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Auto Transformer Question

  1. Nov 20, 2011 #1
    Hello, I have the following question regarding autotransformers and dot convention (please see attached).

    Can anyone give any guidance? I think I have an answer for part (a) but not completely sure if this is correct as I haven't come across a similar example before, sorry about the lack of subscripts on the numbers.

    (a) V1 = jωLI1 + jωM(I2 + I1)

    V2 = jωL2 (I2- I1) + jωMI1

    If the above is correct I think I can do (b) and (c) but if it's incorrect then I would really appreciate some pointers.

    Thanks
     

    Attached Files:

  2. jcsd
  3. Nov 20, 2011 #2

    rude man

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    Your eq. for V2 is correct but that for V1 is not.

    Remember fundamentally: v = jwL1i1 +/- jwMi2
    where v = voltage across L1, i1 = current thru L1 and i2 = current thru L2. So how does that change your first equation?

    Hint: the answer to part (c) does not include R.
     
    Last edited: Nov 20, 2011
  4. Nov 21, 2011 #3
    Thanks for your pointer. So should V1 = jωL1I1 + jωMI2 ? because the current is
    entering at the dotted end of L1 and therefore the mutual inductance is added to the self inductance of L1?
     
  5. Nov 21, 2011 #4

    rude man

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    No, because V1 is not the voltage across L1, and i2 is not the current thru L2!
     
  6. Nov 21, 2011 #5
    Woops! Ok another attempt: V1 = [ jωI1 (L1 + M) ] - [ (I2-I1) (jωL2 + M) ]

    My thinking here is that L1 has self and mutually inductance emf components that are positive with respect to the dot and direction of I1. While L2 has current I2-I1 across it but its self and mutual inductances are opposing L1 and therefore subtracted from L1? Am I close!?
     
  7. Nov 21, 2011 #6

    rude man

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    Oh dear.

    I'm going to give you the equation and you should think about why it's the correct one:

    V1 - V2 = jwL1*i1 + jwM(i1 - i2).

    Maybe there is some confusion here, though. I am assumimng L1 is the self-inductance of the top part of the transformer only, not the entire winding. So, with no load,

    V2/V1 = L2/(L1 + L2), not L2/L1. Did you see it that way? I do believe my interpretation is the intended one.

    Next of course you need to eliminate M. V2/V1 should be a function of L1 and L2 only, since coeff. of coupling = 1.0.
     
    Last edited: Nov 21, 2011
  8. Nov 21, 2011 #7

    rude man

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    BTW I was wrong when I said your other equation was correct. I didn't notice your sign change. It's actually V2 = jwL2(i1 - i2) + jwMi1.
     
  9. Nov 21, 2011 #8

    The Electrician

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    The schematic plainly indicates that L1 is the top inductor. Also, since L1 = 4L2, there are twice as many turns on L1 as on L2. If there are N turns on L2, then there will be 2N turns on L1 and 3N turns for the whole L1+L2 winding. With k=1, the whole thing is wound on a single core, with very tight coupling. Calling it an autotransformer is an apt description.

    Given these facts, I can tell you what the ratio V2/V1 is just from knowing how variacs work.

    But, if L1=4L2, then the ratio V2/V1 = L2/(L1+L2) = 1/5 doesn't seem right.
     
  10. Nov 22, 2011 #9

    rude man

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    I goofed when I said the gain was L2/(L1 + L2). That's definitely wrong. That would have been right if coeff. of coupling k = 0. However, it didn't affect my calculations since I computed the gain in terms of L1, L2 and k.

    Plugging in L1 = 4L2 gets me V2/V1 = 1/3, based on my solving the equations under the assumption that, as you say, L1 is the self-inductance of the top coil. I think that's right. You didn'd state what you believe the gain is - do you agree with 1/3?
     
  11. Nov 22, 2011 #10
    Thank you both for your input. I have found a similar example which may be useful at the bottom of page 77: http://nptel.iitm.ac.in/courses/IIT-MADRAS/Electrical_Machines_I/pdfs/1_11.pdf [Broken]

    It seems to suggest that the current across L2 is I2-I1. In which case the 2 equations would be:

    V1 = jωL1I1 + jωM(I2 - I1) + V2

    V2 = jωL2(I2 - I1) + jωMI1
     
    Last edited by a moderator: May 5, 2017
  12. Nov 22, 2011 #11

    The Electrician

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    Absolutely. V1 is applied to 3N turns and V2 is taken from a winding with N turns, so V2 = V1/3.
     
  13. Nov 22, 2011 #12

    rude man

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    OK Alex. The ball is in your court now - finish the assignment?
     
  14. Nov 22, 2011 #13

    rude man

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    Last edited by a moderator: May 5, 2017
  15. Nov 22, 2011 #14

    The Electrician

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    Alex,

    I think you would do well to set up the equations for a mesh solution to the network using KVL.

    You almost seemed to be heading in that direction in your first post. What you want to end up with is two equations that look like this:

    Z11*I1 + Z12*I2 = V1
    Z21*I1 + Z22*I2 = V2

    The Z11, Z12, Z21 and Z22 are coefficient expressions such as jω(L2+RL). When RL is connected, V2 = 0. If you will do this, all the parts of the problem fall out immediately.
     
  16. Nov 23, 2011 #15

    rude man

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    electrician - I don't think you want terms like that, do you?

    The Z11, Z12, Z21 and Z22 are coefficient expressions such as jω(L2+RL). :uhh:

    "When RL is connected, V2 = 0. " ??? :uhh:
     
    Last edited: Nov 23, 2011
  17. Nov 23, 2011 #16

    rude man

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    To sum up:

    V1 - V2 = jwL1*i1 + jwM(i1 - i2).
    V2 = jwL2(i1 - i2) + jwMi1

    You have 2 equations and 2 unknowns: V1 and i1.

    Remember k=1 so M = √(L1*L2)
     
    Last edited: Nov 23, 2011
  18. Nov 23, 2011 #17

    The Electrician

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    Well, maybe not quite like that; just sort of like that. The right paren got a little too far to the right.


    For part (c), I2 isn't an unknown; it's a given. We don't need 3 equations.
     
  19. Nov 23, 2011 #18

    rude man

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  20. Nov 23, 2011 #19

    The Electrician

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    I had suggested to Alex that he form the mesh equations for the system:

    "What you want to end up with is two equations that look like this:

    Z11*I1 + Z12*I2 = V1
    Z21*I1 + Z22*I2 = V2

    The Z11, Z12, Z21 and Z22 are coefficient expressions such as jω(L2)+RL. When RL is connected, V2 = 0. If you will do this, all the parts of the problem fall out immediately."

    I didn't mean that the value of the voltage across RL would be zero. Since the quantities on the right sides of each equation are the voltage sources in each mesh, the absence of any source in the mesh involving RL means that the right hand side of the second equation would be set to zero (where V2 was shown). It would have been less confusing if I had shown the equations as:

    Z11*I1 + Z12*I2 = E1
    Z21*I1 + Z22*I2 = E2

    and said that E2 would be set to zero for part (c) of the problem. I was trying to steer him toward a formal "Z matrix" solution. The solution will give an expression for I2 and multiplying by RL gives the voltage labeled V2 in the problem.
     
  21. Nov 23, 2011 #20

    rude man

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    I'm a big fan of two-port methods myself but if he hasn't had that yet he's going to be more confused than ever.

    He should run the Kirchhoff equations (nodal or current, I never remember which is which, I always sum currents = 0 at every dependent node, is that KVL or KCL? As in (v2-v1)/R1 = (v2-v3)/r2 + v3/R3?
     
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