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B Confusion about transformer

  1. Dec 7, 2016 #1
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    I don't understand the concept behind equation (1). I think that the equation V=-Nd(flux)/dt is only for magnetic induction. However the power of the primary coil is usually supply by an a.c. power source. (So as to provide a changing B field.) Therefore, it is not an induction process. How can V=-Nd(flux)/dt?
     
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  3. Dec 7, 2016 #2

    cnh1995

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    The voltage across the primary is indeed supplied by the source and it is independent of the construction of the transformer. It's the flux that depends on the supplied voltage as Φ(t)=(1/Np)∫Vp*dt.
     
  4. Dec 7, 2016 #3
    Φ=net flux in the transformer Φp=flux produced by primary coil Φs=flux produced by secondary coil
    Φ = Φp - Φs
    dΦ/dt = dΦp/dt - dΦs/dt
    Isn't Φ(t)=(1/Np)∫Vp*dt account only for dΦp/dt ?
     
  5. Dec 7, 2016 #4

    cnh1995

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    That's not how a transformer works.

    The net flux in the core is fixed by the applied primary voltage, regardless of what is connected across the secondary, how many turns the primary and secondary have or what is the material of the core etc (Edit:but the core material does affect the coupling co-efficient which affects the mutually induced emf). Flux produced by the secondary depends on the secondary current Is which in turn depends on the load (and secondary turns). Primary draws an extra current Ip to cancel the opposing secondary flux. This extra current drawn by the primary is the reflected load current Ip, such that primary mmf=secondary mmf i.e. Np*Ip=Ns*Is. The primary current which establishes the core flux is called as the 'magnetizing' current, which is way smaller than the reflected load current Ip (about 3 to 5 %) and hence, it is often neglected in the basic transformer model.
     
    Last edited: Dec 7, 2016
  6. Dec 7, 2016 #5
    Thanks. Learnt a lot. Why is the 'magnetizing' current so small? As the primary circuit only involves a emf supply and a coil, isn't it short circuit?
     
  7. Dec 7, 2016 #6

    cnh1995

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    The primary circuit is not a short circuit.
    It is an R-L circuit with very small R (often taken as zero in the basic model) and a very large L. The ferromagnetic core makes the magnetizing inductance very high. Hence, the magnetizing current is very small.
    In other words, the permeability of the core is very high and hence, it can be easily magnetized. So, to establish the necessary amount of flux in the core, very small magnetizing current is required. Ferromagnetic core also increases the coupling co-efficient and mutual inductance of the windings.
    If you are studying transformers in detail, do not confuse 'magnetizing' inductance with the 'leakage' inductance of the transformer.

    Good luck!
     
    Last edited: Dec 7, 2016
  8. Dec 7, 2016 #7

    cnh1995

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    I believe this red part is technically incorrect.
    I meant to write "it's the flux linkage of primary that is fixed by the primary voltage." It is irrespective of no of primary turns i.e. the product Np*Φ remains constant for a given input voltage. Similarly, the flux linkage of the secondary is Ns*Φ which decides the induced secondary emf. Here Φ=B*Area of the core and B=μ*n*Imag.
     
    Last edited: Jan 4, 2017
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