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    A cell of negligable internal resistance is connected in series with a microammeter with negligable resistance and two resistors of 10kOhm and 15kOhm. The current is 200microA.

    Where a voltmeter is connected in parallel to the 15kOhm resistor, the current in the microammeter increases to 250microA.

    Calculate the resistance of the voltmeter.

    I did a really bizarre method and got a very strange answer. That, and the fact that I'm not confident at all with electricity questions, and it leads me to think that 18000.8 ohms is totally wrong.
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    I get 30kohms
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    The method is;
    Find voltage using original info.
    Find total resistance of the new circuit.
    Take the 10k from total to find the resistance of the 15k and voltmeter in parallel.
    Use 1/Rt = 1/15k + 1/(R of voltmeter)
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    (Original post by JamesF)
    The method is;
    Find voltage using original info.
    Find total resistance of the new circuit.
    Take the 10k from total to find the resistance of the 15k and voltmeter in parallel.
    Use 1/Rt = 1/15k + 1/(R of voltmeter)
    total voltage = IR = 25000 * 0.0002 = 5V

    total resistance of new circuit = V/I = 5/0.00025 = 20000

    20 000 = 10 000 + 1/(1/15000 + 1/x)
    10 000 = 1/(1/15000 + 1/x)
    10 000/15 000 + 10000/x = 1
    10 000x + 150 000 000 = 15 000 x
    5 000 x = 150 000 000
    x = 30 000

    wohoo

    thanks a lot, I appreciate that you don't just tell me the answer but give me the rough method to find it out myself. Well worthy of a rep.
 
 
 
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