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    MS
    http://core.physicsinfo.co.uk/download.php?file=1509
    QP
    http://www.edexcel.com/migrationdocu...e_20100118.pdf

    I need help on 15b. As temperature decreases, resistance of thermistor increases. By the parallel resistors formula, the resistance of the parallel combination would decrease, but the MS says that the parallel combination resistance increases. This doesn't make any sense to me, so an someone please explain this to me?

    Thanks.
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    (Original post by krisshP)
    MS
    http://core.physicsinfo.co.uk/download.php?file=1509
    QP
    http://www.edexcel.com/migrationdocu...e_20100118.pdf

    I need help on 15b. As temperature decreases, resistance of thermistor increases. By the parallel resistors formula, the resistance of the parallel combination would decrease, but the MS says that the parallel combination resistance increases. This doesn't make any sense to me, so an someone please explain this to me?

    Thanks.
    In the parallel resistor equation, if you increase one of the resistors, the resistance of the combination also increases.

    Try it.

    2 and 4 ohm resistors

    1/R = 1/4 + 1/2 gives R= 4/3 = 1.33 Ohms

    Increase the 2 ohm to 4 and you get

    1/R = 1/4 + 1/4 giving R = 2 Ohms
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    (Original post by Stonebridge)
    In the parallel resistor equation, if you increase one of the resistors, the resistance of the combination also increases.

    Try it.

    2 and 4 ohm resistors

    1/R = 1/4 + 1/2 gives R= 4/3 = 1.33 Ohms

    Increase the 2 ohm to 4 and you get

    1/R = 1/4 + 1/4 giving R = 2 Ohms
    Makes sense

    But when you ADD a resistor to the parallel combination, by the parallel resistor equation, the resistance of the combination DECREASES. I got muddled up

    THANK YOU!
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    Yes. At the start, when you add the extra (3rd) resistor, the total resistance decreases.
    After that, if one of the parallel resistors already in place increases, the total also increases as explained.
 
 
 
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