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

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    (Original post by Mo2351)
    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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    I did the same thing as I thought only the second power question asked for average power! I didn't finish the second one though as I ran out of time so I lost an easy 2 marks there and I also lost 4 marks on the one about 2 wires because I said they were in parallel and and calculated 299.8V or something like that.
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    (Original post by tort 779)
    Calculating peak to peak voltage, which i believe was 650V
    I got 651 because I didn't round when calculation the Vrms so will I get the marks?
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    230 x _/2 = 325.27
    325.27 x 2 (peak to peak) = 650.5 so I rounded up to 651, is everyone saying I'll lose the mark for doing it more accurately?
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    (Original post by ShenanigansTVLV)
    I got 651 because I didn't round when calculation the Vrms so will I get the marks?
    Yeah probably they will allow 250/251 but 250 is preferable as all values are given to 2 s.f
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    (Original post by ShenanigansTVLV)
    230 x _/2 = 325.27
    325.27 x 2 (peak to peak) = 650.5 so I rounded up to 651, is everyone saying I'll lose the mark for doing it more accurately?
    No you will likely get the mark but again 250 is better as in the entire question everything they tell you is to 2 s.f
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    (Original post by tort 779)
    No you will likely get the mark but again 250 is better as in the entire question everything they tell you is to 2 s.f
    Yeah, because they didn't specify to do it to a reasonable number of significant figures I put "To 3 S.F." in brackets as well
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    Alright so I see everyone mainly got 650V, 4400W and 228 ohms for the first few questions in 6.

    But how the hell do you answer the last part? No one's even speaking about it lol. It's as if no one did it. I think I got 34W although my working out was bs so i think my answer is too. I litetally had no idea what i was doing
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    (Original post by tort 779)
    No you will likely get the mark but again 250 is better as in the entire question everything they tell you is to 2 s.f
    Also, I got 8800W for the first power question as I thought it asked for total power and not average power so you would have to use peak PD instead of Vrms? I swear I triple checked the question but obviously I could be wrong. I'm gutted because I only had 3 minutes left and rushed the "actual Vrms" question and accidentally put the two wires in parallel and I also didn't finish the 2nd power question as a result :/ So I've already definitely lost 6 marks so it would be gutting if I lost the others too and lost my A.
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    (Original post by Lawliettt)
    Alright so I see everyone mainly got 650V, 4400W and 228 ohms for the first few questions in 6.

    But how the hell do you answer the last part? No one's even speaking about it lol. It's as if no one did it. I think I got 34W although my working out was bs so i think my answer is too. I litetally had no idea what i was doing
    The last bit was Vrms^2/R with Vrms and R for both the wires combined. I ran out of time to do it so am devastated that I probably lost my A I was resitting for.
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    (Original post by Lawliettt)
    Alright so I see everyone mainly got 650V, 4400W and 228 ohms for the first few questions in 6.

    But how the hell do you answer the last part? No one's even speaking about it lol. It's as if no one did it. I think I got 34W although my working out was bs so i think my answer is too. I litetally had no idea what i was doing
    I got 34.2W
    Working:
    Use current calculated in the part about resistors which was 19.02A
    Then use I^2R (19.02)^2*0.0945
    This gives P=34.2W
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    (Original post by ShenanigansTVLV)
    The last bit was Vrms^2/R with Vrms and R for both the wires combined. I ran out of time to do it so am devastated that I probably lost my A I was resitting for.
    But the voltage in the 2 wires wouldn't have been equal to the given Vrms. Otherwise you'd be doing something like 230^2/(0.015*3.15*2). That answer looks collosal lol. It implies more power was used in the wires than the heating element



    (Original post by tort 779)
    I got 34.2W
    Working:
    Use current calculated in the part about resistors which was 19.02A
    Then use I^2R (19.02)^2*0.0945
    This gives P=34.2W
    You make it look so simple lol. i got that answer but without current somehow. I think i just worked out the proportion of the total rms voltage used in the 2 wires then did v^2/r to get 34.2

    It looks dodgy though.
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    (Original post by Lawliettt)
    But the voltage in the 2 wires wouldn't have been equal to the given Vrms. Otherwise you'd be doing something like 230^2/(0.015*3.15*2). That answer looks collosal lol. It implies more power was used in the wires than the heating element
    Sorry, I meant Vrms of just the wires so Total Vrms - Vrms of heating element = Vrms of 2 wires
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    (Original post by ShenanigansTVLV)
    Also, I got 8800W for the first power question as I thought it asked for total power and not average power so you would have to use peak PD instead of Vrms? I swear I triple checked the question but obviously I could be wrong. I'm gutted because I only had 3 minutes left and rushed the "actual Vrms" question and accidentally put the two wires in parallel and I also didn't finish the 2nd power question as a result :/ So I've already definitely lost 6 marks so it would be gutting if I lost the others too and lost my A.
    I used 230V because thats the voltage that a dc power supply would give.
    The heater cant work with ac (I believe) hence the heater has a method of converting the ac to dc. Therefore V across the heater should be the dc voltage produced by the ac supply, this is Vrms hence i used Vrms but i can be wrong
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    (Original post by tort 779)
    I used 230V because thats the voltage that a dc power supply would give.
    The heater cant work with ac (I believe) hence the heater has a method of converting the ac to dc. Therefore V across the heater should be the dc voltage produced by the ac supply, this is Vrms hence i used Vrms but i can be wrong
    But if something has a Vrms, doesn't it imply that it is from an A.C. current as the Vrms of an "AC" is the value of a DC that gives the same heating effect through the same resistor?
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    (Original post by ShenanigansTVLV)
    But if something has a Vrms, doesn't it imply that it is from an A.C. current as the Vrms of an "AC" is the value of a DC that gives the same heating effect through the same resistor?
    I think it said in the question that it was AC hence the question about the peak to peak voltage?
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    (Original post by txnilxnur)
    But a photon isn't a particle! And it said X was a particle I'm sure
    Actually a photon IS a particle, a particle of light; due to the wave-particle duality. But either way I don't think that a photon was a suitable answer for particle X.
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    (Original post by shuu00)
    Actually a photon IS a particle, a particle of light; due to the wave-particle duality. But either way I don't think that a photon was a suitable answer for particle X.
    Photon is the exchange particle (boson) for electromag interaction. I believe its not a suitable answer because a boson of electromag interaction cant be created in a weak interaction.
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    (Original post by tort 779)
    I think it said in the question that it was AC hence the question about the peak to peak voltage?
    Yeah, so since it's AC then the total power would be equal to the peak voltage^2/resistance. Therefore, 325^2/12 = 8800W?
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    (Original post by shuu00)
    Actually a photon IS a particle, a particle of light; due to the wave-particle duality. But either way I don't think that a photon was a suitable answer for particle X.
    Realistically, a photon could be correct but AQA like to ask ambiguous questions with a variety of possible answers and then only give the marks for a specific one they choose.
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    (Original post by dont leave blank)
    Attachment 537797

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


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    is that the exam paper?
 
 
 
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