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Design a multirange voltmeter

  1. Feb 22, 2009 #1
    1. The problem statement, all variables and given/known data
    "Design a multirange voltmeter capable of full-scale deflection for 20.0V, 50.0V, and 100 V. Assume that the meter movement is a galvanometer that has a resistance of 60.0 ohms and gives a full-scale deflection for a current of 1.00 mA"


    2. Relevant equations
    V = IR
    Kirchoff's Law's

    3. The attempt at a solution
    I know that the current through the galvanometer must be 1.00mA, meaning i'll need a large resistor in series with the galvanometer. I would be able to solve this no problem if only one voltage was needed, it is the variable voltages that throw me. So far, I believe i have a voltmeter (galv + large resistor) in parallel with a small resistor (so that most of the current flows through the small). I've attempted kirchoff's and equating v/r for each voltage, but there are too many variables for me to solve. Help? Thanks in advance

    Example of a similar, simpler problem:
    "A galvanometer requires a current of 1.5mA for full-scale deflection and has a resistance of 75 ohm, may be used to measure currents of much larger values. Calculate the value of the shunt resistor that enables the meter to be used to measure a current of 1 A and full-scale deflection"
    Solution:
    V = IR = (75)(1.5x10^-3) = (1-1.5x10^-3)R
    R = .113 ohms
     
    Last edited: Feb 22, 2009
  2. jcsd
  3. Feb 22, 2009 #2

    Redbelly98

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    Forget about the parallel resistor, that is for current measurements and is not needed for voltage measurements.

    Try doing things 1 voltage at a time. For 20.0V, what does the series resistor need to be?
     
  4. Feb 22, 2009 #3
    Ok, so in order to have full-scale deflection w/ 20.0 V:

    V = IR
    20/(1x10^-3) = 20000 ohms (<--total resistance)
    20000 - 60 = 19940 ohms for the resistor in series with the 60 ohm galvanometer.

    Likewise, for 50 V the resistor would need to be 49940 ohm and for 100 V, 99940 ohm

    I'm not sure how to use that information to help me solve the problem, perhaps I am misinterpreting what the question is asking, I was trying to work the problem so that, as a solution, I have a drawing of a circuit with defined values for each component. Can anyone give further guidance?
     
    Last edited: Feb 22, 2009
  5. Feb 22, 2009 #4

    Redbelly98

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    Well, you have the 60-ohm galvanometer. And you have 3 different series-resistances, but the circuit should use only one of those at any given time, depending on the desired scale.

    All that's missing, it would seem, is some means of switching among the different series resistances.
     
  6. Feb 22, 2009 #5
    OH! Switches! Thank you Redbelly98, that was just the tip I needed. I definitely overcomplicated this problem.

    I will updated this post with the solution presented in class once this is reviewed to compare with. Thanks!
     
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