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# A level physics multiple choice answer to an internal resistance question watch

1. So with the new 2015 specification for AQA i found a specimen paper that had multiple choice. obviously these didn't come with an in depth answer on why they are correct so i was just wondering if anyone had a clue on how to answer this.

p.s. apparently the answers D although in my mind I worked it out as 3V.

Picture: http://s24.postimg.org/lvxw2h7ed/psy1.png

edit: So i've basically looked through my textbooks for an hour found out that the V in V = (emf) + Ir is actually the external voltage supplied to the resistor but using the new calculation it still dosen't work, i'm really stuck here :P.
2. this ones even more fun -_-. Think i got the bit about the gradient being negative and there being a squared relationship for the area in (pi)r2 but jeez they don't make these easy.

3. Why do I click on these links?

They only remind me that I'm such a fail at Physics.

Don't be me.
4. (Original post by profjb)
edit: So i've basically looked through my textbooks for an hour found out that the V in V = (emf) + Ir is actually the external voltage supplied to the resistor but using the new calculation it still dosen't work, i'm really stuck here :P.
Your problem may be that you have this equation the wrong way around. emf = V + Ir, where V is the potential difference across the terminals of the supply (or equivelently, across the external circuit). Does this help?
5. (Original post by profjb)
this ones even more fun -_-. Think i got the bit about the gradient being negative and there being a squared relationship for the area in (pi)r2 but jeez they don't make these easy.

What was the mark scheme answer to this one?
6. (Original post by Kozmo)
What was the mark scheme answer to this one?
D.
7. (Original post by profjb)
this ones even more fun -_-. Think i got the bit about the gradient being negative and there being a squared relationship for the area in (pi)r2 but jeez they don't make these easy.

Discount A and C because if you think logically, as you increase the diameter, you are allowing more electrons to move more freely because there is more area. Hence so there is a lower resistance as less particles collide with each other as they move. Is the answer D? Because a directly proportional drop does not seem feasible but more gradual.
8. Yeah the answers D. I defiantly understand the fact that you can discount the 2 positive graphs now which is great :P, i'm also assuming that the D graph can't cross the y axis really because after some thought if you express the graph as an equation i think it would represent:

r = n(pi)r^2 + C

Where:
r = R/L
(pi)r^2 = Area

this would indicate that the graph is quadratic and so follows that kind of shape but I could be entirely wrong :P.
Still can't understand why it dosen't cross through the origin though.

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Updated: February 19, 2016
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