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autotransformador caso1
Figura 94-04.1

    Item b

    Initially, we will calculate the currents I1 and I2

    I1  =  Sn / V1  =  20,000 / 2,000  =  10   A
    I2  =  Sn / V2  =  20,000 / 200  =  100   A

    The currents calculated above are the currents when the transformer operates conventionally, that is, with two independent windings. When the transformer operates as an autotransformer at full power, the terminal currents IL and IH are:

    IH  =  I2  =  100   A
    IL  =  I1 + I2  =  110   A
    As we know, the power supplied by the autotransformer is equal to the input (or output) power. Therefore:
    Sauto  =  Sin  =  VL IL  =  VH IH

    Replacing the variables with their respective numerical values, we have:

    Sauto  =  Sin  =  2,000 x 110  =  2,200 x 100
    Performing the calculation, we obtain
    Sauto  =  220,000   VA  =  220 kVA
    Note that, in this case, when the transformer operates as an autotransformer, its apparent output power is 1 + a times the nominal power of the conventional transformer, since a = V1 / V2 = 10.

    Item c

    To calculate the transformed power, we will use eq. 94-03 presented below.

    STR  =  IH ( VH - VL )  =  VL ( IL - IH )
    eq.   94-03

    Replacing the variables with their respective numerical values, we have:

    STR  =  100 (2,200 - 2,000 )  =  2,000 (110 - 100 )
    Performing the calculation, we obtain
    STR  =  20,000   VA  =  20   kVA
    To calculate the power per conduction, we will use eq. 94-05 shown below.
    Scon  =  Sin - STR  =  VL IH
    eq.   94-05
    Replacing the variables with their respective numerical values, we have:
    Scon  =  220,000 - 20,000  =  200,000   VA  =  200   kVA