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Circuit LOSt

  1. Feb 27, 2007 #1
    Circuit ....LOSt

    :confused: 1. The problem statement, all variables and given/known data
    A sandard cell or 1080V gives a balance length of 512 mm on a potentiometer. If the p.d across a standard 10.0 (ohm) resistor in a circuir gives a balance length of 784 mm, calculate the current in the circuit with the 10.0 (ohm) resistor. Draw a circuit sketch og the two circuit arrangements required in the above case.
    I am stuck here..am i going right?.please help someone..
    3. The attempt at a solution

    E = k * 512*10^-3 (512 times ten to the power of minus 3)

    V = E - Ir
    V = k * 784*10^-3

    V = IR I = V/R = V/10

    E = V + Ir

    K* 784*10^-3 = k * 512*10^-3 - K*784*10^-3(r)/10

    K's cancel

    784*10^-3(r)/10 = 784*10^-3 - 512*10^-3
    = 0.0784r = 0.272
    = 3.46(ohms)

    .....what else do i do from here?
  2. jcsd
  3. Feb 28, 2007 #2


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    Homework Helper

    The info on the standard cell (I assume you ment 1.080 V) gives the potential gradient of the potentiometer:

    [tex]e = 1.080/0.512\ V/m[/tex]

    With this value you can subsequently calculate the potential over the standard resistor. Which enables you to calc the current through it via Ohm's law.
  4. Feb 28, 2007 #3
    how do you calculate the potential? would it be V=0.784(1.080/0.512) ?
  5. Mar 1, 2007 #4


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    Homework Helper

    Yes. The potential gradient over the wire of the potentiometer is constant (the diameter of the wire is constant).
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