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    Usually when I get something wrong in a past paper, I can understand how the correct answer works after looking at the mark scheme. However, I have just done a past paper question, and I am completely stumped..... help!

    AQA PHYA1 Jan 2011, Q7b:

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    The temperature of the thermistor is increased so that its resistance decreases. State and explain what happens to the pd across the 1200 ohm resistor. (3)

    Answer: Current increases, hence pd across 540 ohm resistor increases, hence pd across 1200 ohm decreases. :confused::confused:
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    (Original post by jazzynutter)
    Usually when I get something wrong in a past paper, I can understand how the correct answer works after looking at the mark scheme. However, I have just done a past paper question, and I am completely stumped..... help!

    AQA PHYA1 Jan 2011, Q7b:

    Name:  circuit.jpg
Views: 280
Size:  8.6 KB

    The temperature of the thermistor is increased so that its resistance decreases. State and explain what happens to the pd across the 1200 ohm resistor. (3)

    Answer: Current increases, hence pd across 540 ohm resistor increases, hence pd across 1200 ohm decreases. :confused::confused:
    You may think about it with the same principles but in a different 'way.'

    Voltage is identical in parallel for any two components. As V= IR This means that, for both components, I is inversely proportional to R. so let (1) denote it's for the variable resistor. this implies that I(1) increases.

    But by Kirchoffs law, the current entering the parallel part of the circuit is conserved so the current for the 1200 ohm R (let this be I2) decreases as I total - I(1) = I2 and (1) has increases but obviously, the total current is constant.

    Finally, the resistance is constant in 1200 ohm resistor, so V is directly proportional to I since I has decreased, the voltage decreases also.
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    (Original post by jazzynutter)
    Usually when I get something wrong in a past paper, I can understand how the correct answer works after looking at the mark scheme. However, I have just done a past paper question, and I am completely stumped..... help!

    AQA PHYA1 Jan 2011, Q7b:

    Name:  circuit.jpg
Views: 280
Size:  8.6 KB

    The temperature of the thermistor is increased so that its resistance decreases. State and explain what happens to the pd across the 1200 ohm resistor. (3)

    Answer: Current increases, hence pd across 540 ohm resistor increases, hence pd across 1200 ohm decreases. :confused::confused:
    You're right. Think of the circuit as a potential divider, with the parallel resistances forming the bottom part of the divider.

    Now the p.d.s across both the thermistor and 1200 ohm resistor are always the same, since they're connected to the same points in the circuit.

    So, if the total resistance of the parallel resistors is R_2, then the voltage V_2 across them is:

    V_2 = 15\frac{R_2}{540+R_2}

    This tends to 0 as R_2 tends to 0. But

    R_2 = \frac{1200R_{th}}{1200+R_{th}}

    which tends to 0 as R_{th} (the thermistor resistance) tends to 0. So V_2 tends to 0 as R_{th} tends to 0.
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    (Original post by atsruser)
    You're right. Think of the circuit as a potential divider, with the parallel resistances forming the bottom part of the divider.

    Now the p.d.s across both the thermistor and 1200 ohm resistor are always the same, since they're connected to the same points in the circuit.

    So, if the total resistance of the parallel resistors is R_2, then the voltage V_2 across them is:

    V_2 = 15\frac{R_2}{540+R_2}

    This tends to 0 as R_2 tends to 0. But

    R_2 = \frac{1200R_{th}}{1200+R_{th}}

    which tends to 0 as R_{th} (the thermistor resistance) tends to 0. So V_2 tends to 0 as R_{th} tends to 0.

    Wow, this makes so much more sense when I look at that section of the circuit as a potential divider!!! I guess I was getting so wound up over the mark scheme banging on about current that I completely overlooked a much simpler alternative.
 
 
 
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