Current and Voltage Meters

Discussion in 'Homework Help' started by theluckyone, Sep 13, 2004.

  1. theluckyone

    Thread Starter New Member

    Sep 13, 2004
    1
    0
    I am looking at the basic design and calculations for current and Voltage meters. The questions give a meter resistance and tells you the full scale deflection current. then it asks you to calculate the resistance needed in parallel to change the full scale deflection current; finally asking for the combined resistance of the meter. I would like to know what Full scale deflection means and what equations I would need based on the information given to be able to calculate the new deflection current. No voltage measurement is given for the current meter and no current is given for the voltage meter.

    I attached the questions (numbers 4 and 5)
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    Hi,

    Back in the old days of Simpson 260's, meters were analog in nature. The meter movement, so called, was actually a weenie DC motor. It was forced back to zero by a hairspring, which exerted a constant force against the motor shaft. There was a needle attached to the shaft that acted as the indicator.

    Full scale deflection occured when the current through the meter movement rotated the needle clockwise to a stop. Most meters had 60 deg. swing, some 90.

    To act as a voltmeter, large-value resistors were switched in series with the movement, such that the resulting current drove the indicator through an angle that corresponded to the applied voltage. A rotary switch selected a path through an elaborate resistor network to scale the current.

    Consider your resistors as voltage-to-current converters in these problems. For number 4, Rx has to shunt current around the 100 ohms of the movement so there's only enough across the meter branch to push 25 microamps fo rthe full scale deflection of the meter. In 5, Rx has to be large enough that only 50 microamps flows through the series circuit. variations on E = IR will get you there.
     
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