The full-load copper loss of a transformer is 1600 W. At half-load, the copper loss will be

A. 6400 W

B. 1600 W

C. 800 W

D. 400 W

Answer: Option D

This Question Belongs to Electrical Engineering >> Transformers

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Comments ( 5 )

  1. Swarup Das
    Swarup Das :
    3 years ago

    Cu loss directly proportional to I^2. So if I become half loss will be reduced to one-fourth.

  2. Umesh Prabhu
    Umesh Prabhu :
    3 years ago

    copper loss is proportional to KVA^2 or I^2 so for half load its is = (1/2)^2*1600 = 400 W

  3. Israfil Rana
    Israfil Rana :
    3 years ago

    In the open circuit test of a 260/130 V single phase transformer, obtained data were: primary current 0.4 A, Power input 25 W. If the resistance of the primary winding is 0.5 ohm. Determine the core loss. Answer pleas?

  4. Supriya Mathew
    Supriya Mathew :
    4 years ago

    for full load, copper loss = I^2 * R = 1600
    For half load, I = I/2
    therefore, half load copper loss = (I/2)^2 * R
    = (1/4)*(I)^2 * R
    = (1/4) * 1600
    = 400

  5. Rahul Kumar
    Rahul Kumar :
    5 years ago

    We know that,
    Pcu= x^2* Pcu(full load)
    Pcu= (1/2)^2 *1600w
    = 400w

Related Questions on Transformers

A good voltage regulation of a transformer means

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B. output voltage fluctuation with power factor is least

C. difference between primary and secondary voltage is least

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