# AQA Physics: Electricity question with areas and efficiency and power and stuff... Watch

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uhhh help

before you ask what i tried, i've already checked the mark scheme and i would have never done what they've written. problem is, i have no idea what they did in the mark scheme/where they're pulling all these numbers and formulas from...

question 7 part a from http://filestore.aqa.org.uk/subjects...2-QP-JUN15.PDF

mark scheme http://filestore.aqa.org.uk/subjects...W-MS-JUN15.PDF

like in part ai, how do you know which point exactly is gonna be the max other than trial and error and knowing that it will be somewhere near the top right?

part aii is okay i think i get that

it's part aiii i really dont get - i dont remember ever having to learn the formula I=P/A... i literally don't know what they did in the markscheme because they haven't put many numbers in lolAttachment 635116635118635100

before you ask what i tried, i've already checked the mark scheme and i would have never done what they've written. problem is, i have no idea what they did in the mark scheme/where they're pulling all these numbers and formulas from...

question 7 part a from http://filestore.aqa.org.uk/subjects...2-QP-JUN15.PDF

mark scheme http://filestore.aqa.org.uk/subjects...W-MS-JUN15.PDF

like in part ai, how do you know which point exactly is gonna be the max other than trial and error and knowing that it will be somewhere near the top right?

part aii is okay i think i get that

it's part aiii i really dont get - i dont remember ever having to learn the formula I=P/A... i literally don't know what they did in the markscheme because they haven't put many numbers in lolAttachment 635116635118635100

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#2

Heya! Okidoke, first off, thanks for providing me with this paper! I should check out the AQA website more often, so that's good to have.

Second off, I'll help with the part a)i). I think it's a BS question. You're 100% right: it's total bull imo, and I would have lost the marks to be fair. MS is implying that they want the voltage and current reading from right before there is a dip, at about 0.55-0.57 volts. The current at this point is about 0.9, and that gives us the answer they want. My best explanation (not great) for this is: since the X*Y is equal to power for a given point, the power keeps increasing as we get a higher voltage; once we reach a certain point and the gradient decreases, voltage may keep increasing but current decreases, and the decrease in current, even with an increasing voltage, yields a smaller power. The increasing voltage can't compensate for the currents decrease, as the current will decrease at a greater factor than the PD. That's barely visible in the graph, but so be it.

For part a)iii), we calculated the power. The answer we may have gotten may not have been correct, but if they don't award ECF marks, then they should go drink a gallon of peach flavoured detergent. I'll explain this in fragments below, I feel as though you hate my wall of text.

Efficiency = useful/total, and we know that for an area of 1 metre, the total is 450W, assuming 100% efficiency. It's not 100%, it's 15%, 0.15. Thus, for an area of 1 metre, we get 15% of 450, about 67.5 if my head is fine with me. Divide your value for power by that, and you get the ratio of your cells area to the area of a metre at 15% efficiency.

How do you use that ratio? You tell me :P

Second off, I'll help with the part a)i). I think it's a BS question. You're 100% right: it's total bull imo, and I would have lost the marks to be fair. MS is implying that they want the voltage and current reading from right before there is a dip, at about 0.55-0.57 volts. The current at this point is about 0.9, and that gives us the answer they want. My best explanation (not great) for this is: since the X*Y is equal to power for a given point, the power keeps increasing as we get a higher voltage; once we reach a certain point and the gradient decreases, voltage may keep increasing but current decreases, and the decrease in current, even with an increasing voltage, yields a smaller power. The increasing voltage can't compensate for the currents decrease, as the current will decrease at a greater factor than the PD. That's barely visible in the graph, but so be it.

For part a)iii), we calculated the power. The answer we may have gotten may not have been correct, but if they don't award ECF marks, then they should go drink a gallon of peach flavoured detergent. I'll explain this in fragments below, I feel as though you hate my wall of text.

Efficiency = useful/total, and we know that for an area of 1 metre, the total is 450W, assuming 100% efficiency. It's not 100%, it's 15%, 0.15. Thus, for an area of 1 metre, we get 15% of 450, about 67.5 if my head is fine with me. Divide your value for power by that, and you get the ratio of your cells area to the area of a metre at 15% efficiency.

How do you use that ratio? You tell me :P

*Again, part a)i) was BS imo. Others can argue it isn't and that the average student will have the logic to note that the current decreases far faster than the increase in p.d can compensate at the point where the current decreases, but still, COME ON*
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(Original post by

Heya! Okidoke, first off, thanks for providing me with this paper! I should check out the AQA website more often, so that's good to have.

Second off, I'll help with the part a)i). I think it's a BS question. You're 100% right: it's total bull imo, and I would have lost the marks to be fair. MS is implying that they want the voltage and current reading from right before there is a dip, at about 0.55-0.57 volts. The current at this point is about 0.9, and that gives us the answer they want. My best explanation (not great) for this is: since the X*Y is equal to power for a given point, the power keeps increasing as we get a higher voltage; once we reach a certain point and the gradient decreases, voltage may keep increasing but current decreases, and the decrease in current, even with an increasing voltage, yields a smaller power. The increasing voltage can't compensate for the currents decrease, as the current will decrease at a greater factor than the PD. That's barely visible in the graph, but so be it.

For part a)iii), we calculated the power. The answer we may have gotten may not have been correct, but if they don't award ECF marks, then they should go drink a gallon of peach flavoured detergent. I'll explain this in fragments below, I feel as though you hate my wall of text.

Efficiency = useful/total, and we know that for an area of 1 metre, the total is 450W, assuming 100% efficiency. It's not 100%, it's 15%, 0.15. Thus, for an area of 1 metre, we get 15% of 450, about 67.5 if my head is fine with me. Divide your value for power by that, and you get the ratio of your cells area to the area of a metre at 15% efficiency.

How do you use that ratio? You tell me :P

**Callicious**)Heya! Okidoke, first off, thanks for providing me with this paper! I should check out the AQA website more often, so that's good to have.

Second off, I'll help with the part a)i). I think it's a BS question. You're 100% right: it's total bull imo, and I would have lost the marks to be fair. MS is implying that they want the voltage and current reading from right before there is a dip, at about 0.55-0.57 volts. The current at this point is about 0.9, and that gives us the answer they want. My best explanation (not great) for this is: since the X*Y is equal to power for a given point, the power keeps increasing as we get a higher voltage; once we reach a certain point and the gradient decreases, voltage may keep increasing but current decreases, and the decrease in current, even with an increasing voltage, yields a smaller power. The increasing voltage can't compensate for the currents decrease, as the current will decrease at a greater factor than the PD. That's barely visible in the graph, but so be it.

For part a)iii), we calculated the power. The answer we may have gotten may not have been correct, but if they don't award ECF marks, then they should go drink a gallon of peach flavoured detergent. I'll explain this in fragments below, I feel as though you hate my wall of text.

Efficiency = useful/total, and we know that for an area of 1 metre, the total is 450W, assuming 100% efficiency. It's not 100%, it's 15%, 0.15. Thus, for an area of 1 metre, we get 15% of 450, about 67.5 if my head is fine with me. Divide your value for power by that, and you get the ratio of your cells area to the area of a metre at 15% efficiency.

How do you use that ratio? You tell me :P

*Again, part a)i) was BS imo. Others can argue it isn't and that the average student will have the logic to note that the current decreases far faster than the increase in p.d can compensate at the point where the current decreases, but still, COME ON**exotic*questions like that but idk why my teacher threw it in our question pack lol

and i just noticed

you go to a sixth form pretty damn close to me. like it's not my sixth form cos i go to the one most local to me but. if i for some reason didnt go my current one i couldve easily been in yours

spooky

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#4

450 * area is the amount of power hitting the cell (i.e., power in).

From a(i) we know the maximum power out is 0.52W.

So power out / power in = efficiency

0.52/(450*a) = 0.15, a = 0.52/(0.15*450) = 0.0077...

From a(i) we know the maximum power out is 0.52W.

So power out / power in = efficiency

0.52/(450*a) = 0.15, a = 0.52/(0.15*450) = 0.0077...

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#5

(Original post by

Second off, I'll help with the part a)i). I think it's a BS question. You're 100% right: it's total bull imo, and I would have lost the marks to be fair. MS is implying that they want the voltage and current reading from right before there is a dip, at about 0.55-0.57 volts.

**Callicious**)Second off, I'll help with the part a)i). I think it's a BS question. You're 100% right: it's total bull imo, and I would have lost the marks to be fair. MS is implying that they want the voltage and current reading from right before there is a dip, at about 0.55-0.57 volts.

*Again, part a)i) was BS imo. Others can argue it isn't and that the average student will have the logic to note that the current decreases far faster than the increase in p.d can compensate at the point where the current decreases, but still, COME ON*From the graph it's a simple observation that the cell power output is also therefore defined by the area bounded by the co-ordinates at any point on the curve.

Maximum power is therefore achieved when the area is at a maximum.

The logic is very straightforward.

The mark scheme gives a reasonably generous allowance for error.

This is why it's important to READ THE WHOLE QUESTION FIRST then you would notice that part a)ii also wants the power curve plotted.

From part ii) the maximum power point can be read off the second graph reinforcing the first part.

Please also note that the mark scheme is for examiners use and is absolutely not intended to be used as a model answer.

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(Original post by

450 * area is the amount of power hitting the cell (i.e., power in).

From a(i) we know the maximum power out is 0.52W.

So power out / power in = efficiency

0.52/(450*a) = 0.15, a = 0.52/(0.15*450) = 0.0077...

**setsdfseeet**)450 * area is the amount of power hitting the cell (i.e., power in).

From a(i) we know the maximum power out is 0.52W.

So power out / power in = efficiency

0.52/(450*a) = 0.15, a = 0.52/(0.15*450) = 0.0077...

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