Ampere-turns and Resistivity

Discussion in 'Homework Help' started by RJElec, Jan 28, 2014.

  1. RJElec

    Thread Starter New Member

    Apr 28, 2013
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    Is it possible to separate ampere-turns into N and I in the following question

    A coil of copper wire produces 18000 ampere-turns when connected across 220v d.c. Mean length of each coil 0.75m resistivity of the copper is 0.01725μΩm. When the coil is connected across a 200V d.c supply it requires a 22Ω resistor to be connected in parallel to maintain the original supply current.

    i) find the number of turns on the coil
    ii) diameter of the copper wire

    Im not sure how to find the original supply current i looked up some things and found that F=NI however it doesn't look like i can break NI up into single parts.

    Any pointers would be much appreciated!
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    A starting point is to set down the equations which satisfy the conditions that the unknown coil resistance in parallel with 22Ω across 200 V gives the same current draw as the unknown coil resistance alone across 220V. Since the only unknown is the coil resistance and you have two conditions to consider, you should be able to resolve the matter. Everything else will follow once you have the unknown resistance.
     
    Last edited: Jan 28, 2014
  3. RJElec

    Thread Starter New Member

    Apr 28, 2013
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    I got the answers to be a) 180turns and b)1.16mm using the relationship 1/R2 = 1/Rc + 1/22

    is there an easier way to solve this as the maths took a little while to figure out.
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    Simpler than what? You're not giving us much to work with. Unless you show us how you did it, how can we determine if you could have done it a simpler way?

    Using the relationship you've given, it's about three lines of math to get the coil resistance. Then two lines to get the number of turns. Those can be gotten without doing any math harder than 220/200. Then three lines to get the diameter, but this involves math that is best evaluated with a calculator.

    If your work is considerably more involved that this, please post it and we can show you where you went for a self-guided tour the long way and what the short way would have been.

    Have you checked your answers? What is the resistance of a 0.75m length of copper wire that has a diameter of 1.16mm?

    http://en.wikipedia.org/wiki/American_wire_gauge

    This is basically 17 AWG wire, or roughly what you would find in a lamp cord. Does it make sense that a length of wire that size and that long (roughly two feet) would have the resistance you got for that coil?

    I haven't cranked the numbers, but a mental estimate puts the diameter at somewhere around 0.1mm, which would be somewhere around 39 AWG or 38 AWG. The resistance of #39 wire is about 2.73Ω/m. So a 0.75m length of this wire would have a resistance of about 2Ω. How does that compare with the calculated resistance of the coil?

    But this brings up another point for practical consideration.

    How much current did you determine is flowing in this wire?

    Consider that the fusing current (the current at which the wire will melt) at 10s is 99A for #17 wire and 2A for #39 wire.

    The ampacity rating for #16 insulated wire for a 90°C temperature rise (i.e., pretty darn hot!) wire is 18A.

    What do you think is going to happen to this coil when it is energized with 220V?

    To get to wire that is rated for the current you are finding you would have to use at least #3 wire. A 0.75m length of that wire has a resistance of ~0.5mΩ.

    Now let's consider the size of this coil we are talking about. It is 0.75m long with 180 turns, or 4.2mm per turn, giving a mean diameter of 1.3mm. Does it make sense that this would be made out of wire that is, itself, 1.6mm in diameter? Does it make sense at all, even for #39 wire?

    How much power is the coil dissipating? I get a number that is in the tens of kilowatts. In a coil that is about the size of an inch long piece of large drafting pencil lead?

    Now -- the problem is clearly ignoring this issue and you have to work the problem that is given. But it might be worth mentioning this issue along with your work. You might even get a point or two of extra credit.

    But I gotta wonder just who thought up this problem!

    Unless either you messed up presenting it or I messed up reading it -- both are possible.
     
    RJElec likes this.
  5. #12

    Expert

    Nov 30, 2010
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    This is how I interpret the instructions.
    A bit of algebra gives me amps and turns.
    Maybe this is enough to get you started.
     
    Last edited: Feb 3, 2014
  6. WBahn

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    Mar 31, 2012
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    Yep. Same here, although the inductance is immaterial and the reactance is zero. The resistances of both inductors are the same (not mentioned in your diagram) and the currents are the supply currents, not the inductor currents (not indicated in your diagram). But I think we have the same set up in mind.
     
  7. #12

    Expert

    Nov 30, 2010
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    I labeled it Xl when I should have labeled it Rl
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    Ah. Easy goof to make.

    Do you come to the same conclusions that I did?
     
  9. #12

    Expert

    Nov 30, 2010
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    I didn't even look into the size of the wire. I just took it far enough to determine that solving the problem is possible.
     
  10. RJElec

    Thread Starter New Member

    Apr 28, 2013
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    Ok so heres the maths i used

    R2 = ( 1/Rc + 1/22)^-1

    V1 = IxRc V2 = IxR2

    V1/Rc = V2 = V1/Rc = V2 (1/Rc + 1/R2)

    So V1 x Rc/Rc = V2 (1/Rc + 1/22) Rc

    V1/V2 = 1+ Rc/22 = 22V1/V2 -22 = Rc

    22 x 220/200 - 22 = Rc Giving 2.2ohms

    the rest was just using formulas known.

    Thank you for all of the previous replies!
     
  11. WBahn

    Moderator

    Mar 31, 2012
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    You got it!
     
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