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

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Original post by Lawliettt
How would you get 4040? It was literally (230^2)/12=4400. And even if it was 4040, you would be wrong because you had to use 2 significant figures


Read first line, Correct add etc. That was my typing mistake
For the emf question I put 6 without the unit, do you reckon I will lose the mark as it only asked for the value not the units...? 🤕


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Original post by txnilxnur
I agree with you here I don't understand why its 5v across the 18ohm resistor and not 1... anyone care to explain why its 5v not 1v!!

also anyone know of an unofficial mark scheme anywhere?


6.0V was your emf. When current flows through the circuit, volts are lost inside the battery due to the internal resistance. The oscilloscope acts much like a voltmeter, and when both switches were closed the oscilloscope measured the terminal pd (which obviously does not account for the volts lost inside the battery). The resistance of the external resistor was 18 ohms, and the pd across it was 5.0V (i.e. the terminal pd), hence current is worked out using V=RI so I=V/R =5/18 =0.277777... A and so on
(edited 7 years ago)
Original post by matty9
For the conservation to find particle X I put photon, it doesn't have a charge, lepton number = 0, baryon number = 0, doesn't violate anything so would I by any chance get the mark?


But a photon isn't a particle! And it said X was a particle I'm sure
Guys do you think I will get any marks for working out on the power question I got 8800W
Original post by dont leave blank
For the emf question I put 6 without the unit, do you reckon I will lose the mark as it only asked for the value not the units...? 🤕


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It was one mark so i imagine you will get the mark for simply stating the magnitude without the unit
what did everyone draw for the circuit for the 6 market question??
Original post by Imtheish
Guys do you think I will get any marks for working out on the power question I got 8800W


Maybe one mark for writing down the equation: P=V^2/R
But no other marks
So did I because I thought it asked for total power but everyone else has calculated average power so are we wrong?
Original post by tort 779
Maybe one mark for writing down the equation: P=V^2/R
But no other marks

Okay thanks do you remember if there was another question after that one before the 228V
Original post by tort 779
Maybe one mark for writing down the equation: P=V^2/R
But no other marks


I thought that particular question had marks for using an appropriate sfs..?


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Original post by tort 779
It was one mark so i imagine you will get the mark for simply stating the magnitude without the unit


Yh..hopefully


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Original post by ShenanigansTVLV
So did I because I thought it asked for total power but everyone else has calculated average power so are we wrong?


Apparently we are
Original post by txnilxnur
what did everyone draw for the circuit for the 6 market question??


ImageUploadedByStudent Room1464119133.125621.jpg

I think they'd expect something like this..but forgot to do the data logger..


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I got 651V for my peak to peak as I didn't round when calculating Vtotal? Will I still get the mark?
Reply 475
Original post by txnilxnur
But a photon isn't a particle! And it said X was a particle I'm sure


First line of wikipedia: "A photon is an elementary particle" so your point is invalid xD
Reply 476
In question 6 bi I thought it asked for total power and not average power so would it not be wrong to use V rms instead of the peak voltage ??


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Original post by Imtheish
Apparently we are


I thought the power question was asking for the total power and not the mean power supplied so wouldn't it be 8800W? Also, are the wires in parallel or series? I was running out of time and assumed parallel so will I lose all 4 marks?
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 got 651V and 4480W and then 228.2V so pretty close, fingers crossed its right.
Original post by Imtheish
Okay thanks do you remember if there was another question after that one before the 228V


Calculating peak to peak voltage, which i believe was 650V

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