Oxford PAT 2013, Questions 11 and 12



10 thoughts on “Oxford PAT 2013, Questions 11 and 12

  1. i just drew a timeline and started with 2x for A and x for B and then applied the rules to about 15 days and saw when my ratio is reversed? found this to be much easier, what do you think?

    although might be a problem if the values were large

        1. I can’t find the question now but it was a question on reflection of plane mirrors, but I can’t find the paper. Could you tell me where I can study this topic because I haven’t really dork any calculations with mirrors

  2. Could you please explain why you let the decay equations equal one another? I can’t quite understand why it’s 1/2B2. Probably missing something very obvious. Thanks in advance

    1. Hi Gio, thanks for your question.

      The answer is a bit cryptic, so I’ll explain the details. Let’s suppose that initially the amount of A is A and the amount of B is B, and after time t the amount of A remaining is a and the amount of B remaining is b. Because the half life of A is 3 days and the half life of B is 6 days, we can say that a = A 2^(-t/3) and b = B 2^(-t/6).

      The question tells us that initially there is twice as much A as B, so A = 2B. And we want to know the value of t when a = b/2, which we can rewrite as A 2^(-t/3) = (1/2) B 2^(-t/6), and (substituting 2B for A) we get 2 B 2^(-t/3) = (1/2) B 2^(-t/6).

  3. Sorry to keep asking questions! I don’t quite understand question 11 here, the bitesize website says that primary voltage / secondary voltage = turns on primary / turns on secondary which would mean that the answer was 200, but you seem to have used another method to get the answer of 50! I’m not sure if the Vp/Vs = Tp/Ts is the right equation to use here though?
    Thanks again! And yes I am about to take it, thanks – I think I’ll need all the luck I can get. If it wasnt for this website though my revision would have been hundreds of times harder as nobody can help me with them apart from 1 physics teacher who’s always busy!

    1. To start with, let me clarify a bit of terminology: I have written OUT for secondary and IN for primary in the answer above.

      Now let’s look at the first line of the answer:

      Vout = 100 x 2.4 / 4.8 = 50

      This is using the fact that the transformer is an ‘ideal’ transformer. In an ideal transformer energy is conserved, so the power delivered at the secondary is the same as the power put in at the primary. The power at the primary equals voltage times current which equals 100 x 2.4, so the voltage across the secondary equals power divided by current, which equals 100 x 2.4 / 4.8, which equals 50V.

      Now the second line of the answer:

      Tout = Tin Vout / Vin = 100 x 50 / 100 = 50

      This is using the ‘voltage ratio equals turns ratio’ property that bitesize correctly describes: Vp/Vs = Tp/Ts. Therefore Ts = TpVs/Vp, or in my terminology, Tout = Tin Vout/Vin, which equals 100 x 50 / 100, which equals 50 turns.

      The thing that you need to remember is that a transformer doesn’t just multiply (or divide) a voltage according to the turns ratio, but it also conserves energy (and hence power), so if it doubles the voltage it must halve the current, and if (as in this case) it doubles the current, it must halve the voltage.

      I hope that helps. Let me know if you need more explanation. And best of luck in the exam!

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