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# AQA Physics PHYA1 - 24 May 2016 - RESIT [Exam Discussion Thread]

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Why bother with a post grad course - waste of time? 17-10-2016
1. (Original post by TheLifelessRobot)
X was definitely a neutral pion as all baryons decay through the weak interaction and therefore strangeness is not conserved. A very similar question came up in the June 2014. I put K^0 so I definitely lost a mark there
It wasn't just similar, I think it was exactly the same!
2. (Original post by Jonah1958)
For the heater question, the wires were in parallel right???
from that i worked out the rms voltage across the heating component to be 229.5V
Why would the wires be in parallel? The circuit would be a cell and a resistor (filament) with each side of the resistor connected to the cell therefore it is in series.
3. For the last part of the heater question how did you work out the power produced by the heating component???
4. (Original post by cowie)
What did people put for the "how can this characteristic be used to work out the resistance of the diode between A and B" or something like that? I actually left it blank because I couldn't think of a thing!

Posted from TSR Mobile
I thought the wording was odd, because it referred to O to B as the "characteristic" of the question earlier on. I assumed they were talking about how the resistance changes from O to A and talked about how the resistance decreases as the gradient increases or something along those lines
5. (Original post by TheLifelessRobot)
Why would the wires be in parallel? The circuit would be a cell and a resistor (filament) with each side of the resistor connected to the cell therefore it is in series.
Damn, didnt think about it like that, yours seems more likely
6. (Original post by TheLifelessRobot)
Why would the wires be in parallel? The circuit would be a cell and a resistor (filament) with each side of the resistor connected to the cell therefore it is in series.
The way I saw it was, two wires coming out of a plug going onto the same component - they must have been in parallel? Very vague wording in the question to be fair
7. For 6 I got 650V, then 4400W, then 228V (that one was a disgusting question) and for the last one i just wrote down BS
8. (Original post by gasman95)
I thought the wording was odd, because it referred to O to B as the "characteristic" of the question earlier on. I assumed they were talking about how the resistance changes from O to A and talked about how the resistance decreases as the gradient increases or something along those lines
The graph was I against V and resistance is calculated by V/I I put that the resistance at a given point is calculated by 1/gradient and that resistance decreases as voltage increases.
9. (Original post by TajwarC)
The way I saw it was, two wires coming out of a plug going onto the same component - they must have been in parallel? Very vague wording in the question to be fair
It said the two wires made a cable.

Cable implies a single (series) cord.
10. (Original post by Jonah1958)
For the last part of the heater question how did you work out the power produced by the heating component???
I used the V^2/R with the voltage being the RMS Voltage (minus 2 x wire pd) as it is by definition the equivalent voltage in dc circuit to provide the same heating effect
11. (Original post by Lawliettt)
For 6 I got 650V, then 4400W, then 228V (that one was a disgusting question) and for the last one i just wrote down BS
I put 651V instead of 650V I agree with the rest.
230 * root2 * 2 = 650.5382...... which rounds up to 651V?
12. (Original post by TheLifelessRobot)
The graph was I against V and resistance is calculated by V/I I put that the resistance at a given point is calculated by 1/gradient and that resistance decreases as voltage increases.
Sounds right! I hope they weren't asking how to determine the exact numerical resistance
13. the circuit diagram and 6 marker - what did everyone put
14. (Original post by micycle)
It said the two wires made a cable.

Cable implies a single (series) cord.
June 15 - Q7

There's a cable that's made from other cables, each with their own resistance. When calculating the resistance of the entire cable, you had to assume they were in parallel
15. That's what I thought
16. (Original post by TheLifelessRobot)
I put 651V instead of 650V I agree with the rest.
230 * root2 * 2 = 650.5382...... which rounds up to 651V?
I did 2sf (650) due to 230 being 2sf.
17. (Original post by TajwarC)
June 15 - Q7

There's a cable that's made from other cables, each with their own resistance. When calculating the resistance of the entire cable, you had to assume they were in parallel
In this paper they made it obvious they were in parallel. In this it was not..? (Hoping)

Posted from TSR Mobile
18. This is how I answered the heater question:

R = 12

V = 230

Resistor:

R = 0.015 (x2)

New current = 230/(12+2*0.015) = 19.11

V lost over cable (2 wires) = 19.11*0.015*2 = 0.574V

New V = 230 - 0.574 = 229.4V
19. (Original post by TajwarC)
June 15 - Q7

There's a cable that's made from other cables, each with their own resistance. When calculating the resistance of the entire cable, you had to assume they were in parallel
Yeah but 229.5v to the same sig figs as question is 230, so mustve been series to get an acceptable different answer of 228v
20. (Original post by cowie)
In this paper they made it obvious they were in parallel. In this it was not..? (Hoping)

Posted from TSR Mobile
I don't know. But I imagined that the "cable" in this paper was similar to the one in last year's

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