculate the current again. 
blobbybill
 Follow
 2 followers
 8 badges
 Send a private message to blobbybill
 Thread Starter
Offline8ReputationRep: Follow
 14
 26012017 23:12
(Original post by Joinedup)
Absolutely!
you'll do I=V/(R_{fixed}+R_{LDR}) each time (resistances are added because they're in series)
then
V_{LDR}=I x R_{LDR}
then on to the next value of LDR resistance and calculate the current again.
Hypothetically speaking, the LDR resistance could go up (according to V=IR) if the current increased by a bit but the resistance only dropped by a tiny tiny amount. But that wouldn't be possible, as then you would have more total voltage than the EMF of the power supply. Can you explain that to me? Or should I not worry about that? Am I just overthinking it? 
 Follow
 15
 26012017 23:21
(Original post by blobbybill)
Thanks!
Hypothetically speaking, the LDR resistance could go up (according to V=IR) if the current increased by a bit but the resistance only dropped by a tiny tiny amount. But that wouldn't be possible, as then you would have more total voltage than the EMF of the power supply. Can you explain that to me? Or should I not worry about that? Am I just overthinking it?
Battery fixed... fixed resistor fixed... LDR variable 
 Follow
 16
 27012017 00:08
(Original post by blobbybill)
Hereis the question:
I get how the resistance decreases in the LDR, and I would have gotten all the marks. I have a question though.
The mark scheme says When the LDR resistance is reduced, that decreases the pd across the LDR, OR increases the pd across the fixed resistor.
I get how a lower LDR resistance means a lower pd across the LDR, as V=IR so a smaller value of R would mean a smaller value of V.
However, I don't get how a lower LDR resistance would mean the pd would increase across the fixed resistor. I know the current would be the same everywhere in the circuit (series circuit)., Why would the pd increase across the fixed resistor? i don't get that, as R doesn't change, so V=IR, why would that increase the value of V?
ThanksLast edited by leafeater; 27012017 at 00:10. 
blobbybill
 Follow
 2 followers
 8 badges
 Send a private message to blobbybill
 Thread Starter
Offline8ReputationRep: Follow
 17
 27012017 09:30
(Original post by Joinedup)
Absolutely!
you'll do I=V/(R_{fixed}+R_{LDR}) each time (resistances are added because they're in series)
then
V_{LDR}=I x R_{LDR}
then on to the next value of LDR resistance and calculate the current again.
However, I have a question. Since I would increase when R decreases, how come that it never happens where the voltage increases (for example, in V=IR where R only decreases a little and I increased a lot)? I can't figure out why that never happens (even in normal circuits).
Is it because you are using R in the calculation of the current (I=V/R)? I don't understand why that isn't possible, as the current always increases when resistance falls and voltage falls, so why is it not possible for V=IR that V could increase, why doesn't it ever happen where I increases a lot when R decreases only a little (and in that case, make the value of V increase)? Why does that not occur? Surely it could?
In theory, if I were to increase by a lot, and R were to only decrease by a tiny amount, then V would increase (rather than decrease). Am I right in thinking that this scenario isn't possible because you have to calculate the value of I using the total resistance, V=IR, I=V/R, R=V/I all use each others numbers in their calculations, so you don't end up with a scenario like the one I just said?Last edited by blobbybill; 27012017 at 09:44. 
blobbybill
 Follow
 2 followers
 8 badges
 Send a private message to blobbybill
 Thread Starter
Offline8ReputationRep: Follow
 18
 27012017 09:30
(Original post by Laurenx)
Haha, good idea  you're more likely to accept it if you figure it out yourself, compared to simply accepting whatever someone else tells you!!
@blobbybill I hope it makes sense soon
However, I have a question. Since I would increase when R decreases, how come that it never happens where the voltage increases (for example, in V=IR where R only decreases a little and I increased a lot)? I can't figure out why that never happens (even in normal circuits).
Is it because you are using R in the calculation of the current (I=V/R)? I don't understand why that isn't possible, as the current always increases when resistance falls and voltage falls, so why is it not possible for V=IR that V could increase, why doesn't it ever happen where I increases a lot when R decreases only a little (and in that case, make the value of V increase)? Why does that not occur? Surely it could?
In theory, if I were to increase by a lot, and R were to only decrease by a tiny amount, then V would increase (rather than decrease). Am I right in thinking that this scenario isn't possible because you have to calculate the value of I using the total resistance, V=IR, I=V/R, R=V/I all use each others numbers in their calculations, so you don't end up with a scenario like the one I just said?Last edited by blobbybill; 27012017 at 09:44. 
uberteknik
 Follow
 104 followers
 21 badges
 Send a private message to uberteknik
 Study Helper
Online21ReputationRep:Study Helper Follow
 19
 27012017 10:18
(Original post by blobbybill)
Plugged in numbers about 20 times now, I see that the same otucome always occurs.
However, I have a question. Since I would increase when R decreases, how come that it never happens where the voltage increases (for example, in V=IR where R only decreases a little and I increased a lot)? I can't figure out why that never happens (even in normal circuits).
Is it because you are using R in the calculation of the current (I=V/R)? I don't understand why that isn't possible, as the current always increases when resistance falls and voltage falls, so why is it not possible for V=IR that V could increase, why doesn't it ever happen where I increases a lot when R decreases only a little (and in that case, make the value of V increase)? Why does that not occur? Surely it could?
Current is the flow (rate) at which charge (electrons) move through the circuit. Think of electrons in a conductor like water moving in a pipe as an analogy.
The supply voltage is a measure of the energy available to the circuit per quantity of electrons.
Resistors do what they say on the tin: they 'resist' the flow of electrons (current). In doing so, they act like a radiator in a central heating system by causing electrons to give up the energy supplied by the battery.
The energy given up when current flows in the resistance is also measured as a voltage  this time the voltage is developed across the resistance and is known as a potential difference. (pd)
In other words, the voltage of the supply is a source of energy, the voltage across the resistance is energy lost.
In the LDR example, the total resistance in the path of the current is R_{fixed} + R_{LDR}
The supply voltage V_{supply} provides the pressure to drive current around the circuit. The current is limited by the resistance in the electrons path.
I_{total} = V_{supply }/ (R_{fixed} + R_{LDR}) .....................(1)
Because the current flows through a single series path, the value of that current is the same at all points in the circuit.
The energy given up when the current passes through the LDR is measured in volts. And this is simply the current I_{total} x R_{LDR }
In other words, V_{LDR} = I_{total} x R_{LDR}......................(2)
If the LDR resistance changes, then because the supply voltage is the same, the total current in the circuit must also change. If R_{LDR} decreases, then I_{total} must increase since the fixed resistor does not change and the total resistance as fallen. If R_{LDR} increases, then I_{total} must decrease since the fixed resistor does not change and the total resistance has increased. (See eq (1))
Critically, because the total current has changed, then the energy lost across the resistances must also change:
If the LDR resistance increases, then the total current must fall (the supply voltage has not changed eq 1). And because the current is the same at all points in the circuit, the voltage developed (energy lost) across the LDR must also fall. (See eq (2))
If the LDR resistance decreases, then the total current must increase (the supply voltage has not changed eq 1). And because the current is the same at all points in the circuit, the voltage developed (energy lost) across the LDR must also increase. (See eq (2))
You must make the conceptual leap that the supply voltage is energy provided while the voltage (pd) across the resistor is energy lost.Last edited by uberteknik; 27012017 at 14:41. 
blobbybill
 Follow
 2 followers
 8 badges
 Send a private message to blobbybill
 Thread Starter
Offline8ReputationRep: Follow
 20
 27012017 12:43
(Original post by uberteknik)
I think you are missing some important concepts:
Current is the flow (rate) at which charge (electrons) move through the circuit. Think of electrons in a conductor like water moving in a pipe as an analogy.
The supply voltage is a measure of the energy available to the circuit per quantity of electrons.
Resistors do what they say on the tin: they 'resist' the flow of electrons (current). In doing so, they act like a radiator in a central heating system by causing electrons to give up the energy supplied by the battery.
The energy given up when current flows in the resistance is also measured as a voltage  this time the voltage is developed across the resistance and is known as a potential difference. (pd)
In other words, the voltage of the supply is a source of energy, the voltage across the resistance is energy lost.
In the LDR example, the total resistance in the path of the current is R_{fixed} + R_{LDR}
The supply voltage V_{supply} provides the pressure to drive current around the circuit. The current is limited by the resistance in the electrons path.
I_{total} = V_{supply }/ (R_{fixed} + R_{LDR}) .....................(1)
Because the current flows through a single series path, the value of that current is the same at all points in the circuit.
The energy given up when the current passes through the LDR is measured in volts. And this is simply the current I_{total} x R_{LDR }
In other words, V_{LDR} = I_{total} x R_{LDR}......................(2)
If the LDR resistance changes, then because the supply voltage is the same, the total current in the circuit must also change. If R_{LDR} decreases, then I_{total} must also decrease since the fixed resistor does not change. If R_{LDR} increases, then I_{total} must also increase since the fixed resistor does not change. (See eq (1))
Critically, because the total current has changed, then the energy lost across the resistances must also change:
If the LDR resistance increases, then the total current must fall (the supply voltage has not changed eq 1). And because the current is the same at all points in the circuit, the voltage developed (energy lost) across the LDR must also fall. (See eq (2))
If the LDR resistance decreases, then the total current must increase (the supply voltage has not changed eq 1). And because the current is the same at all points in the circuit, the voltage developed (energy lost) across the LDR must also increase. (See eq (2))
You must make the conceptual leap that the supply voltage is energy provided while the voltage (pd) across the resistor is energy lost.
eq 1). And because the current is the same at all points in the circuit, the voltage developed (energy lost) across the LDR must also fall. (See eq (2))
When the LDR resistance increases, I understand that I decreases because I=V/R, and V is the supply voltage and R is the total resistance in that equation. For the voltage across the LDR, I know that is given by V=IR. So why does the voltage across the LDR fall/decrease just because the current is the same everywhere in the circuit?
Can you explain that to me please?
EDIT: Ohh, do you mean because for both resistors (LDR and fixed), the value of I will be the same. So in V=IR for both resistors, the highest resistor size will have the most voltage across it, and the smaller resistor would have a smaller voltage through it? Is that the right reason of explaining it? If that's not correct, please explain why it is. Thanks!Last edited by blobbybill; 27012017 at 14:06.
Reply
Submit reply
Related discussions:
 AQA Physics Unit 1 PHYA1 20th May 2013
 AQA Physics Unit 1 PHYA1 20th May 2014 OFFICIAL
 AS Physics PHYA1  17/05/12
 OCR Gateway Physics P4P5P6  18/06/13
 Physics: OCR A G482 Electrons, Waves and Photons  4/06 ...
 OCR A G482 9th June 2014
 AQA Physics: PHYA1 Tuesday 19th May 2015 (AM) and ...
 The Physics ASLevel Thread
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
This forum is supported by:
 charco
 Mr M
 TSR Moderator
 Nirgilis
 usycool1
 Changing Skies
 James A
 rayquaza17
 Notnek
 RDKGames
 randdom
 davros
 Gingerbread101
 Kvothe the Arcane
 The Financier
 The Empire Odyssey
 Protostar
 TheConfusedMedic
 nisha.sri
 Reality Check
 claireestelle
 Doonesbury
 furryface12
 Amefish
 harryleavey
 Lemur14
 brainzistheword
 Rexar
 Sonechka
 LeCroissant
 EstelOfTheEyrie
 CoffeeAndPolitics
 an_atheist
 Moltenmo
Updated: January 29, 2017
Share this discussion:
Tweet