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A Multirange Coil Ammeter Galvonmeter (?) resistance question

  1. May 17, 2008 #1

    TFM

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    [SOLVED] A Multirange Coil Ammeter Galvonmeter (???) resistance question

    1. The problem statement, all variables and given/known data

    A Multirange Ammeter. The resistance of the moving coil of the galvanometer G in the figure (See Below) is 48 Ohms, and the galvanometer deflects full scale with a current of 0.0200 A. When the meter is connected to the circuit being measured, one connection is made to the post markedand the other to the post marked with the desired current range.

    Part A

    Find the magnitude of the resistance R1 required to convert the galvanometer to a multirange ammeter deflecting full scale with currents of 10.0 A, 1.00 A, and 0.100 A.

    Part B

    Find the magnitude of the resistance R2 required to convert the galvanometer to a multirange ammeter deflecting full scale with currents of 10.0 A, 1.00 A, and 0.100 A.

    Part C

    Find the magnitude of the resistance R3 required to convert the galvanometer to a multirange ammeter deflecting full scale with currents of 10.0 A, 1.00 A, and 0.100 A.

    2. Relevant equations

    (I'm not sure which are actually relevant, so I will put a few down)

    P=IV

    V=IR

    Kirchoff Rules:

    Current Going into Junction equals total current Out of Junction

    Total Voltage over a loop = 0

    Resistors in Series:

    [tex] R_{equv} = R_1 + R_2 + R_3 [/tex]

    3. The attempt at a solution

    The question has confused me as to what I should do!!! Are we supposed to assume that the other resistors aren't there when find the Resistance?

    Any Ideas will be greatly appreciated,

    TFM (:confused:)
     
  2. jcsd
  3. May 17, 2008 #2

    alphysicist

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    Hi TFM,

    I did not see a diagram, so I'm not sure how these are all hooked up; perhaps I'm thinking of the problem setup completely wrong! However, I think the general idea is that you have resistors hooked up along with your galvanometer and that entire setup (galvanometer+resistor) is your ammeter.

    Then, if you want the ammeter to deflect full scale at 10 A, you choose the resistors so that when 10 A enters the ammeter, only 0.02A enters the galvanometer coil.

    If this doesn't answer your question, perhaps you could post a picture of the circuit or give more description?
     
  4. May 18, 2008 #3

    TFM

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    I meant to put the image on :redface:, Sorry, Anyway, here it is, from the MP Question

    TFM
     

    Attached Files:

  5. May 18, 2008 #4

    TFM

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    Any ideas now that the there is the diagram?

    TFM
     
  6. May 19, 2008 #5

    alphysicist

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    My last post had the idea to get the resistances of the three resistors. Did you try it? The point was that, for example when being used as a 10 A ammeter, you know the current through the galvanometer and r2 and r3 are all the same (because they are in series); you also know the numerical values of this current and therefore also the current through r1.

    Writing a Kirchoff loop equation for all three situation should give you your answer.
     
  7. May 19, 2008 #6

    TFM

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    I still seem to be slightly confused (Sorry)

    When we need to work out the resistance f R1, what happens to the resistanmce in R2 and R3? Also, we are not given a voltage?

    TFM
     
  8. May 19, 2008 #7

    alphysicist

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    I believe the way the question is written is rather confusing. It seems to be suggesting that you'll find three resistance values for all of the resistors, but I do not believe that is correct; also, you can't find R1 by itself. Depending on which of the terminals you are using (10A, 1A, or 0.1A) you essentially have three different circuits.

    For the 10 A circuit, just ignore the 1A and 0.1A leads. This means you have the galvanometer, R2, and R3 in series, and they are in parallel with R1. For the case of maximum deflection, you know the currents through each branch. If you write a Kirchoff's loop equation (on just the loop), you don't need the voltage. (Although you can find the maximum voltage for the 10A case right away with Ohm's law.) What loop equation do you get for the 10A case?

    Using only the 1A lead gives a different circuit, and a different loop equation; and the same for the 0.1A lead. With three equations and three unknowns, you can solve for all three at once. What do you get?
     
  9. May 19, 2008 #8

    TFM

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    Would these diagrams look right to find the first Resistance. If so, would the first equation be:

    [tex] 10R_1 + I(R_2 + R_3) + 48I_{Galvometer} [/tex]

    TFM
     

    Attached Files:

  10. May 19, 2008 #9

    alphysicist

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    I can't see the attachments yet, but that's the right kind of idea for the first; but there are several problems.

    The current is going to enter the ammeter through the + terminal and leave through the one marked 10A. The current in each branch will therefore go to the right through r1 and the galvanometer in the diagram. So as you follow the loop around, when you go against the current the voltage difference will be negative. So if you write your loop equation down clockwise, the IR term for R1 will be negative. (If you go counterclockwise, the other three terms will be negative.)

    Also, there are 10A coming into the + terminal, but that splits up in the two branches. We want the galvanometer to be at full deflection when 10A comes in, and we know what current in the galvanometer gives full deflection (it's given in the problem statement). So that's the current in the galvanometer, and it's also the current in R2 and R3 when its set up to be a 10A ammeter. What is that current?

    So if 10A is coming in, and you know how much is going into the galvanometer branch, how much is going through the branch with R1?

    Once you make those changes and set your expression to zero, you have an equation with three unknowns. What do you get?
     
  11. May 19, 2008 #10

    TFM

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    Would the current be 0.020 Amps?

    Using Kirchoffs Law, current in = Current out,

    would the current be 10 - 0.02 = 9.98 Amps

    Would the changes make the equation:

    [tex] -9.98R_1 + 0.02(R_2 + R_3) + 480.02 [/tex]

    Edit: Equation should have been:

    [tex] -9.98R_1 + 0.02(R_2 + R_3) + 48*(0.02) [/tex]

    Sorry


    TFM

    (I have reattached the diagrams in a more visible form - I didn't realise Bitmap images didn't show well!)
     

    Attached Files:

  12. May 19, 2008 #11

    alphysicist

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    TFM,

    Your final equation for the 10A case (once you set it to zero) looks right to me.

    Did you create those diagrams, or did they give them to you?

    The way I was thinking of designing it was, for the 10A case, just ignore the 1A and 0.1A leads. They won't be connected to anything, so they are not affecting the circuit. If you are drawing a picture, just erase them!

    Do the same for the 1A case; you're connecting this device to a circuit using only the + lead and the 1A lead; then the 10A and 0.1A lead are not doing anything. In that case, the galvanometer is in series with R3; also R1 and R2 are in series with each other; and (galvanometer+R3) is in parallel with (R1+R2). What loop equation do you get for the 1A case?
     
  13. May 20, 2008 #12

    TFM

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    Yeah I drew them myself, but what you said about ignoring the other leads makes more sense.

    So:

    [tex] -9.98R_1 + 0.02(R_2 + R_3) + 48*(0.02) = 0[/tex]

    So for circuit 2:

    [tex] 0.02(48 + R_3) - (1-0.02)(R1 + R2) = 0[/tex]

    And for Circuit 3:

    [tex] 0.02(48) - (0.1-0.02)(R_1+R_2+R_3)[/tex]

    Do these equations look right?

    TFM
     
  14. May 20, 2008 #13

    alphysicist

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    The loop equations look right to me; they are all equal to zero and now it's a matter of algebra.
     
  15. May 20, 2008 #14

    TFM

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    I have rearranged the equations to give:

    [tex] 0.02R_2 + 0.02R_3 + 0.96 = 9.98R_1 [/tex] - (1)

    [tex] R_3 = 49R_1+49R_2 - 48 [/tex] - (2)

    [tex] 0.96-0.92R_1 - 0.92R_3 = R_2 [/tex] - (3)

    Putting (2) into (3)

    [tex] R_2 = -0.9375 - 0.985R_1 [/tex] - (4)

    PUtting (4) into (1):

    [tex] 0.9412 + 0.02R_3 = 9.96R_1 [/tex] - (R1.5)

    putting (3) into (2):

    [tex] R_3 = 499R_1 - 470.6 [/tex] - (5)

    Putting (5) into (1.5)

    [tex] R_1 = 423.54 [/tex]

    This seem slightly high, though?

    Edit: and according to MP, is wrong?

    TFM
     
  16. May 20, 2008 #15

    alphysicist

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    I think this Eq. 3 is incorrect.
     
  17. May 20, 2008 #16

    TFM

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    Does this look better?

    [tex] 0.077-0.0064R_1 - 0.0064R_3 [/tex] - (3)

    TFM
     
  18. May 20, 2008 #17

    alphysicist

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    I don't think so, because in your last post, you were trying to solve Eq. 3 for R2. What is this expression equal to? I don't think it is equal to R2. Remember that the original equation for circuit 3 is:

    [tex] 0.02(48) - (0.1-0.02)(R_1+R_2+R_3)=0[/tex]

    I would combine all of the numerical factors first and then solve for R2.
     
  19. May 20, 2008 #18

    TFM

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    Lets See:

    [tex] 0.02(48) - (0.1-0.02)(R_1+R_2+R_3) = 0 [/tex]

    [tex] (0.96) - (0.08)(R_1+R_2+R_3) = 0 [/tex]

    [tex] (0.96) - (0.08R_1 + 0.08R_2 + 0.08R_3) = 0 [/tex]

    [tex] (0.96) - 0.08R_1 - 0.08R_2 - 0.08R_3 = 0 [/tex]

    [tex] (0.96) - 0.08R_1 - 0.08R_3 = 0.08R_2 [/tex]

    [tex] 12 - R_1 - R_3 = R_2 [/tex]

    Is this better?

    TFM
     
  20. May 20, 2008 #19

    alphysicist

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    Yes, that's looks right.
     
  21. May 20, 2008 #20

    TFM

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    Okay, so:

    Put (3) into (2):

    [tex] R_3 = -2.04R_1 - 0.075 [/tex] (4)

    put (4) into (3):

    [tex] 12.075 + 1.04R_1 = R_2 [/tex] (5)

    Put (5) and (4) into (1)

    [tex] 0.02(-2.04R_1 - 0.075 ) + 0.02(12.075 + 1.04R_1 ) + 0.96 = 9.98R_1 [/tex]

    Gives me

    R_1 = 0.12 Ohms

    Does this look right?

    TFM
     
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