Resistance and Voltage Question A Level Watch

romina23
Badges: 9
Rep:
?
#1
Report Thread starter 2 years ago
#1
hiya, me and physics again,

I've had a look through the mark scheme but please could somebody explain to me Q3)c)ii) on http://www.ocr.org.uk/Images/175458-...nd-photons.pdf

why it decreases through the 750 Ohm resistor, surely it would increase because passes through the LDR, so more would go to the 750 one??

thank you!!!
0
reply
username1973239
Badges: 16
Rep:
?
#2
Report 2 years ago
#2
(Original post by romina23)
hiya, me and physics again,

I've had a look through the mark scheme but please could somebody explain to me Q3)c)ii) on http://www.ocr.org.uk/Images/175458-...nd-photons.pdf

why it decreases through the 750 Ohm resistor, surely it would increase because passes through the LDR, so more would go to the 750 one??

thank you!!!
An LDR is a semiconductor is the first point I would make. You need to explain how light intensity increases/decreases conductivity of the LDR.
0
reply
username1973239
Badges: 16
Rep:
?
#3
Report 2 years ago
#3
(Original post by romina23)
hiya, me and physics again,

I've had a look through the mark scheme but please could somebody explain to me Q3)c)ii) on http://www.ocr.org.uk/Images/175458-...nd-photons.pdf

why it decreases through the 750 Ohm resistor, surely it would increase because passes through the LDR, so more would go to the 750 one??

thank you!!!
It's a potential divider circuit as well so think about ratios.
1
reply
romina23
Badges: 9
Rep:
?
#4
Report Thread starter 2 years ago
#4
(Original post by TSlayerr)
It's a potential divider circuit as well so think about ratios.
ah true, so with potential dividers is that involving 1/Rtotal = 1/R1 + 1/R2 etc, and if so then where from there?

((thank you btw!!))
0
reply
username1973239
Badges: 16
Rep:
?
#5
Report 2 years ago
#5
(Original post by romina23)
ah true, so with potential dividers is that involving 1/Rtotal = 1/R1 + 1/R2 etc, and if so then where from there?

((thank you btw!!))
I would use Voltage out = (Resistance2)/(Total resistance) x Voltage In
0
reply
username1973239
Badges: 16
Rep:
?
#6
Report 2 years ago
#6
(Original post by romina23)
ah true, so with potential dividers is that involving 1/Rtotal = 1/R1 + 1/R2 etc, and if so then where from there?

((thank you btw!!))
As it's a semiconductor, greater light intensity = greater energy for vibrations of atoms which free more electrons.
0
reply
romina23
Badges: 9
Rep:
?
#7
Report Thread starter 2 years ago
#7
(Original post by TSlayerr)
I would use Voltage out = (Resistance2)/(Total resistance) x Voltage In

ok cool, so if you used that then R2 = 400 and total would = 6000/23, or is this for the series part of the circuit??

for the second part of the reply, that means more current can flow if there are more free electrons, right?
0
reply
RogerOxon
Badges: 21
Rep:
?
#8
Report 2 years ago
#8
(Original post by romina23)
I've had a look through the mark scheme but please could somebody explain to me Q3)c)ii) on http://www.ocr.org.uk/Images/175458-...nd-photons.pdf

why it decreases through the 750 Ohm resistor, surely it would increase because passes through the LDR, so more would go to the 750 one??
As the light intensity on the LDR increases, its resistance falls. Therefore, the resistance of the 750 Ohm + LDR (in parallel) section decreases.

Consider the LDR+750 Ohm as a sinlge (decreasing resistance) in series with R_1. We see that less of the 45V drop will occur over the 750 Ohm + LDR section, as its resistance is decreasing.
0
reply
RogerOxon
Badges: 21
Rep:
?
#9
Report 2 years ago
#9
P.S. The question does say 'without calculation', so you just have to explain how the voltage divided between the (essentially) two resistances changes.
1
reply
username1973239
Badges: 16
Rep:
?
#10
Report 2 years ago
#10
(Original post by romina23)
ok cool, so if you used that then R2 = 400 and total would = 6000/23, or is this for the series part of the circuit??

for the second part of the reply, that means more current can flow if there are more free electrons, right?
More free electrons = lower resistance.
As the LDR has lower resistance, the ratio changes. The 750 ohm resistor now takes a LARGER chunk of the voltage than before.
0
reply
romina23
Badges: 9
Rep:
?
#11
Report Thread starter 2 years ago
#11
(Original post by RogerOxon)
As the light intensity on the LDR increases, its resistance falls. Therefore, the resistance of the 750 Ohm + LDR (in parallel) section decreases.

Consider the LDR+750 Ohm as a sinlge (decreasing resistance) in series with R_1. We see that less of the 45V drop will occur over the 750 Ohm + LDR section, as its resistance is decreasing.
ahh ok thank youu!! so the voltage R_1 resistor stays the same (because it's resistance remains constant) meaning the parallel loop voltage falls, so both the 750 ohm resistor and LDR fall in voltage also?
0
reply
romina23
Badges: 9
Rep:
?
#12
Report Thread starter 2 years ago
#12
(Original post by TSlayerr)
More free electrons = lower resistance.
As the LDR has lower resistance, the ratio changes. The 750 ohm resistor now takes a LARGER chunk of the voltage than before.
ok, so the decrease in total resistance means both component's resistances decrease but the more voltage still passes through the resistor than the LDR?
0
reply
username1973239
Badges: 16
Rep:
?
#13
Report 2 years ago
#13
(Original post by romina23)
ok cool, so if you used that then R2 = 400 and total would = 6000/23, or is this for the series part of the circuit??

for the second part of the reply, that means more current can flow if there are more free electrons, right?
Remember that this is a series circuit. Current is the same throughout. If a resistor has a higher resistance than another, a larger voltage is needed to maintain the same current throughout

as I=V/R
0
reply
RogerOxon
Badges: 21
Rep:
?
#14
Report 2 years ago
#14
(Original post by romina23)
ahh ok thank youu!! so the voltage R_1 resistor stays the same (because it's resistance remains constant) meaning the parallel loop voltage falls, so both the 750 ohm resistor and LDR fall in voltage also?
You have R_1 in series with the combination of 750 Ohm and the LDR (which are in parallel) - let's call that R_2.

As R_2 reduces in resistance (as the LDR does), more current flows in the circuit. As R_1 doesn't change, it accounts for more of the voltage drop than R_2.

Another way of thinking about it is that R_1 and R_2 form a voltage divider. As R_2 reduces, it takes less of the voltage drop.
0
reply
RogerOxon
Badges: 21
Rep:
?
#15
Report 2 years ago
#15
(Original post by romina23)
ok, so the decrease in total resistance means both component's resistances decrease but the more voltage still passes through the resistor than the LDR?
Voltage doesn't "pass through" anything - it's across. It is the force that pushes a current through a resistance.
0
reply
username1973239
Badges: 16
Rep:
?
#16
Report 2 years ago
#16
(Original post by romina23)
ok, so the decrease in total resistance means both component's resistances decrease but the more voltage still passes through the resistor than the LDR?
The LDR resistor's resistance has decreased because of higher light intensity.
The fixed resistor has the same resistance as it is a 'fixed' resistor.

Overall the resistance has decrease, but this isn't really the point.

The point is that the ratio has changed.
0
reply
username1973239
Badges: 16
Rep:
?
#17
Report 2 years ago
#17
(Original post by romina23)
ok, so the decrease in total resistance means both component's resistances decrease but the more voltage still passes through the resistor than the LDR?
Eg. Ratio before = LDR 1500 : 750 Fixed resistor
Ratio after = LDR 400 : 750 Fixed resistor

Voltage out = (Fixed resistor resistance/Total resistance) x Voltage in
0
reply
RogerOxon
Badges: 21
Rep:
?
#18
Report 2 years ago
#18
(Original post by romina23)
ok, so the decrease in total resistance means both component's resistances decrease but the more voltage still passes through the resistor than the LDR?
This statement makes me think that electricity hasn't been explained very well to you, so I'll try.

Think of the circuit as plumbing. The battery is a pump - it pushes the water around the circuit. The resistors are valves that are more or less closed, resisting the flow of the water.

The LDR is a valve that is effectively being opened by the light - it's resistance decreases, so the flow of water through it will increase.

The voltage difference between any two points can be thought of as the pressure difference. It's always a difference - not an absolute value. By convention (unless otherwise stated) it's measured between the negative terminal of the battery and the point of interest.

The current is the rate at which water is flowing.

By opening the LDR valve, more water (current) flows through the pump (battery), as the total resistance has decreased. Less pressure (voltage) is needed to push water (current) through the parallel combination of the LDR and 750 Ohm resistor, so more of the pump's pressure (voltage) is dropped over R_1.
0
reply
romina23
Badges: 9
Rep:
?
#19
Report Thread starter 2 years ago
#19
(Original post by RogerOxon)
You have R_1 in series with the combination of 750 Ohm and the LDR (which are in parallel) - let's call that R_2.

As R_2 reduces in resistance (as the LDR does), more current flows in the circuit. As R_1 doesn't change, it accounts for more of the voltage drop than R_2.

Another way of thinking about it is that R_1 and R_2 form a voltage divider. As R_2 reduces, it takes less of the voltage drop.
ahh i see thank you, I think I'm thinking of voltage too much like current and that's confusing me, but thank you so much!
0
reply
romina23
Badges: 9
Rep:
?
#20
Report Thread starter 2 years ago
#20
(Original post by RogerOxon)
This statement makes me think that electricity hasn't been explained very well to you, so I'll try.

Think of the circuit as plumbing. The battery is a pump - it pushes the water around the circuit. The resistors are valves that are more or less closed, resisting the flow of the water.

The LDR is a valve that is effectively being opened by the light - it's resistance decreases, so the flow of water through it will increase.

The voltage difference between any two points can be thought of as the pressure difference. It's always a difference - not an absolute value. By convention (unless otherwise stated) it's measured between the negative terminal of the battery and the point of interest.

The current is the rate at which water is flowing.

By opening the LDR valve, more water (current) flows through the pump (battery), as the total resistance has decreased. Less pressure (voltage) is needed to push water (current) through the parallel combination of the LDR and 750 Ohm resistor, so more of the pump's pressure (voltage) is dropped over R_1.
this is an amazing way off putting it thank you so much
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • University of East Anglia
    PGCE Open day Postgraduate
    Tue, 3 Mar '20
  • University of Bradford
    Postgraduate Open day/Evening Postgraduate
    Tue, 3 Mar '20
  • Queen's University Belfast
    Postgraduate LIVE Masters & PhD Study Fair Postgraduate
    Wed, 4 Mar '20

Do you get study leave?

Yes- I like it (511)
59.77%
Yes- I don't like it (44)
5.15%
No- I want it (242)
28.3%
No- I don't want it (58)
6.78%

Watched Threads

View All
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise