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Potential divider and potentiometer principle

  1. Aug 16, 2005 #1
    does anyone knows how to do it? help and pls explain to me... thanks!

    1) The slide-wire of a simple potentiometer is 2 meters long and has a resistance of 5 ohms. The emf of the working battery is 6 V and its internal resistance is 0.20 W. What resistance must be added to the working-battery circuit in order for the wire to have a direct reading of 1 mV/mm?

    2) A slide-wire potentiometer is balanced against a 1.0182-V standard cell when the slide wire is set at 40.2 cm out of a total length of 100 cm. For an unknown source, the setting is 13. 8 cm. What is the emf of the unknown?

    3) Under what condition is a voltmeter more desirable than a potentiometer for the measurement of voltages?

    4) Prove that the potentiometer measures the emf of the test cell and not its TPD (hint: use the equation TPD=EMF-Ir).
  2. jcsd
  3. Aug 16, 2005 #2


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    Staff: Mentor

    Assuming the resistance is uniform along the wire, then the resistance per unit length is given by 5 ohms / 2 m = 2.5 ohms/m.

    Now remember V = IR or V/m = I * R/m.

    Also remember 1 mV/mm = 1 V/m.

    I think you can handle it now.
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