# Oxford PAT 2016, Question 20

## 8 thoughts on “Oxford PAT 2016, Question 20”

1. alex says:

Why is the voltage across each heater 42V ?

1. You mean the voltage across B and C. It’s a symmetrical circuit (you can draw a vertical line catting through A and the LHS of the line is a mirror image of the LHS). Since the voltage across the whole circuit is 84V, this means that the voltage across the two halves is 42V.

2. George says:

Hey Sir. heat capacity has been removed from the new specifications. I am really confused why it occurred again on 2016 paper. Does that mean that even material which is not listed on the specification could be assumed?

Thank you!

1. Good point. I guess that the examiners just thought that it was a sufficiently trivial use of the concept, and everybody who’s done AS knows what specific heat capacity means. But technically I think you’re right, strictly speaking they shouldn’t have asked about SHC.

However still I don’t think you should go into any great depth on the things deleted from the syllabus.

3. Teymour Beydoun says:

Thank you for your quick reply! Do you offer by any chance private tutoring for the PAT?

1. I don’t offer private tutoring. But my son has just finished his undergraduate Physics at Magdalen College, and he would be happy to do something: email matthew.steggles@magd.ox.ac.uk if you’re still interested.

4. Teymour Beydoun says:

Hello,

Could you explain why Va = 0 ?

Thank you

1. That’s a good question. In circuits like this, it is a well-known principle that if you have a line of symmetry then the potentials at the points on that line are all the same, but it’s not immediately obvious why, and actually what I have written here is a bit slapdash, so I’ll illustrate in more detail.

Suppose the voltage across A is not zero, so that there is a current flowing downwards through A. Then, by Kirchhoff’s law, the current in B must be greater than the current in A, and so the voltage across B is greater than the voltage across A, so the voltage across B is greater than 42V. But, if we consider the two top resistors and use Kirchhoff’s law again, the current in the LHS resistor must be less than the current in the RHS resistor and so by similar reasoning the voltage across the LHS resistor is less than 42V. But this would mean that the potential at the top of A would be less than the potential at the bottom of A, which is inconsistent with our original assumption that the current was flowing downwards through A. If we made the opposite assumption, that the current was flowing upwards through A, then we could prove that the potential at the bottom of A must be less than the potential at the top of A, which again is inconsistent with the original assumption. So current can’t be flowing up or down through A, so the voltage across A must be zero.

You can derive this more formally by just using Kirchhoff’s and Ohm’s law to solve for the current in each resistor, and you’ll find that the result gives Ia == 0, (and therefore Va = 0) but it’s a lot of simultaneous equations, which is quite a hassle to write down and solve, so it’s definitely easiest to use symmetry-based arguments.