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AQA physics Jan 2013

Is it me or is this exam a bastard? I can achieve A's in every other one but I just break down on this one for some reason!

Anyway, I'm hoping someone can tell me how they would go about doing the following questions:

6(b) - C-D

6(c) i and ii

7(c)

http://filestore.aqa.org.uk/subjects/AQA-PHYA1-QP-JAN13.PDF

Thanks!
Original post by jf1994
Anyway, I'm hoping someone can tell me how they would go about doing the following questions:

6(b) - C-D


Assuming you produced the correct answers for A-C and D-F?

Work out the voltage potential wrt the negative terminal of the battery at point C and then point D.

Then subtract D from C which gives the potential difference between the two points.

jf1994

6(c) i and ii


6(c)i: Apply Kirchoffs Current Laws at branch A. The current in path AE stay the same because nothing has changed (resistances in the current path and pd between AE have not changed). Also remember that the pd across the common junction of all parallel paths is the same.

6(c)ii: The thermistor forms a potential divider with the 10K ohm resistor.

The pd across the thermistor is then:

Vtherm = Vsupply * (Rtherm / R10K + Rtherm)

As the thermistor resistance falls then since the 10K resistor remains constant, more voltage must be dropped across the 10K and less across the thermistor.

jf1994


The current drawn by the circuit will produce a pd across the internal resistance of the new power supply.

But the new supply emf is the same as the old supply. From KVL (volts dropped around the circuit must sum to the supply), therefore since volts are now dropped across the internal resistance, the pd developed across the lamp must fall.

Power = V2/R and since the lamp resistance has not changed but the pd across it has fallen, then less power will be developed and the lamp will therefore be dimmer.
(edited 9 years ago)

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