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    Hey guys, so I was doing this past paper and I got stuck on one question as the mark scheme has an answer I can't understand

    It is this paper:

    http://store.aqa.org.uk/qual/gce/pdf...-EMPASASQP.PDF

    The question I am stuck on is on Section B (the last section), question 5 a)i) and ii).

    The mark scheme says the following:

    5) a)i) When contact is at T, pd across X = E,
    When contact is at L, pd across X = 0.

    ii) When contact is at T, pd across X = E,
    When contact is at L, pd across X > 0.

    Thanks!
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    Do you understand why it's zero when the contact is at L in the 1st one?
    The pd across the voltage divider resistor goes from E (the voltage of the supply) at the top, to zero at the bottom. (I've put it equal to 6V just for the sake of having an actual number to talk about). The point S "taps off" a proportion of that voltage which depends on where it is. At the mid point, for example, it would be at 3 volts and the pd across X would be 3V.
    The circuit is such that the bottom horizontal wire is taken to be at 0 volts for reference purposes.
    When S is at L it is at the zero reference point.

    In the 2nd circuit, the point L is not connected to the 0 volt reference, and the pd at the point S, which I've drawn at the bottom, will depend on the resistance of the rheostat and the resistance of X.
    The current in this circuit has to flow through the rheostat but also through X. They are in series, so part of the 6 volts will be across the rheostat and part across X. It means that the pd at L, the pd across X, will not be zero.
    In fact the pd across X plus the pd across the rheostat will equal 6V.
 
 
 
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