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Electricity: ammeter range calculation

  1. Apr 7, 2017 #1
    1. The problem statement, all variables and given/known data

    An ammeter has a range of 10mA or 0.01A and a resistor of 20Ω.
    Three shunts are used:
    R1 = 4Ω ; R2 = 2Ω ; R3 = 4Ω
    What will the different ranges be (m3, m2, m1)?
    shunt.png

    3. The attempt at a solution

    I started by calculating the voltage:
    Ua = Ia ⋅ Ra
    Ua = 0.01A ⋅ 20Ω = 0.2V
    voltage.png

    Then I started calculating the current for each resistor
    shunt formula.png

    I1 = Ia ⋅ Ra / R1 = 0.01A ⋅ 20Ω / 4Ω = 0.05A
    I2 = Ia ⋅ Ra / R2 = 0.01A ⋅ 20Ω / 2Ω = 0.1A
    I3 = Ia ⋅ Ra / R3 = 0.01A ⋅ 20Ω / 4Ω = 0.05A

    And here is where I get lost.
    On my answer sheet it says that:
    m3 = 0.075A ; m2 = 0.05A ; m1 = 0.03A

    Can someone explain me what I'm missing here?
    Thanks for reading up until here.
     
    Last edited: Apr 7, 2017
  2. jcsd
  3. Apr 7, 2017 #2

    BvU

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    Science Advisor
    Homework Helper
    Gold Member

    Hello Epoh, :welcome:

    For example: m1: if 10 mA goes through the meter, how much goes through the shunts ?
     
  4. Apr 7, 2017 #3
    I don't know if I'm right but I found something:
    If we take m1 that equals 0.03A and put it in the ohm's law we get:
    R = Ua / m1 = 0.2 / 0.03 = 6.667Ω

    If we find the equivalent resistor for serial we get:
    Rt = R1 + R2 + R3 = 10Ω

    Then the equivalent resistor for parallel with the serial above:
    Rt = 20 ⋅ 10 / 20 + 10 = 6.667Ω

    Then again the ohm's law:
    m1 = Ua / Rt = 0.2 / 6.667 = 0.0299A
    Then indeed it is 0.03A, but this doesn't work with the rest.
    A coincidence or am I getting close to the answer?

    I'm seriously stuck.
    Thanks for your time.
     
  5. Apr 7, 2017 #4

    gneill

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    Staff: Mentor

    For a full scale reading on the ammeter its current is specified to be 10 mA, so that's a fixed quantity for each of the cases.

    For each case determine which resistances are carrying only this 10 mA current (obviously the ammeter 's 20 Ω resistance is always one of them). You may want to make a sketch of the circuit and draw in the currents. Here's an example for the "m2" case. The others can be worked in a similar fashion, but this "middle" case will show the method:
    upload_2017-4-7_21-36-38.png

    Note the path of the meter current (red) and which components it flows through. Effectively, those components are in series. The remaining resistors then make up the shunt resistance for this case. It has a separate current (blue).

    Does that give you any ideas? Can you work from there for this case?
     
  6. Apr 8, 2017 #5
    So I tried, I don't know if I'm right,

    I1 (blue) = 0.05A
    Red through R2 = 0.01A
    Ia = 0.01

    I1 + red through R2 and R1 + Ia = 0.05 + 0.01 + 0.01 = 0.07
    m3 = 0.07
    m3.jpg


    R = R1 + R2 = 4 + 2 = 6Ω
    I blue = U / R = 0.2 / 6 = 0.03A
    Red through R3 = 0.01A
    Ia = 0.01

    I blue + red through R2 + Ia = 0.03 + 0.01 + 0.01 = 0.05
    m2 = 0.05
    m2.jpg


    R = R1 + R2 + R3 = 4 + 2 + 4 = 10Ω
    I (blue) = U / R = 0.2 / 10 = 0.02A
    Ia = 0.01

    Ia + I (blue) = 0.01 + 0.02 = 0.03A
    m1 = 0.03
    m1.jpg

    It's close to my answer sheet, maybe I'm something missing.
    Thanks for your time sir.
     
    Last edited: Apr 8, 2017
  7. Apr 8, 2017 #6

    gneill

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    Staff: Mentor

    Okay, the problem is that you're assuming that the voltage across the shunt resistor(s) in each case is the same 0.2 V. It is not.

    The voltage across the shunt is the same as the voltage that occurs across the combined meter and series resistance that forms the "red" current. Here's a modified view of the middle case, m2, where the resistance that is part of the meter's path has been moved to highlight their relationship with the shunt resistance:

    upload_2017-4-8_11-14-46.png
    Now, look at the red path where you know the given current is 10 mA: What's the potential drop from node 0 to m2?
     
  8. Apr 8, 2017 #7

    I think I got it.
    I calculated m2 and it seems that It's right.
    Since the voltage in parallel is the same in all resistors that are in parallel I came with this answer.
    Answer m2.jpg

    I also tried m3 with the same method:
    Answer m3.png
     
  9. Apr 8, 2017 #8

    gneill

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    Staff: Mentor

    Excellent! Well done. :smile:
     
  10. Apr 8, 2017 #9
    Sir, I thank you for your help.
     
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