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  • 4 years ago

An ammeter deflects full scale with a current of 5 amperes and has a total resistance of 0.5 ohms what sunt resistance must be connected to it to measure 25 amperes full scale?

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  1. y2o2
    • 4 years ago
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    \[Rs = {I_g . R_g \over I-I_g} = {5 \times 0.5 \over 25-5} = {1 \over 8} \ \ ohm \]

  2. radar
    • 4 years ago
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    Lets think about this. If we are attempting to measure 25 amps, and the ammeter is full scale at 5 amps, then the shunt must accomodate the excess 20 amps. Now the voltage drop this shunt must equal the drop as when 5 amps goes through the .5 ohm meter resistance which would be .5 X 5 or 2.5 volt. What resistance would produce 2.5 volts when 20 amps is flowing through it? E=IR 2.5=20R R=2.5/20=.125 ohm or 1.8 as y2o2 posted above.

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