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Single phase transformer EMF

  1. Nov 7, 2013 #1
    1. The problem statement, all variables and given/known data
    A step down transformer, 2400/240 V, 50Hz has the following resistance and leakage parameters.
    Z1 = R1 + jX1 = 4 + j12 Ω
    Z2 = R2 + jX2 = 0.04 + j0.12 Ω

    The transformer is operating at 80% of its rated load. The power factor of the load is 0.866 leading.
    Determine the induced EMF in the primary side

    2. Relevant equations

    E = 4.443 f N B A
    Ni = Hl = R phi
    phi = BA
    E = N d(phi)/dt

    3. The attempt at a solution
    I don't know how to solve for N and phi. It's as if I'm leading nowhere.
    So far, I've got

    Attempt 1:
    a = V1/V2 = 2400/240 = 10
    i1 = 2400/(4 + j12) = 60 - j180 = 189.7 cis (-71.57)
    but then if I try to use Ni=Hl I get stuck on finding H from B=uH

    Attempt 2: Try solving for N with L
    L = N^2/R
    N = sqrt(LR)
    R = 4 + j12
    Substitute L = N^2/R into N = sqrt(LR) you just end up with N = N (nothing useful)

    Attempt 3: Transfer Z2 from the secondary side over to the primary side

    a= 2400/240 = 10
    a^2 = 100
    Z2' = 100 (0.04 + j0.12) = 4 + j12
    Zeq = Z1 + Z2' = 2 (4 + j12) = 8 + j24
    Then using Ni = R phi
    i = 2400/(8+j24) = 30 - j90 A
    R = 8 + j24
    Then get stuck on finding phi...

    Attempt 4: using V1/V2 = N1/N2
    2400/240 = N1/N2
    Assume 100 turns in N1, then N2 = 10 turns.
    and let phi = i
    then E = 4.443 (50) (100) (60-j180) = some massive number in MegaVolts

    I've searched and searched in lecture notes for anything to do with the turns ratio, but can't find anything useful.

    Please help.
     
  2. jcsd
  3. Nov 8, 2013 #2

    rude man

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    Need to know the transformer's rated load.

    Then I would compute mutual inductance M which leads directly to emf_1 due to current_2.

    There may be another way to do this using equivalent circuits, but keep in mind we want the emf at the primary generated by the current in the secondary.
     
  4. Nov 9, 2013 #3

    NascentOxygen

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    Definitely we need to know the transformer's rated load.
     
  5. Nov 9, 2013 #4
    Ok for the purposes of the exercise, I'll try 24KVA and if I get stuck, I'll post back. Thanks
     
  6. Nov 10, 2013 #5

    rude man

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    Think 'out of the box' on this one.

    Think about what "the emf induced in the primary" really means.

    If I have a single coil L and apply a voltage V, what is the "induced emf"?

    Hint: a voltage across an inductor is an emf. Why? Because the voltage across the inductor is produced by its magnetic field, so one form of energy (magnetic) is transformed into another (electrical). That is the definition of an emf.

    By contrast, the voltage across a capacitor is not an emf.
     
    Last edited: Nov 10, 2013
  7. Nov 15, 2013 #6
    Thanks rude man. I worked on this problem shortly after your last post and never got round to replying.
    I found I2 = 24000(0.8)/240 = 80 Amps --> arccos 0.866 = 30 degrees --> I2 = 80/_30 Amps
    Then E2 = V2 + I2Z2 = 240 + (80/_30)(0.04 + j0.12) = 237.97 + j9.91 Volts
    E1/E2 = a = 2400/240 = 10
    So E1 = aE2 = 10(237.97 + j9.91) = 2379.7 + j99.1 Volts

    Was it really that simple?
     
  8. Nov 15, 2013 #7

    rude man

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    Your answer is probably very close to right. But it seems you did not include the effects of R1 and X1.

    As I see it, the voltage induced in the primary winding is just V1 - I1R1

    so use your equiv. ckt. to compute I1 and you're home. The answer is going to be close to V1 which = 2400V, and therefore to your answer also.


    My rationale is as follows:
    Basic equations for a loaded transformer (load = ZL) are
    V1 = I1(R1 + sL1) + sMI2
    V2 = -I2ZL = I2(R2 + sL2) + sMI1

    where s = jω for sinusoids, L1 and L2 are the open-circuit primary and secondary winding inductances, R1 and R2 are the primary and secondary winding resistances, and M = mutual inductance. I1 and I2 are defined as currents flowing into the respective windings and it's assumed that dotted ends face each other.

    So the "voltage induced in the primary" by this picture is in my opinion
    V1(induced) = sL1I1+ sMI2. Note that X1 and X2 are included in M.

    There may be room for interpretation here as to what exactly is "the EMF induced in the primary side".
     
    Last edited: Nov 15, 2013
  9. Nov 16, 2013 #8

    NascentOxygen

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    Probably not. :smile:

    In power engineering, there are conventionally accepted approximations. You need to be guided in this by worked problems you have encountered in your course.

    I doubt that you can assume the turns ratio here is 10:1 because if it were then with rated supply voltage (24000V) the transformer would not be delivering rated output voltage (240V) at rated load (24kVA). The buyer would be upset at being short-changed on his voltage. :frown:

    So when delivering full rated output (240V, 100A) the turns ratio must be such that it overcomes the voltage drops in the transformer model's non-idealities.

    I calculate the necessary turns ratio to be 9.62 but I am rusty on this so that may not be exact. But it supports the intuitive expectation that the voltage divide across the turns must be less than 10:1 so my answer is around what I'd expect.

    That completes half of the working, the hardest part.

    You don't happen to know the answer, or have a worked example as a guide?
     
  10. Nov 16, 2013 #9

    NascentOxygen

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    How will you determine the value of M here?
     
  11. Nov 16, 2013 #10
    I didn't assume. The turns ratio is V1/V2 = N1/N2 = I2/I1 = 2400/240 = 10/1

    Where did you get 24000 Volts from? Do you mean 24000 VA? They are two completely different quantities

    Where did you get 9.62 from?

    Nope.
     
  12. Nov 16, 2013 #11

    NascentOxygen

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    That looks like an assumption to me.

    That's 2400V with a bouncy 0 key. :smile:

    I calculated it, on the basis that the rated voltage of a transformer must be delivered at rated load. You will get 240V out of a precisely 10:1 transformer only at zero load.
     
  13. Nov 16, 2013 #12
    Heh, that's what our lecturer taught us :smile:

    :smile:

    Ok, I'm satisfied this problem is sufficiently solved.

    Thanks guys.
     
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