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AQA Physics PHYA4 - Thursday 11th June 2015 [Exam Discussion Thread]

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Could someone please help me with question 6b on this paper?
http://filestore.aqa.org.uk/subjects/AQA-PHYA1-QP-JAN13.PDF
Thanks in advance
Original post by ubisoft


I got C:
By using C=Q/V rearranged to V=Q/C you can see that V is proportional to Q, and if Q becomes 1.5Q, V also becomes 1.5V.
With E=(Q^2)/2C you can see that E is proportional to Q^2, so (1.5Q)^2=2.25V
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Original post by ubisoft
I mean like ums total out of both sections? on average. I got 110 on June 2011 lol but I fluked it badly, got lots of educated guesses correct on MC, and that one had a lot of mechanics which I'm good at. I'm getting around 20 on mc now, and about 35-40 on written. That's about 100-108 ums / 120 for most papers


That's still really good! I'm not sure what I'm getting in terms of UMS. I'll just do the paper and revise from my mistakes really.

I believe I need around 110 on both papers for an A* which I think is unrealistic. I need 207 overall for an A though.


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Original post by l1lvink
I got C:
By using C=Q/V rearranged to V=Q/C you can see that V is proportional to Q, and if Q becomes 1.5Q, V also becomes 1.5V.
With E=(Q^2)/2C you can see that E is proportional to Q^2, so (1.5Q)^2=2.25V


When I did the paper yesterday I also got that answer with the same reasoning :smile:


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Original post by l1lvink
Could someone please help me with question 6b on this paper?
http://filestore.aqa.org.uk/subjects/AQA-PHYA1-QP-JAN13.PDF
Thanks in advance


Been a while since I did unit 1, but:

Using the fact that the PD is split into the same ratio as the resistances of the components in each branch (as they act as potential dividers):

The PD across each branch is 12V as the two branches are in parallel.

Between A-C, there is a
Unparseable latex formula:

20k \Ohm

resistor, which accounts for 12\frac{1}{2} the total resistance (and therefore half the PD) of that branch.

122\frac{12}{2} gives 6.0V6.0V across A-C.

Between D-F, there is a
Unparseable latex formula:

5.0k \Ohm

thermistor, which accounts for 13\frac{1}{3} the total resistance (and therefore a third of the PD) of that branch.

123\frac{12}{3} gives 4.0V4.0V across D-F.

Between C-D, the first branch has a PD of 6.0V6.0V at C and the second branch has a PD of 4.0V4.0V at D.

Hence, 6.0V−4.0V=2.0V6.0V - 4.0V = 2.0V between C-D.

Does that help?


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Original post by l1lvink
I got C:
By using C=Q/V rearranged to V=Q/C you can see that V is proportional to Q, and if Q becomes 1.5Q, V also becomes 1.5V.
With E=(Q^2)/2C you can see that E is proportional to Q^2, so (1.5Q)^2=2.25V


I don't understand which energy equation to use? some equations say E is prop. to V and some say V^2, same with Q and Q^2, it's confusing me
Original post by CD223
That's still really good! I'm not sure what I'm getting in terms of UMS. I'll just do the paper and revise from my mistakes really.

I believe I need around 110 on both papers for an A* which I think is unrealistic. I need 207 overall for an A though.


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110 is not too crazy, if I got that then you could too, you know more physics than me :biggrin: I need 114 on each paper though, so that's a bit too much
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Original post by ubisoft
I don't understand which energy equation to use? some equations say E is prop. to V and some say V^2, same with Q and Q^2, it's confusing me


[br]Q=CV[br][br]Q=CV[br]

and

[br]E=12QV[br]⇒E=12CV2[br]⇒E=12Q2C[br][br]E=\dfrac{1}{2}QV[br]\Rightarrow E=\dfrac{1}{2}CV^{2}[br]\Rightarrow E=\dfrac{1}{2}\dfrac{Q^2}{C}[br]

As such, if V becomes 1.5V, E becomes
[br](1.5)2E[br]⇒2.25E[br][br](1.5)^{2}E[br]\Rightarrow 2.25E[br]


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Original post by ubisoft
110 is not too crazy, if I got that then you could too, you know more physics than me :biggrin: I need 114 on each paper though, so that's a bit too much


I really don't know more physics :wink:


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Original post by CD223
[br]Q=CV[br][br]Q=CV[br]

and

[br]E=12QV[br]⇒E=12CV2[br]⇒E=12Q2C[br][br]E=\dfrac{1}{2}QV[br]\Rightarrow E=\dfrac{1}{2}CV^{2}[br]\Rightarrow E=\dfrac{1}{2}\dfrac{Q^2}{C}[br]

As such, if V becomes 1.5V, E becomes
[br](1.5)2E[br]⇒2.25E[br][br](1.5)^{2}E[br]\Rightarrow 2.25E[br]


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So is E prop. to V or V^2? What's the 'original' equation if that makes sense?
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Original post by ubisoft
So is E prop. to V or V^2? What's the 'original' equation if that makes sense?


Well, as above,

[br]E=12QV[br]⇒E=12CV2[br]⇒E=12Q2C[br][br]E=\dfrac{1}{2}QV[br]\Rightarrow E=\dfrac{1}{2}CV^{2}[br]\Rightarrow E=\dfrac{1}{2}\dfrac{Q^2}{C}[br]

The top one is provided on the formulae sheet I believe.

But as Q=CVQ=CV,

[br]EαV2[br][br]E \alpha V^2[br]


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Original post by ubisoft
I don't understand which energy equation to use? some equations say E is prop. to V and some say V^2, same with Q and Q^2, it's confusing me


Ok, so you've got 3 equations (0.5QV, 0.5CV^2 and 0.5(Q^2/C). Just for simplicity I'll refer to them as QV, CV and QC as defined by their variables. So you know that Q is the variable being increased, so the equation that we want must include Q; this removes CV. Now we also know that V will change (the 'other' half of the question is about this). Now this means that if we were using QV, you would have two variables that are changing (both Q and V), and this would make things a heck of a lot harder. Therefore, we use QC as C remains the same, but Q changes, meaning that we can calculate the change in E.
Original post by CD223
Been a while since I did unit 1, but:

Using the fact that the PD is split into the same ratio as the resistances of the components in each branch (as they act as potential dividers):

The PD across each branch is 12V as the two branches are in parallel.

Between A-C, there is a
Unparseable latex formula:

20k \Ohm

resistor, which accounts for 12\frac{1}{2} the total resistance (and therefore half the PD) of that branch.

122\frac{12}{2} gives 6.0V6.0V across A-C.

Between D-F, there is a
Unparseable latex formula:

5.0k \Ohm

thermistor, which accounts for 13\frac{1}{3} the total resistance (and therefore a third of the PD) of that branch.

123\frac{12}{3} gives 4.0V4.0V across D-F.

Between C-D, the first branch has a PD of 6.0V6.0V at C and the second branch has a PD of 4.0V4.0V at D.

Hence, 6.0V−4.0V=2.0V6.0V - 4.0V = 2.0V between C-D.

Does that help?


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Grrrr, you're so good that I'm not allowed to rate you anymore :angry:
Thanks so much :biggrin:
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Original post by l1lvink
Grrrr, you're so good that I'm not allowed to rate you anymore :angry:
Thanks so much :biggrin:


Hahaha I'm really not good. I remember being stuck for ages on that question last year. :wink:

How's unit 1 going?


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What's the hardest past paper to try
Original post by Abdullah07
What's the hardest past paper to try


Try the June 2015 paper, lol


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Original post by l1lvink
Ok, so you've got 3 equations (0.5QV, 0.5CV^2 and 0.5(Q^2/C). Just for simplicity I'll refer to them as QV, CV and QC as defined by their variables. So you know that Q is the variable being increased, so the equation that we want must include Q; this removes CV. Now we also know that V will change (the 'other' half of the question is about this). Now this means that if we were using QV, you would have two variables that are changing (both Q and V), and this would make things a heck of a lot harder. Therefore, we use QC as C remains the same, but Q changes, meaning that we can calculate the change in E.


Your understanding of physics is good! How come you have difficulty in some questions?


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Original post by CD223
Between C-D, the first branch has a PD of 6.0V6.0V at C and the second branch has a PD of 4.0V4.0V at D.

Hence, 6.0V−4.0V=2.0V6.0V - 4.0V = 2.0V between C-D.

I don't understand how this works: Why are we allowed to simply take the difference of potential between the 20k resistor and 5k thermistor? (Or between the 10k and 20k resistors)
Hm, this took me a while (not the problem itself, but why subtracting the potential differences works) to figure out, but I guess you can think of the branches in the circuit as two different stairs. Each stairs has a height of 12V, and then you just figure out the height difference between the two stairs across the voltmeter:



So yes, the answer would just be 2V.

stairs.png2.0KB
arrangement.png0.7KB
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Original post by PotterPhysics
I don't understand how this works: Why are we allowed to simply take the difference of potential between the 20k resistor and 5k thermistor? (Or between the 10k and 20k resistors)


Essentially if you think of the resistors as buckets with holes in them to represent voltage drops, each branch has 12.0V12.0V at the top, and has dissipated 6.0V6.0V of that through the 20K resistor, leaving a PD of 6.0V6.0V at C.

A similar story at D, where 12.0V12.0V is at the top and in this case 8.0V8.0V has been dropped across the 10K resistor, leaving a PD of 4.0V4.0V at D.

This means if you connect C to D, the net PD between the two points is 6.0V−4.0V=2.0V6.0V - 4.0V = 2.0V


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