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

Think about it this way. It's a power source. 2 cables on either side of it. The other end of the cables connect to the heating system. How is it possible to make a parallel circuit from that set up?

I did 2sf (650) due to 230 being 2sf.

Couldn't 230V be 3 sf too? E.g. 230.323...V = 230V to 3 sf?

Personally I found that exam extremely easy and thrilling to read. I agree with most of the answers given in this post, however it must be said that some of the answers given here do not quite stand up to my par.
I was aiming to achieve 80UMS, but after sitting it I expect myself to get- at least- 98UMS. Question 6+7 were rather quite easy, and a lot of the people on this post seem to completely blow the questions out of proportion.

I'm praying to the Gods for you to all do well, however we all know that if you failed the exam then you yourself have no one to blame other than yourself- and no amount of attribution bias that is present will remove this fact.

At the end of the day I will be achieving an A*, whereas most of you commoners will, maybe, scrap a C overall.
See you all in the next post I grace you with my presence.

I put 651V instead of 650V I agree with the rest.
230 * root2 * 2 = 650.5382...... which rounds up to 651V?

June 15 - Q7There'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

But I thought that was a different kind of question. It talks about a high-voltage transmission cable, most likely regarding power grid stuff.This one was just a simple circuit in series? It even said a wire on either side if I remember correctly.

Ye that's what I thought. When you give your answer as 650, it's as if you rounded your answer from 230*root 2 during your working which is something we should avoid?

That's what I thought.

Forget about question 6 for a minute, everyone got 89 electrons in question 1 right? And im hoping no one accidentally worked out specific charge for the question before it

Forget about question 6 for a minute, everyone got 89 electrons in question 1 right? And im hoping no one accidentally worked out specific charge for the question before it

This is how I imagined the circuit to be with a 2 wires connected to either side of the resistor/cell.

First of all, 98 UMS is only 2 UMS above an A grade, the paper is out of 120. 80 is a C grade. You were aiming for a C grade. Commoner.

Yh man there was no circuit diagram and there must be a wire going to 0V...

I thought the resistance would increase if the voltage increased for diode...
V=IR V proportial to R... was a bit confused because in terms of the graph I guess this would not be true only if correct is constant will V is being increased.

But the graph was I against V. Not V against I like it usually is. And the curved upwards.

The gradient would be 1/R. So doesn't that mean as V increases, R decreases? Because the higher the gradient (aka the higher V), the higher the 1/R value. Which means the lower the R value?

Maybe I messed up lol

Forget about question 6 for a minute, everyone got 89 electrons in question 1 right? And im hoping no one accidentally worked out specific charge for the question before it

But the graph was I against V. Not V against I like it usually is. And the curved upwards.

The gradient would be 1/R. So doesn't that mean as V increases, R decreases? Because the higher the gradient (aka the higher V), the higher the 1/R value. Which means the lower the R value?

Maybe I messed up lol

But the graph was I against V. Not V against I like it usually is. And the curved upwards.

The gradient would be 1/R. So doesn't that mean as V increases, R decreases? Because the higher the gradient (aka the higher V), the higher the 1/R value. Which means the lower the R value?

Maybe I messed up lol