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Determine shunt resistance

  1. Jul 28, 2013 #1
    1. The problem statement, all variables and given/known data
    DC circuit
    2 resistors in parallel Rm and Rs
    one resitance will be a meter (Rm =0.1 ohm), the other will be the shunt
    Max current is 200A
    Im (meter current is 1mA)



    2. Relevant equations
    I= Is + Im
    V=IRt



    3. The attempt at a solution
    I am thinking first to find shunt current which Is
    Is = I - Im
    =199.9A ?
     
  2. jcsd
  3. Jul 28, 2013 #2

    gneill

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    Staff: Mentor

    Hi damon669, Welcome to Physics Forums.

    Yes, the shunt current will likely come in handy. What is it you're trying to find? Your problem description doesn't explicitly pose a question...
     
  4. Jul 28, 2013 #3
    The question is
    Determine the required value of the shunt resistance if the maximum value of the current I is 200A. The meter can read a maximum of 1 mA and has a resistance of 0.1Ω
     
  5. Jul 28, 2013 #4

    gneill

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    Okay. So you know the currents that are flowing through each of the components (you might want to hang on to a few more decimal places for the shunt current for intermediate steps, since you are dealing with a difference between a large number and a tiny one). What else can you determine from the given values?
     
  6. Jul 28, 2013 #5
    V=IR

    so 0.1 x 0.00001
    = 0.0000001
     
  7. Jul 28, 2013 #6

    gneill

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    Yes, you can determine the potential across the meter resistance. However, be careful of your decimal places... check the value you've used for current. (it's often advantageous to use scientific notation, particularly when metric prefixes are given). Also be sure to include units on results.

    So then, what's the potential across the shunt resistance?
     
  8. Jul 28, 2013 #7
    Rs = V/Is

    10^(-6)/199.9
     
  9. Jul 28, 2013 #8
    I read you can use product over sum rule to get V
    V=I x RsRm/Rs+Rm
    can I use this? if so how to make Rs the subject?
     
  10. Jul 28, 2013 #9
  11. Jul 28, 2013 #10

    gneill

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    Your result will be out by a few orders of magnitude. Did you check the current value that you used for Im as I suggested?
     
  12. Jul 28, 2013 #11

    gneill

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    You could do that. It's just algebra, rearranging the expression. Start by getting rid of the fraction (multiply both sides by the denominator expression).

    There's a more straightforward option, too. Look up "current divider". It would go well with the starting values you've been given.
     
  13. Jul 28, 2013 #12

    gneill

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    What is that supposed to be???
     
  14. Jul 28, 2013 #13
    current divider gives

    I = Im + Is
    so Is = I - Im (already done)
    I - V/Rm
     
  15. Jul 28, 2013 #14
    thought i worked out the voltage to be 5mA, which would be the same in all parts of this parallel circuit
     
  16. Jul 28, 2013 #15

    gneill

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    That doesn't look like the current divider rule, and you're not given V ahead of time...

    But before going on with a new method you need to fix the one you've used first. Check the numerical value you've used for Im!
     
  17. Jul 28, 2013 #16

    gneill

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    While it is true that the voltage will be the same across both resistances, that voltage is not 5mV !
    (and mA is a current: milli-Amps).

    Redo your voltage calculation and pay particular attention to the magnitudes of the values!
     
  18. Jul 28, 2013 #17
    V = IR
    = 0.001 x 0.1
    =0.0001
    = 10^(-4)
     
  19. Jul 28, 2013 #18

    gneill

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    That looks better :smile: Be sure to add units to your results! Always! You will lose marks if you fail to include units!

    So, with an appropriate metric prefix that would be 0.1 mV.
     
  20. Jul 28, 2013 #19
    So Rs = V/Is
    =0.0001/199.9
    =0.0000005003 ohms
     
  21. Jul 28, 2013 #20

    gneill

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    Yup. Now's the time to use scientific notation and round results to the appropriate number of significant figures.
     
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