OCR B G491, 17th May 2012 Morning.
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i know total current entering a junction = total current leaving it, but does it just split equally in a parallel circuit? or is it the same for each 'arm' of the circuit?
Oh sorry I got confused my self, V stays constant, current and resistance are the ones that change
Oh sorry I got confused my self, V stays constant, current and resistance are the ones that change
(edited 11 years ago)
b) there are 70 pixels per mm on the CCD
i) calculate the number of pixels across the image of the 67 mm diameter ball [1]. << I think that's a trick question as it gives the 67 people would think to use that but it asks for the image so I would use the .4mm from the earlier question
ii) calculate the least distance te ball must move sideways for is image to move one pixel [1]
c) the position of the ball on 2 consecutive images can be used to determine the distance it has moved between images. Two such images give a value do the distance moved of 0.080m. State and explain the max possible value for this measurement based on your answer to bii.
i) calculate the number of pixels across the image of the 67 mm diameter ball [1]. << I think that's a trick question as it gives the 67 people would think to use that but it asks for the image so I would use the .4mm from the earlier question
ii) calculate the least distance te ball must move sideways for is image to move one pixel [1]
c) the position of the ball on 2 consecutive images can be used to determine the distance it has moved between images. Two such images give a value do the distance moved of 0.080m. State and explain the max possible value for this measurement based on your answer to bii.
@rrfc can you post q 11 please
Original post by KoalaKim
I know exactly what you mean!!! The only good thing is that the grade boundaries are normally quite low 
much lower than chem A anyway!!!
much lower than chem A anyway!!!Anyone else seen the grade boundaries?
For the grade boundaries on the OCR website, I worked out the average % needed for an A in both papers:
paper 1 (tomorrow) : 61%
paper 2 (next week) : 67%
The percentages vary by around 5% from 60%. (55% to 75%)
Low, considering the paper can be seen as "easy".
sure thing, found some of these last ones the easiest...
11 This question is about electrical light fittings and plugs and the materials used in their
construction.
(a)
(i) A 12 V halogen lamp is rated at 25 W.
Calculate the operating current for the lamp.
current = ...................................................... A [1]
(b) Explain in terms of their microscopic structure why metals are good electrical conductors [3]
(ii) Calculate the conductance of the lamp.
conductance = ...................................................... S [1]
c)
The pins of a 12 V halogen lamp slot into metal sockets held by ceramic fittings, but plastic
plugs with metal contacts are used to connect appliances to the mains.
(i) Ceramics and plastics are suitable materials for lamp fittings and plugs because of their
good electrical insulation properties.
Suggest a reason why, unlike metals, ceramics and plastics are good insulators.
[1]
(ii) Suggest and explain in terms of their material properties why ceramics are preferred to
plastics for halogen lamp fittings.
[2]
(iii) Suggest a reason, in terms of their material properties, why plastics are preferred to
ceramics for electric plugs.
[1]
anyone to answer all would be grateful, and thats the paper finished
11 This question is about electrical light fittings and plugs and the materials used in their
construction.
(a)
(i) A 12 V halogen lamp is rated at 25 W.
Calculate the operating current for the lamp.
current = ...................................................... A [1]
(b) Explain in terms of their microscopic structure why metals are good electrical conductors [3]
(ii) Calculate the conductance of the lamp.
conductance = ...................................................... S [1]
c)
The pins of a 12 V halogen lamp slot into metal sockets held by ceramic fittings, but plastic
plugs with metal contacts are used to connect appliances to the mains.
(i) Ceramics and plastics are suitable materials for lamp fittings and plugs because of their
good electrical insulation properties.
Suggest a reason why, unlike metals, ceramics and plastics are good insulators.
[1]
(ii) Suggest and explain in terms of their material properties why ceramics are preferred to
plastics for halogen lamp fittings.
[2]
(iii) Suggest a reason, in terms of their material properties, why plastics are preferred to
ceramics for electric plugs.
[1]
anyone to answer all would be grateful,
and thats the paper finishedOriginal post by ser00
i know total current entering a junction = total current leaving it, but does it just split equally in a parallel circuit? or is it the same for each 'arm' of the circuit?
i know total current entering a junction = total current leaving it, but does it just split equally in a parallel circuit? or is it the same for each 'arm' of the circuit?
I know this has already been explained but hopefully I will be able to explain it more from first principles so that you understand it better.
(Using the diagram I uploaded above BTW).
The first thing to note is that:
The voltage of the cell = The voltage across the first resistor = The voltage across the second resistor.
This is because each coulomb of charge enters each junction with the same energy and leaves its junction with the same (but different to the starting value) energy.
Knowing that Voltage is constant across each 'arm', you can now use the formula:
I=V/R
to find the current at each arm and before the junction.
It is easier to find it for each arm before finding the current before.
So the current flowing in the first arm (in that diagram) is equal to V/R1 (R1 is the resistance of the first resistor - forgot to label it! V is the voltage of the cell).
The current flowing in the second arm is equal to V/R0 (same as above).
As you said above, the current before the junction is the same as the current after, so to find the current before, you just add these results up. So to address your question, neither of the scenarios you described above are (necessarily) correct. In order to work out the current in any 'arm' you can either do V/R or just subtract the currents in the other arms from the current before the junction, dependin on what you know.
BTW you can use the fact that current before junction = current after to derive how resistance behaves in parallel (just in case you forget in the exam). You know that:
V/Rtotal = V/R1 + V/R0
V is constant, therefore:
1/Rtotal = 1/R1 + 1/R0
You know that conductance is the opposite of resistance, therefore:
Gtotal = G1 + G2
So, in parallel, conductances add up.
Original post by ser00
why does a voltmeter have to be high resistance? wouldnt that just take all the voltage and stop the circuit working?
why does a voltmeter have to be high resistance? wouldnt that just take all the voltage and stop the circuit working?
As explained above (briefly and not very well, let me know if you want me to clarify), voltage is the same in each 'arm' of a parallel circuit. Voltmeters are connected in parallel. Therefore having a high resistance affects the voltage across the voltmeter in a different way (a beneficial way) to what you suggested (although you would be correct if it was connected in series, which is why they are connected in parallel!).
The voltage in each arm of that part of the circuit (by that I mean the voltage across the voltmeter and across the component) is given by:
V = IR (where I is the current before the junction, or the total current of the mini-parallel circuit and R is the total resistance of the mini-parallel circuit).
In parallel, conductances add up, therefore:
V = I * 1/(1/Rcomponent +1/Rvoltmeter)
Therefore:
V= I/(1/Rcomponent + 1/Rvoltmeter)
As the resistance of the voltmeter is really high, its reciprocol may as well be 0, therefore:
V~~I/(1/Rcomponent +0) (can't do 'roughly equals' on computer)
This simplifies to give:
V~~IRcomponent
Therefore, when the resistance of the voltmeter is high, the voltage across it is roughly the same as the voltage across the component would be if it didn't have a voltmeter attatched to it. Therefore, having a high resistance ensures that it has the smallest possible effect on the voltage it reads (if that makes sense).
I hope you can see that if Rvoltmeter was much lower then its reciprocol would be much higher, therefore it would read a voltage lower than what it should be.
I will answer your other questions after. Need to do a past paper!
Anyway let me know if I'm not explaining it well enough, I will try to simplify it. Also, I can provide a better proof that voltage stays the same in parallel circuits if you like.
Original post by hihihihihi
i know total current entering a junction = total current leaving it, but does it just split equally in a parallel circuit? or is it the same for each 'arm' of the circuit?
I didn't write it clearly enough but yes current is equally split because of that equation
I didn't write it clearly enough but yes current is equally split because of that equation
What?
b) there are 70 pixels per mm on the CCD
i) calculate the number of pixels across the image of the 67 mm diameter ball [1]. << I think that's a trick question as it gives the 67 people would think to use that but it asks for the image so I would use the .4mm from the earlier question
70x 0.37 = 26 pixels
ii) calculate the least distance te ball must move sideways for is image to move one pixel [1]
3.7x10^-4/26 =1.42x10^-6m
c) the position of the ball on 2 consecutive images can be used to determine the distance it has moved between images. Two such images give a value do the distance moved of 0.080m. State and explain the max possible value for this measurement based on your answer to bii.
i) calculate the number of pixels across the image of the 67 mm diameter ball [1]. << I think that's a trick question as it gives the 67 people would think to use that but it asks for the image so I would use the .4mm from the earlier question
70x 0.37 = 26 pixels
ii) calculate the least distance te ball must move sideways for is image to move one pixel [1]
3.7x10^-4/26 =1.42x10^-6m
c) the position of the ball on 2 consecutive images can be used to determine the distance it has moved between images. Two such images give a value do the distance moved of 0.080m. State and explain the max possible value for this measurement based on your answer to bii.
Original post by When you see it...
What?
current is split equally in parallel circuits
Original post by hihihihihi
current is split equally in parallel circuits
Not necessarily. I've explained it above. To summarise:
I = V/R
and V is constant.
Therefore the 'arms' with higher resistances take lower proportions of the total current.
oh yea your right and i cant answer this question:
c) the position of the ball on 2 consecutive images can be used to determine the distance it has moved between images. Two such images give a value do the distance moved of 0.080m. State and explain the max possible value for this measurement based on your answer to bii.
c) the position of the ball on 2 consecutive images can be used to determine the distance it has moved between images. Two such images give a value do the distance moved of 0.080m. State and explain the max possible value for this measurement based on your answer to bii.
Original post by When you see it...
I know this has already been explained but hopefully I will be able to explain it more from first principles so that you understand it better.
(Using the diagram I uploaded above BTW).
The first thing to note is that:
The voltage of the cell = The voltage across the first resistor = The voltage across the second resistor.
This is because each coulomb of charge enters each junction with the same energy and leaves its junction with the same (but different to the starting value) energy.
Knowing that Voltage is constant across each 'arm', you can now use the formula:
I=V/R
to find the current at each arm and before the junction.
It is easier to find it for each arm before finding the current before.
So the current flowing in the first arm (in that diagram) is equal to V/R1 (R1 is the resistance of the first resistor - forgot to label it! V is the voltage of the cell).
The current flowing in the second arm is equal to V/R0 (same as above).
As you said above, the current before the junction is the same as the current after, so to find the current before, you just add these results up. So to address your question, neither of the scenarios you described above are (necessarily) correct. In order to work out the current in any 'arm' you can either do V/R or just subtract the currents in the other arms from the current before the junction, dependin on what you know.
BTW you can use the fact that current before junction = current after to derive how resistance behaves in parallel (just in case you forget in the exam). You know that:
V/Rtotal = V/R1 + V/R0
V is constant, therefore:
1/Rtotal = 1/R1 + 1/R0
You know that conductance is the opposite of resistance, therefore:
Gtotal = G1 + G2
So, in parallel, conductances add up.
As explained above (briefly and not very well, let me know if you want me to clarify), voltage is the same in each 'arm' of a parallel circuit. Voltmeters are connected in parallel. Therefore having a high resistance affects the voltage across the voltmeter in a different way (a beneficial way) to what you suggested (although you would be correct if it was connected in series, which is why they are connected in parallel!).
The voltage in each arm of that part of the circuit (by that I mean the voltage across the voltmeter and across the component) is given by:
V = IR (where I is the current before the junction, or the total current of the mini-parallel circuit and R is the total resistance of the mini-parallel circuit).
In parallel, conductances add up, therefore:
V = I * 1/(1/Rcomponent +1/Rvoltmeter)
Therefore:
V= I/(1/Rcomponent + 1/Rvoltmeter)
As the resistance of the voltmeter is really high, its reciprocol may as well be 0, therefore:
V~~I/(1/Rcomponent +0) (can't do 'roughly equals' on computer)
This simplifies to give:
V~~IRcomponent
Therefore, when the resistance of the voltmeter is high, the voltage across it is roughly the same as the voltage across the component would be if it didn't have a voltmeter attatched to it. Therefore, having a high resistance ensures that it has the smallest possible effect on the voltage it reads (if that makes sense).
I hope you can see that if Rvoltmeter was much lower then its reciprocol would be much higher, therefore it would read a voltage lower than what it should be.
I will answer your other questions after. Need to do a past paper!
Anyway let me know if I'm not explaining it well enough, I will try to simplify it. Also, I can provide a better proof that voltage stays the same in parallel circuits if you like.
(Using the diagram I uploaded above BTW).
The first thing to note is that:
The voltage of the cell = The voltage across the first resistor = The voltage across the second resistor.
This is because each coulomb of charge enters each junction with the same energy and leaves its junction with the same (but different to the starting value) energy.
Knowing that Voltage is constant across each 'arm', you can now use the formula:
I=V/R
to find the current at each arm and before the junction.
It is easier to find it for each arm before finding the current before.
So the current flowing in the first arm (in that diagram) is equal to V/R1 (R1 is the resistance of the first resistor - forgot to label it! V is the voltage of the cell).
The current flowing in the second arm is equal to V/R0 (same as above).
As you said above, the current before the junction is the same as the current after, so to find the current before, you just add these results up. So to address your question, neither of the scenarios you described above are (necessarily) correct. In order to work out the current in any 'arm' you can either do V/R or just subtract the currents in the other arms from the current before the junction, dependin on what you know.
BTW you can use the fact that current before junction = current after to derive how resistance behaves in parallel (just in case you forget in the exam). You know that:
V/Rtotal = V/R1 + V/R0
V is constant, therefore:
1/Rtotal = 1/R1 + 1/R0
You know that conductance is the opposite of resistance, therefore:
Gtotal = G1 + G2
So, in parallel, conductances add up.
As explained above (briefly and not very well, let me know if you want me to clarify), voltage is the same in each 'arm' of a parallel circuit. Voltmeters are connected in parallel. Therefore having a high resistance affects the voltage across the voltmeter in a different way (a beneficial way) to what you suggested (although you would be correct if it was connected in series, which is why they are connected in parallel!).
The voltage in each arm of that part of the circuit (by that I mean the voltage across the voltmeter and across the component) is given by:
V = IR (where I is the current before the junction, or the total current of the mini-parallel circuit and R is the total resistance of the mini-parallel circuit).
In parallel, conductances add up, therefore:
V = I * 1/(1/Rcomponent +1/Rvoltmeter)
Therefore:
V= I/(1/Rcomponent + 1/Rvoltmeter)
As the resistance of the voltmeter is really high, its reciprocol may as well be 0, therefore:
V~~I/(1/Rcomponent +0) (can't do 'roughly equals' on computer)
This simplifies to give:
V~~IRcomponent
Therefore, when the resistance of the voltmeter is high, the voltage across it is roughly the same as the voltage across the component would be if it didn't have a voltmeter attatched to it. Therefore, having a high resistance ensures that it has the smallest possible effect on the voltage it reads (if that makes sense).
I hope you can see that if Rvoltmeter was much lower then its reciprocol would be much higher, therefore it would read a voltage lower than what it should be.
I will answer your other questions after. Need to do a past paper!
Anyway let me know if I'm not explaining it well enough, I will try to simplify it. Also, I can provide a better proof that voltage stays the same in parallel circuits if you like.
wow your amazing. thank you so much - i think i sort of get it (as well as im ever going to anyway haha!)...good luck tomorrow!! (although you clearly dont need it since your a genius
) thanks hihihi as well <310) some crap about tennis balls and cameras...
a) a tennis ball has a diameter of 67mm. When the ball is 10m from the camera, a sharp image is formed on a CCD 55mm behind the lens.
i) calculate the magnification of the image [1]
m= v/u, so 55x10^-3/67x10^-3 = 0.82
ii) show that the power of the lens in the fixed focus camera is about 18 D [2]
1/v=1/u+1/f so 1/f=1/v-1/u, 1/f= 1/55x10^-3 -(1/-10) = 18.3D
iii) show that the diameter of the image of the ball on the CCD is about 0.4mm. Make your method clear [2]
i don't know the exact method for this but i did (55x10^-3/10)x66x10^-3 = 3.69x10^-4 = 0.37mm
b) there are 70 pixels per mm on the CCD
i) calculate the number of pixels across the image of the 67 mm diameter ball [1]. << I think that's a trick question as it gives the 67 people would think to use that but it asks for the image so I would use the .4mm from the earlier question
70x 0.37 = 26 pixels
ii) calculate the least distance the ball must move sideways for the image to move one pixel [1]
67x10^-3/26=2.58x10^-3m
c) the position of the ball on 2 consecutive images can be used to determine the distance it has moved between images. Two such images give a value do the distance moved of 0.080m. State and explain the max possible value for this measurement based on your answer to bii.
0.080+ (2x 2.58x10^-3) = 0.0852 m
= 0.085m (2s.f.)
a) a tennis ball has a diameter of 67mm. When the ball is 10m from the camera, a sharp image is formed on a CCD 55mm behind the lens.
i) calculate the magnification of the image [1]
m= v/u, so 55x10^-3/67x10^-3 = 0.82
ii) show that the power of the lens in the fixed focus camera is about 18 D [2]
1/v=1/u+1/f so 1/f=1/v-1/u, 1/f= 1/55x10^-3 -(1/-10) = 18.3D
iii) show that the diameter of the image of the ball on the CCD is about 0.4mm. Make your method clear [2]
i don't know the exact method for this but i did (55x10^-3/10)x66x10^-3 = 3.69x10^-4 = 0.37mm
b) there are 70 pixels per mm on the CCD
i) calculate the number of pixels across the image of the 67 mm diameter ball [1]. << I think that's a trick question as it gives the 67 people would think to use that but it asks for the image so I would use the .4mm from the earlier question
70x 0.37 = 26 pixels
ii) calculate the least distance the ball must move sideways for the image to move one pixel [1]
67x10^-3/26=2.58x10^-3m
c) the position of the ball on 2 consecutive images can be used to determine the distance it has moved between images. Two such images give a value do the distance moved of 0.080m. State and explain the max possible value for this measurement based on your answer to bii.
0.080+ (2x 2.58x10^-3) = 0.0852 m
= 0.085m (2s.f.)
11 This question is about electrical light fittings and plugs and the materials used in their
construction.
(a)
(i) A 12 V halogen lamp is rated at 25 W.
Calculate the operating current for the lamp.
P=IV so I=P/V, I=25/12=2.08
current = ...............................2.1 (2s.f.) ...................... A [1]
(b) Explain in terms of their microscopic structure why metals are good electrical conductors [3]
Metals have positive ions in a crystalline lattice with a sea of delocalised electrons. The electrons carry charge so metals have high charge density.
(ii) Calculate the conductance of the lamp.
V=IR so R=V/I which means G=I/V , G=2.08/12=0.173 S
conductance = ................................0.17 (2s.f.) ...................... S [1]
c)
The pins of a 12 V halogen lamp slot into metal sockets held by ceramic fittings, but plastic
plugs with metal contacts are used to connect appliances to the mains.
(i) Ceramics and plastics are suitable materials for lamp fittings and plugs because of their
good electrical insulation properties.
Suggest a reason why, unlike metals, ceramics and plastics are good insulators.
[1]
Ceramics and plastics have low charge density so they don't conduct much or very little.
(ii) Suggest and explain in terms of their material properties why ceramics are preferred to
plastics for halogen lamp fittings.
[2] Plastics can melt at the temperature of the operating lamp. ceramics have high melting point compared to plastic because ceramics need more energy to break than plastic
(iii) Suggest a reason, in terms of their material properties, why plastics are preferred to
ceramics for electric plugs.
[1] They have lower charge density than ceramics
construction.
(a)
(i) A 12 V halogen lamp is rated at 25 W.
Calculate the operating current for the lamp.
P=IV so I=P/V, I=25/12=2.08
current = ...............................2.1 (2s.f.) ...................... A [1]
(b) Explain in terms of their microscopic structure why metals are good electrical conductors [3]
Metals have positive ions in a crystalline lattice with a sea of delocalised electrons. The electrons carry charge so metals have high charge density.
(ii) Calculate the conductance of the lamp.
V=IR so R=V/I which means G=I/V , G=2.08/12=0.173 S
conductance = ................................0.17 (2s.f.) ...................... S [1]
c)
The pins of a 12 V halogen lamp slot into metal sockets held by ceramic fittings, but plastic
plugs with metal contacts are used to connect appliances to the mains.
(i) Ceramics and plastics are suitable materials for lamp fittings and plugs because of their
good electrical insulation properties.
Suggest a reason why, unlike metals, ceramics and plastics are good insulators.
[1]
Ceramics and plastics have low charge density so they don't conduct much or very little.
(ii) Suggest and explain in terms of their material properties why ceramics are preferred to
plastics for halogen lamp fittings.
[2] Plastics can melt at the temperature of the operating lamp. ceramics have high melting point compared to plastic because ceramics need more energy to break than plastic
(iii) Suggest a reason, in terms of their material properties, why plastics are preferred to
ceramics for electric plugs.
[1] They have lower charge density than ceramics
Original post by matty b
In bits/bytes, do we give a Megabyte as 1 x10^6 bytes, or as 1024^2 bytes? they both exist.
this question here says:
A page contains 30 lines of text, with an average of 15 characters per line. Each character is represented by 4 bits. How many megabytes will a book containing 650 pages take up in a computer hard disk?
One answer is 14.8 (using standard form)
The Other is 14.1 (using the 1024^2)
Original post by ser00
wow your amazing. thank you so much - i think i sort of get it (as well as im ever going to anyway haha!)...good luck tomorrow!! (although you clearly dont need it since your a genius ) thanks hihihi as well <3
) thanks hihihi as well <3We all need luck for this exam beause OCR are dickheads and can't set a fair exam.
Anyway if you forget all that stuff about parallel circuits just remember that an 'arm' with a low resistance will take a high proportion of the total current in the circuit and vice versa.
Quick Reply
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