Question from text book

Discussion in 'Homework Help' started by gte, Feb 5, 2010.

  1. gte

    Thread Starter Active Member

    Sep 18, 2009
    347
    4
    The question is:

    So
    2E = I * R
    E = I * (R + 1Mohm)

    ^ Is that correct so far? I was trying to plug in values to see if this actually worked, but I'm not having any luck?


    10v = .00001A * 1Mohm
    20v = .00001A * 2Mohm


    Should the current be a constant in this? Cause when you add resistance in the equation, the voltage number goes up, not down, if the current stays the same? What am I doing wrong?
     
  2. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    E=battery voltage
    r1=1 Mohm
    r2=meter input impedance


    E=Vr1+Vr2=E/2+Vr2

    Thus,

    Vr2=E-E/2=E/2=Vr1

    Vr1=I*r1
    Vr2=I*r2

    since I is the same through both resistances

    Vr1/r1=Vr2/r2

    since Vr1=Vr2

    r1=r2=1 Mohm

    An alternative way to solve it is by using the voltage divider rule.
     
  3. gte

    Thread Starter Active Member

    Sep 18, 2009
    347
    4
    Hi Mik3,

    Thanks for the response, I'm still a little confused with the equations having multiple = signs in them, could your equation also be stated as the following? How can I plug real numbers into it with multiple equals in the equation?


    2E = I * R2

    E = I * (R1+R2)




     
  4. n1ist

    Active Member

    Mar 8, 2009
    171
    16
    The multiple equals just means that all things are equal.

    E=Vr1+Vr2=E/2+Vr2
    is the same as
    E=Vr1+Vr2 and E=E/2+Vr2

    Just view this problem as a simple voltage divider. You have a power supply and two resistors in series. One is the 1M resistor. The other is the input impedance of the meter. You know the voltage across the meter, so solve for its resistance.

    /mike
     
  5. gte

    Thread Starter Active Member

    Sep 18, 2009
    347
    4
    That makes sense, thank you for the response


     
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