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D.C circuits question

  1. Jan 10, 2012 #1
    Determine the required value of the shunt resistance if the maximum value of the current I is 200 A. The meter can read a maximum of 1 mA and has a resistance of 0.1 Ω.


    My answer is;

    Rsh:

    0.1*(0.1x10^-3)/200-(0.1x10^_3)

    =0.1/(200/0.1x10^-3)-1

    =1/19999990

    =0.0000005 ohms

    Is this correct or do I need to return to the drawing board?
     
  2. jcsd
  3. Jan 10, 2012 #2
    diagram which consist's of a meter and a shunt resistor in parallel.
     
  4. Jan 10, 2012 #3

    vk6kro

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    Yes, I get the same answer.

    There is a simple trick to make this easier.

    You calculate the voltage across the meter when it is reading full scale.

    So, in this case the voltage is (0.001 amps times 0.1 ohms) or 0.0001 volts.

    This must also be the voltage across the shunt.
    So, the voltage across the shunt is 0.0001 volts and the current is 200 amps (minus 1 mA which we can ignore).

    So, the shunt resistance must be (0.0001 volts / 200 amps) or 0.0000005 ohms.
     
  5. Jan 10, 2012 #4
    Ash I see.. That would save the messing around.. Thanks very much for confirming that
     
  6. Jan 11, 2012 #5
    -If the shunt is made of copper and has a cross-sectional area of 25 cm2 calculate its required length.
    (For copper take ρ as 1.7 × 10–8 Ωm.)

    R=ρl/A

    0.0000005000025=1.7*10^-8*l/(π*0.25^2)
    l=17.3 cm so require dlength = 0.173 m
    or l=17.3 cm so length is 17.3 cm

    This seem correct?
     
  7. Jan 11, 2012 #6

    vk6kro

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    25 sq cm is the area, not the radius.
     
  8. Jan 12, 2012 #7
    Good catch

    Length = [Resistance X cross-sectional area] / Resistivity

    (5x10^-7*.25)/1.7*10^-8

    Giving

    7.352 cm
     
  9. Jan 12, 2012 #8

    vk6kro

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    I calculated it with everything converted to meters, but I still got the same answer as you did.

    So, yes, it looks OK.
     
  10. Jan 12, 2012 #9
    Thanks very much for spotting that and letting me know :)
     
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