Calculating current flow on secondary side of transformer?

Discussion in 'Homework Help' started by electronice123, Jul 6, 2010.

  1. electronice123

    Thread Starter Senior Member

    Oct 10, 2008
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    Ok, I have had a difficult time understanding how one would calculate current flow on the secondary side of a transformer.


    In a transformer which has a turns ratio of 1:10, 12V and 2 amps is applied to the primary, meaning the seondary has 120V and .2 Amps with no load.

    The secondary circuit has an inductor whose inductive reactance equals 100 Ohms and a series resistance which equals 5000 Ohms.

    How do I calculate the current in the secondary side of this circuit?

    is it just 120V divided by 5,100 Ohms=23mA?
     
    Last edited: Jul 6, 2010
  2. hgmjr

    Moderator

    Jan 28, 2005
    9,030
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    Though a bit of a simplification, one way to think about a transformer is as a power out equal to power in device. Of course there is some inefficiency that must be considered based on the losses that are inevitable in the transformer. Also remember that the inductor in the primary as well as the inductor in the secondary have a finite DC resistance that must be considered when calculating the transformers performance when loaded.

    hgmjr
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    On the basis of the information you provide, I would think the secondary current would be

    Isec=120/(5000+j100)

    where 'j' is the complex operator
     
  4. GetDeviceInfo

    Senior Member

    Jun 7, 2009
    1,571
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    You would use the classic volts over impedance, however the transformer will be limited by it's VA rating.

    At no load, no current of course. On a short circuit side of things, one must consider the percent impedance to calc the current flow.
     
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