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Homework Help: Measuring Range Extension "homework"

  1. Dec 5, 2017 #1
    • Moved from a technical forum, so homework template missing
    dimension the moving-coil movement so that it indicates full deflection at 25 V
    given Values:

    Voltage Source: U2Max = 25V;
    moving-coil movement: IM = 100µA; UM = 270mV

    Rv = ( U2Max - UM ) / IM = (25V - 270mV) / 100 µA = 247,3kΩ

    b) the internal resistance is increased by 20 % due to tolerances in production. What does the moving coil movement indicate at an output voltage of 25V

    My approach is the following...

    RM = UM / IM = 270mV / 100µA = 2700Ω

    now: it is 20% higher --> RM.Tol = RM *1,2 = 3240Ω


    The Answer is : The moving coil movement indicates 24.945 V
  2. jcsd
  3. Dec 5, 2017 #2


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    Hallo Apo, :welcome:

    Nor clear what your dots are representing....
    You found a new value for RM and the question is: how far does the meter that you built with this coil and the 247.3 k##\Omega## series resistance go on the scale, when the actually applied voltage is 25 V ?

    (and you already know that 100 ##\mu##A gives full scale)
  4. Dec 5, 2017 #3
    Hi BvU
    The dots are representing the missing "method" to get to the answer.

    In task-part a) i came to the result, that i have to add an other resistance Rv with 247,3kOhm in series so that the moving coil movement shows full deflection.

    In b) the internal resistance increases by 20% and voltage source supplies 25 V.
    The question is what does the moving coil movement shows?
    The answer has to be 24.945 V.

    My Idea:

    The inner resistance increase by 20%...
    --> Rm = Um/Im --> 270mV * 100µA = 2700Ohm
    --> RM.Tol = Rm * 1,2 = 3240Ohm

    Now the current strength changes? or the voltage drop? "Over moving coil movement"

    --> UM.Tol = RM.Tol * IM = 3240Ohm * 100µA = 324mV

    --> Rv = ( U2Max.Tol - UM.Tol ) / IM = > 247,3kOhm = ( U2Max.Tol - 324mV ) / 100µA => (247,3kOhm * 100µA ) + 324mV = U2Max.Tol
    => U2Max.Tol = 25,054 (not true)
  5. Dec 5, 2017 #4


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    Hehe, with the answer given you should be able to draw your own conclusion ....:rolleyes:

    The intention of he exercise is
    for an applied voltge of 25 V the current is now a bit lower. The ideal coil showed full scale at 100 ##\mu##A, but the coil with 20% more resistance will cause a smaller current, so it does not go all the way to full scale. With full scale (= 100 ##\mu##A ) marked as 25V for the ideal coil, the non-ideal coil shows ...

    extra exercise: how big is now the voltage drop over the coil ?
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