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AQA Physics PHYA4 - Thursday 11th June 2015 [Exam Discussion Thread]

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Reply 2460
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CD223
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Original post by Mehrdad jafari
ive got one today at 2 and tomorrow morning another :frown::frown::frown:


1 tomorrow, 1 on Monday, 1 on Tuesday THEN phya4 :frown:


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My biggest worry is that I'm really prone to making silly mistakes that I'd otherwise avoid. I drop around 3-5 marks alone on these mistakes or just by not reading the question properly

Also I'd advise getting the written paper out of the way first, because with the questions you either know it or you don't, compared to the multiple choice in which you may need more time to work out possible answers.
(edited 8 years ago)
Reply 2462
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CD223
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Is anyone revising AS stuff?


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Original post by CD223
Is anyone revising AS stuff?


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Not really - a huge amount of the synoptic content that comes up in phya4 is on the data sheet, I've yet to come across anything that I couldn't remember so it doesn't seem worth it!
Reply 2464
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CD223
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Original post by Lau14
Not really - a huge amount of the synoptic content that comes up in phya4 is on the data sheet, I've yet to come across anything that I couldn't remember so it doesn't seem worth it!


That's true - there was a question on fundamental nodes the other year though :/


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I keep getting these ones wrong so if someone could help




Reply 2466
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CD223
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Original post by AR_95
I keep getting these ones wrong so if someone could help






ImageUploadedByStudent Room1433427468.727500.jpg


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Original post by CD223


If you're running out of time remember transformers are quite efficient but not 100% so it's usually the value under 98%


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Reply 2468
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Original post by gcsestuff
If you're running out of time remember transformers are quite efficient but not 100% so it's usually the value under 98%


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Lol yeah you say that then AQA read this and stick an efficiency of 7% in this year's paper :wink:


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Original post by CD223


How did you get Is = 0.5?
Reply 2470
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CD223
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Original post by AR_95
How did you get Is = 0.5?


It gave values for the current in the question :smile:


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Original post by CD223
It gave values for the current in the question :smile:


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****ing hell like I said before I lose easy marks because i dont read the question

I didnt even read the part about 0.5 , just stopped at 0.26
Original post by CD223
Lol yeah you say that then AQA read this and stick an efficiency of 7% in this year's paper :wink:


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Yeah someone somewhere will have a need for a inefficient transformer and they will base it on that


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Reply 2473
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CD223
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Original post by AR_95
****ing hell like I said before I lose easy marks because i dont read the question

I didnt even read the part about 0.5 , just stopped at 0.26


Just let that be a lesson for next week :smile:

Original post by gcsestuff
Yeah someone somewhere will have a need for a inefficient transformer and they will base it on that


Haha, true! :biggrin:


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part 2? Sometimes its the easy questions that get to me
Reply 2475
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CD223
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Original post by AR_95


part 2? Sometimes its the easy questions that get to me


It's a uniform electric field so the potential varies linearly.

[br]E=ΔVΔd=2.0V0.6m[br][br]E = \dfrac{\Delta V}{\Delta d} = \dfrac{2.0V}{0.6m}[br]

[br]⇒E=3.33Vm−1 (3 sf)[br][br]\Rightarrow E = 3.33 Vm^{-1}\ \text{(3 sf)}[br]


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Hi, can someone please shed some light on Jun14 MC QUESTION 24. Why is the answer A?
Reply 2477
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CD223
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Original post by TALAKHAN
Hi, can someone please shed some light on Jun14 MC QUESTION 24. Why is the answer A?


The general equation for the induced emf is:

[br]ϵ=BANωsin⁡(ωt)[br][br]\displaystyle \epsilon = BAN \omega \sin (\omega t)[br]

Due to the nature of the sinusoidal variation in emf (when sin⁡(ωt)=1\displaystyle \sin (\omega t) = 1), its peak value is:

[br]ϵ0=BANω[br][br]\displaystyle \epsilon_0 = BAN \omega[br]

As ω=2πf\displaystyle \omega = 2 \pi f, and flux linkage is equivalent to =BAN\displaystyle = BAN:

Dividing both sides by omega gives the max flux linkage equal to:

[br]BANMax=ϵ0ω[br][br]\displaystyle BAN_{Max} = \dfrac{\epsilon_0}{\omega}[br]

[br]∴BANMax=ϵ02πf[br][br]\displaystyle \therefore BAN_{Max} = \dfrac{\epsilon_0}{2 \pi f}[br]


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Original post by CD223
The general equation for the induced emf is:

[br]ϵ=BANωsin⁡(ωt)[br][br]\displaystyle \epsilon = BAN \omega \sin (\omega t)[br]

Due to the nature of the sinusoidal variation in emf (when sin⁡(ωt)=1\displaystyle \sin (\omega t) = 1), its peak value is:

[br]ϵ0=BANω[br][br]\displaystyle \epsilon_0 = BAN \omega[br]

As ω=2πf\displaystyle \omega = 2 \pi f, and flux linkage is equivalent to =BAN\displaystyle = BAN:

Dividing both sides by omega gives the max flux linkage equal to:

[br]BANMax=ϵ0ω[br][br]\displaystyle BAN_{Max} = \dfrac{\epsilon_0}{\omega}[br]

[br]∴BANMax=ϵ02πf[br][br]\displaystyle \therefore BAN_{Max} = \dfrac{\epsilon_0}{2 \pi f}[br]


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thanks
Reply 2479
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CD223
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Original post by MsFahima
thanks


No problem! Good luck for next week :smile:


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