Aqa physics isa

sincostanxxx

ISA

(edited 9 years ago)

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Reply 1

myaman

i am doing the same thing as you and i would recommend that you learn the percentage uncertainty, they pop up a lot

Reply 2

Rugzii

Im doing mines tomorrow ... How did it go? Was it easy?

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Reply 3

luna178

I recommend you do Jun 2013 ISA (for the AS one) if your title is the forces in equilibrium one.

Unfortunately, it was absolutely shocking and I did terribly so I have to now do the emf ISA. :frown:

I wish you luck!

Reply 4

Rugzii

Im doing a2 ... I got an A last year but i dont feel confident for this one

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Reply 5

sincostanxxx

Original post by Rugzii

Im doing mines tomorrow ... How did it go? Was it easy?

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I did the experiment yesterday it was fine, really similar to the June 2011 experiment ^^

Reply 6

sincostanxxx

Original post by luna178

I recommend you do Jun 2013 ISA (for the AS one) if your title is the forces in equilibrium one.

Unfortunately, it was absolutely shocking and I did terribly so I have to now do the emf ISA. :frown:

I wish you luck!

Ah no, I'm doing the A2 one but thank you anyway
Good Luck with your ISA's ^.^

Reply 7

sincostanxxx

Original post by myaman

i am doing the same thing as you and i would recommend that you learn the percentage uncertainty, they pop up a lot

Ohh I remember how to do percentage uncertainty, we went over that so often in AS haha thank you ^^

Reply 8

myaman

the experiment was a ruler horizontally attached to two vertical spring of same spring constant and you get result T^2 against mass and good luck BTW

Reply 9

loooooool

what is the y intercept and gradient for a 1/Current against resistance graph with equation EMF = I(R+r)?

Reply 10

chococaramel

Are the questions in SHM harder that those about capacitors?

Reply 11

ahlam H

Original post by loooooool

what is the y intercept and gradient for a 1/Current against resistance graph with equation EMF = I(R+r)?

Gradient is 1/v since 1/I÷R = 1/I*1/R=1/V the negative intercept on the x axis is the internal resistance -r

Reply 12

jaydeefisher

Inbox me wanting to discuss the EMF/RESISTANCE Isa/

Reply 13

Fanatical Geek

Original post by loooooool

what is the y intercept and gradient for a 1/Current against resistance graph with equation EMF = I(R+r)?

I'm doing this for my ISA at the moment, I'll show you:

EMF =I(R+r)
EMF/I = R+r
1/I =(R+r)/EMF
1/I =R/EMF +r/EMF
1/I=(1/EMF)*R +r/EMF
y=mx + c (constant gradient line)
y=1/A
x=R
m(gradient)= 1/EMF
+c (intercept) = r/EMF.

Hope this helps.

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Reply 14

Fanatical Geek

Original post by ahlam H

Gradient is 1/v since 1/I÷R = 1/I*1/R=1/V the negative intercept on the x axis is the internal resistance -r

What? Please explain - I don't think you are right (see my response above.)

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Reply 15

ahlam H

Original post by Fanatical Geek

What? Please explain - I don't think you are right (see my response above.)

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Yea ur right I got the intercept wrong but ur method looks great,1/v is the same as 1/Emf right?

Reply 16

ahlam H

did u do the written paper

Reply 17

Fanatical Geek

Original post by ahlam H

Yea ur right I got the intercept wrong but ur method looks great,1/v is the same as 1/Emf right?

Not quite, V is the voltage through the components (in this case the resistor) while EMF is the total energy supplied by the battery (before wasted power through internal resistance.)

Basically the same in this case though

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Reply 18

Xetter

Doing my ISA today and just interested as to how you got 1/V as the gradient? My graph is 1/I in the Y axis and R in the X axis.

If you do change in Y over change in X to find the gradient you end up with 1/I all over R.. Which isn't the same as 1/V or 1/emf.

Can anyone explain what I'm doing wrong?