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Potential Dividers and Voltmeters

  1. Sep 25, 2011 #1

    I have a simple arrangement: Two 500 kΩ resistors connected in series. The teacher illustrates what it would be like to measure the p.d across one of these 500 kΩ resistors with a voltmeter with 50 kΩ resistance. To do so, he calculated the resistance in parallel of the voltmeter and one of the resistors. He got 45.5 kΩ. So far so good, but what he did next however is what I don't understand. He divides 45.5 kΩ by the total 545 kΩ and multiplies it by the e.m.f. Why would the voltmeter read (45.5 kΩ / 545 kΩ) x 12 V? I thought it would measure 6V out of the 12 of the source since both resistors had the same resistance. The voltmeter accounts for itself?

    Peter G.
  2. jcsd
  3. Sep 25, 2011 #2


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    What he did is correct. Voltmeters give erroneous readings if you use them to measure the PD across a resistor with resistance of a similar , or larger, value to the meter itself.
    Actually they give accurate readings, but the inclusion of the meter distorts the circuit.
    To measure the PD in the circuit mentioned you would need a Voltmeter of much higher resistance. A CRO would be better.
  4. Sep 25, 2011 #3
    But then why when we set up the same arrangement, but, instead of a voltmeter we use a resistor the p.d across a resistor would be 6 V in this example?
  5. Sep 25, 2011 #4


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    If you set up a voltage divider where the two parts were:
    #1 a 500 000 Ohm resistor
    #2 a 500 000 Ohm and a 50 000 Ohm resistor in parallel

    Then the voltage would not be divided 6 - 6

    it would be divided 1 - 11, since one part of the divider has 11 times the resistance of the other.
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