[KING2011BOB]

Hey i was wondering about this question

H D^2 l/4 pi R -which value had the most uncertainty

My thinking was d^2 because if you square a value the uncertainty on the top decreases,over a small uncertainty on the bottom.

I know a lot of people put R but does anyone know what the right answer is here

[KING2011BOB]

(Original post by sfalcini1)
in our class half of us did your way and the other did a different way. I ended up doing what you did finding 2.3V across other resistor , and as we had the resistance and V across one resistor , i worked out the current as it is the same across both components (series circuit) , then knowing V for both , i worked out R for the resistor which the resistance we were not given and got something like 1300 ohms ??

I reckon sadly however you worked out the ohm rating for the 3.7V across the resistor,I think you needed to work out the 2.3/6 v resistor using the graph

1300/2100 x 6 = 3.7


I got 500 personally although that could quite obviously be wrong

500/1300 x 6=2.3

[Joel R]

(Original post by KING2011BOB)
Hey i was wondering about this question

H D^2 l/4 pi R -which value had the most uncertainty

My thinking was d^2 because if you square a value the uncertainty on the top decreases,over a small uncertainty on the bottom.

I know a lot of people put R but does anyone know what the right answer is here

I think it was D, but because it had the largest percentage uncertainty in it's measurement, rather than it being squared. Not 100% sure though, the numbers threw me off a little bit.

[sfalcini1]

(Original post by KING2011BOB)
I reckon sadly however you worked out the ohm rating for the 3.7V across the resistor,I think you needed to work out the 2.3/6 v resistor using the graph

1300/2100 x 6 = 3.7


I got 500 personally although that could quite obviously be wrong

500/1300 x 6=2.3

Yep your way sounds right , Hopefully i still get marks for working out , and thanks for the clearing up.

[Matt G]

(Original post by KING2011BOB)
I reckon sadly however you worked out the ohm rating for the 3.7V across the resistor,I think you needed to work out the 2.3/6 v resistor using the graph

1300/2100 x 6 = 3.7


I got 500 personally although that could quite obviously be wrong

500/1300 x 6=2.3

So did you get like 497 Ohms for your answer because that is why got, I think I used the same method as you.

[KING2011BOB]

(Original post by Matt G)
So did you get like 497 Ohms for your answer because that is why got, I think I used the same method as you.

I got 511 but it was one of those questions which will have a boundary from 485-515 I would say-either way still get working marks

Did anyone get for the question just after the emf prove question

The value of the resistor to be 6 ohms-I reckon this was rite

[jjwhitty]

can someone please post the paper with the mark scheme??

[Speccysteve]

hey,

can someone email me the paper and the mark skeme please?

My email address is [Mod edit Email address removed. Please use PM's]

[Aish_CoolKid]

(Original post by Speccysteve)
hey,

can someone email me the paper and the mark skeme please?

My email address is [Mod edit Email address removed. Please use PM's]

Did you manage to get hold of the paper?
Im tryna make sure i can get all the past papers done in preperation for my exam , ,would be really grateful if anyone could send it to me, thanks

[cloudhang]

grade boundaires ?

[When you see it...]

(Original post by cloudhang)
grade boundaires ?

They are on the OCR website for this exam.