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Aqa physics spec paper 1 question 3 help!!!

So I did the spec paper a couple of days ago and still can't quite wrap my head around this. It's question 3 on spec paper 1 (set 1) and I have no idea how to do any of the questions apart from 3.1.

For 3.2 I assumed the total p.d. in the circuit would be 22v and then divided it 3:1 between R3 and R1 and that kind of messed me up for the rest of the qs.

If you have any idea you are an angel <3

Reply 1

Ihatephys

First, the idea of potential difference is the work done per unit charge.
So the electrons from V1 has potential=10V, electrons from V2 has potential=12V, you can't add those two together.

potential at the junction below R3 must always be 0 V because you assume the wire has resistance = 0
the potential at the junction between R1 and R2 must be 10 V since no work is done on R1
Therefore, pd across R3 is 10V when R1=0, and pd across R2=12-10=2V

Once you increase the resistance of R1, some of the energy is dissipated on R1, so the potential at the junction between R1 and R2 decreases.
Then follow the same idea above, you can answer part 03.4

Reply 2

noise_lemon

i'm not an expert on the electricity stuff by any means, but i treated the whole thing almost like 2 separate circuits, ie. the left side (V1, R1 and R3) as one circuit, and the right side (V2, R2 and R3) as another, with R3 as the only component shared by both. if R1 = 0 ohms, then there will be no pd measured across it whatsoever (since V is proportional to R etc.), so considering just the left hand side components, R3 must have all the pd in that circuit across it, ie. 10V (the emf of V1). i think this should be true regardless of what's going on with V2 and R2.

if i'm wrong or anyone else can explain better, please correct me :smile: