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AQA Physics Unit 1 PHYA1 20th May 2014 OFFICIAL

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Reply 280
Original post by TheRAG
For the 6 marker, did anyone else say that to measure room temp they had to plot a graph, measure pd without a water bath, then read off the graph?


Yes! I did that too. Thank god someone's got the same answer as me for something 😂


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Reply 281
For the internal resistance q when they asked to use graph to calculate power then current did anyone do P=VI using the emf(6) as V and power which was 2W from the graph? Some people did P=I^2R
Original post by Me123456789
For the 5 mark question what did you write about the graph temperature against voltage or voltage against temperature?


It was potential difference (voltage) against temperature, because they want to see how p.d changed with temperature.
Original post by JPeters
Did anybody get 0.325A for the current on the first circuit question...?


do you know how many marks we would get as theres error carried forward?
Reply 284
Easy paper, although the style of questions was a bit different


I^2R
Original post by JackTeh96
The last question on the 2nd question: about the forces between the protons, what do you put?


Electromagnetic?
Reply 286
Original post by JackTeh96
The last question on the 2nd question: about the forces between the protons, what do you put?


Electromagnetic, Virtual photon
Original post by Razzamataz179
You just put read the voltage and then use that voltage to get a temp from your graph...


There was voltage? Dude the axis was power in WATTS against Resistance in OHMS, where did you get volts from?
Original post by JackTeh96
The last question on the 2nd question: about the forces between the protons, what do you put?



I went for electrostatic and photons.
What did people put for the power question about when internal resistance is negligible, what would change? :s-smilie:
Reply 290
Original post by lad123
You used P=VI 1.95/6.0, but you cant use the emf as the emf wasn't the potential difference across the resistor, as energy was lost across the internal resistor, you had to use P=I^2R


Ah, okay. Thanks.
Someone help please?! I did the resistivity question wrong then corrected it at the LAST SECOND. I put 1.1x10-7 or something... was this correct?
Original post by Ryejd
Electromagnetic, Virtual photon


I thought it is strong force?? by gluons??
Original post by CityWok
For the internal resistance q when they asked to use graph to calculate power then current did anyone do P=VI using the emf(6) as V and power which was 2W from the graph? Some people did P=I^2R


Thought it was just 0.8ohms and 1.95w and then rearragne P=I^2R?
Original post by CharleyJay
What did people put for the power question about when internal resistance is negligible, what would change? :s-smilie:

resistance would be proportional to power dissipated, i.e straight line
had 10 mins to do whole of electricity , ****ed it up , results day >parents get mad > cry > suicide> retake year
GGG **** YOU ELECTRICITY YOU WAAAAAANKER
Particle was ok , QUatumn was ok ( no photoelectic lololol *****) and electicity was stupid as ****
They really need get rid of the writing question and electricity entirely. So many people just don't get electricity, and having a random write-y question in an exam is having your cake and trying to eat it too. Either find a better way of explaining electricity in English or bump it up to uni level.
Original post by deadmau_5
There was voltage? Dude the axis was power in WATTS against Resistance in OHMS, where did you get volts from?


Because the experiment is to plot a graph of voltage against tempereature. So you put the circuit in a room, take a reading of the voltage, then using your graph, find the temperature.
Reply 298
Original post by Randomer96
I went for electrostatic and photons.


You can say electrostatic, not sure if just ''photons'' will suffice though?
Original post by team-punishment
resistance would be proportional to power dissipated, i.e straight line


Ffs I crossed this out. I was hoping for a mark for the general statement that the power would still increase

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