AQA Physics Unit 3 Exam

How did everyone find it?

I thought it was okkk I guess. I messed up on the resistors in parallel and series bit.
I stupidly thought power was inversly proportional to resistance when it clearly isn't from the formula power=current [squared] x resistance.
So yeh there was a larger resistance in the series circuit therefore giving it more power.

I need an A or a B on this paper. Hope I get it :confused:

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Reply 1

username22

I thought it was easier than unit 2 but i'm not so optimistic. :frown:

I don't see how the young modulus question could have been 8 marks - i couldn't think of much to write. What did you get as the voltage between X and Y? didn't understand that one.

Reply 2

theamurleopard

Yes that question may have caught me out. I'm trying to remember..hmm...I think I put 4V,6V or 8V...although I can't remember which one. Do they sound like sensible values? lol. Hopefully everyone finds it hard so the grade boundaries are low. Although saying that most people in my physics class said it was quite easy. Damn.

Reply 3

username22

oh, well i put 12V but have no idea. do you know what the grade boundaries were last year? i reeaally hope they are low for this one.

Reply 4

theamurleopard

Hmm...you may be right because I wasn't confident at all on that question.
I hope they're low aswell. :smile:

Hopefully some more people will post on this thread with some more answers.
What force did you calculate for the tension in the wire at the end? I got 60 something newtons...

Reply 5

username22

yes, i got 60N too. i've got one right! i didn't check my answers though so i am a bit worried. did you get 0.86m for the length of the wire?

and yeah, where are all the aqa people? i need answers - i can hardly remember mine.

Reply 6

theamurleopard

Yup I got 0.86m for the length of the wire...I think. It was definitely 0.8 something or other lol.

With regards to the grade boundaries I'm afraid I don't know what they were last time round. :frown:

They are on the examiners reports which can be found on the AQA website usually. But my computer is misbehaving and taking a few lightyears to load each website so I can't really look and find out. You could though? :rolleyes:

Reply 7

theamurleopard

The gradient of the line on the first graph was less steap wasn't it? I got a mental block for ages on that question.
I just remembered another question I didn't like aswell. The one where they asked how you could get the EMF and internal resistance using the circuit they gave you.

Reply 8

hanh01

well i agree that it was the easiest module of 3.
i mean you could do all the wordy bits if you actually learnt everything.
the questions were very straight forward and i got 60 somefing for the last one... and the pd across X and Y i had no idea so i guessed 12v
:tsr2:

Reply 9

theamurleopard

For the p.d across X and Y i worked out the voltage over X which was the 30 ohm resistor I think...then worked out the p.d over y which was the 20 ohm resistor...and subtracted the two. But as I said I can't remember exactly what answer I got or if I adopted the right method.
Did people just state V=IR for their reason on the first graph?

Reply 10

username22

In June 2006, 38/50 was an A, 33 a B, 28 a C, 24 a D and 20 an E. Hopefully they won't have changed.

Reply 11

saxabojed

theamurleopard

How did everyone find it?
So yeh there was a larger resistance in the series circuit therefore giving it more power.

Hmm sure i got the parallel one giving 48w and the series giving 12w, although probably made a mistake somewhere, anyone got any ideas ?

Reply 12

Richie90

saxabojed thats what i got..it was confusign due to the arrangement of the resistors - if resistance goes up power SHOULD but not in this case..I think we're right.

Reply 13

calcium878

Yeah I got 48W for Figure 7 and 12W for Figure 8.

Reply 14

saxabojed

phew thats good then, i was hoping for an A in physics, revised that to around a B-ish, but A should still be capeable with one resit, and i've got computing to fall back on.

Also for the X Y one i put 10 V ( as Vout1 was 6 and Vout2 was 4) again like other people i was not too sure on it.

Tension in the wire i think i got 61N (2sf) tho may have made an sf error, but yeah 60-61 ish, I don't remember having to work out the length of the wire unless that was the question where you were given the resistivity, I had a got at that but got something completely rediculous lol

Internal resistance one i got something like 2.xx ohms but i did that in the last couple of mins was abit of a rush.

Reply 15

Richie90

Internal resistance was less then 0.5 i think

Reply 16

theamurleopard

I got 0.5 for the internal resistance aswell. :smile:

I know I've lost 8 marks near enough. Probably a few more here and there aswell.
The question regarding the heating effect and the resistors. Then the one where they gave you the circuit diagram and asked how you could calculate values for the emf and internal resistance. My friend said something about drawing a graph and the emf being the gradient and the internal resistance being the y intercept.