Simple circuit question involving an LDR and a fixed resistor - I have a question.Watch

#1
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
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1 year ago
#2
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
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#3
(Original post by Lauren-x-)
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
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.
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1 year ago
#4
(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.
If the resistance of the LDR changes, R for the LDR isn't constant
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#5
(Original post by h3rmit)
If the resistance of the LDR changes, R for the LDR isn't constant
i get that, I know why the resistor voltage decreases (as V=IR, and R decreases, so V also decreases). It's the fixed resistor that I am confused about, as R doesn't change for the fixed resistor.
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1 year ago
#6
(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?
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.
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#7
(Original post by Lauren-x-)
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.
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?

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1 year ago
#8
(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?

You'd get free force if the total voltage increased without the power supply emf increasing.
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1 year ago
#9
(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?

(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 VLDR + Vresistor = 6V.

Or you can think of it like 6V - Vresistor = VLDR

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!!
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1 year ago
#10
(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?

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
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#11
(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
Will do. I'd need to work out new values of I (the current) when I decrease the LDR resistance wouldn't I?

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?
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1 year ago
#12
(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
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
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1 year ago
#13
(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?
Absolutely!

you'll do I=V/(Rfixed+RLDR) each time (resistances are added because they're in series)

then

VLDR=I x RLDR

then on to the next value of LDR resistance and calculate the current again.
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#14
(Original post by Joinedup)
Absolutely!

you'll do I=V/(Rfixed+RLDR) each time (resistances are added because they're in series)

then

VLDR=I x RLDR

then on to the next value of LDR resistance and calculate the current again.
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?
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1 year ago
#15
(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?
LDR resistance going up means current goes down - don't try thinking about what happens if you vary everything at the same time, just vary one thing at once and calculate the effect.

Battery fixed... fixed resistor fixed... LDR variable
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1 year ago
#16
(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?

Thanks
The pd across the fixed resistor would increase because both of the resistors pd or voltage has to equal that of the battery.
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#17
(Original post by Joinedup)
Absolutely!

you'll do I=V/(Rfixed+RLDR) each time (resistances are added because they're in series)

then

VLDR=I x RLDR

then on to the next value of LDR resistance and calculate the current again.
I've done it about 15 times now, and you are right. With V=IR, when R decreases, V also decreases.

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?
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#18
(Original post by Lauren-x-)
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
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?

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?
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1 year ago
#19
(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?
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 Rfixed + RLDR

The supply voltage Vsupply provides the pressure to drive current around the circuit. The current is limited by the resistance in the electrons path.

Itotal = Vsupply / (Rfixed + RLDR) .....................(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 Itotal x RLDR

In other words, VLDR = Itotal x RLDR......................(2)

If the LDR resistance changes, then because the supply voltage is the same, the total current in the circuit must also change. If RLDR decreases, then Itotal must increase since the fixed resistor does not change and the total resistance as fallen. If RLDR increases, then Itotal 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.
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#20
(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 Rfixed + RLDR

The supply voltage Vsupply provides the pressure to drive current around the circuit. The current is limited by the resistance in the electrons path.

Itotal = Vsupply / (Rfixed + RLDR) .....................(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 Itotal x RLDR

In other words, VLDR = Itotal x RLDR......................(2)

If the LDR resistance changes, then because the supply voltage is the same, the total current in the circuit must also change. If RLDR decreases, then Itotal must also decrease since the fixed resistor does not change. If RLDR increases, then Itotal 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.
It's this bit that confuses me, I get the rest of it now, thanks.
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))
I understand that the current is the same at all points in the circuit, that is because it is a series circuit. However, why must the voltage across the LDR fall because the current is the same everywhere in the circuit?

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!
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