In transformers, which of the following statements is valid

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ISRO (VSSC) Technical Assistant Electrical 2017 Official Paper
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  1. In an open circuit test, copper losses are obtained while in short circuit test, core losses are obtained
  2. In an open circuit test, current is drawn at high power factor
  3. In a short circuit test, current is drawn at zero power factor
  4. In an open circuit test, current is drawn at low power factor

Answer (Detailed Solution Below)

Option 4 : In an open circuit test, current is drawn at low power factor
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Detailed Solution

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Open circuit test:

F1 J.P 18.8.20 Pallavi D22

  • It is done by keeping one of the windings open (without load, usually high voltage winding is open) and applying rated voltage to other winding (usually low voltage winding because it is easier to apply rated voltage).
  • The current drawn from this terminal is the no-load current at a low power factor corresponding to the core loss component. Since the no-load current is very small it doesn’t contribute to the copper loss. Core loss is calculated by multiplying the applied voltage and no-load current.
  • As the secondary side is open, the entire coil will be purely inductive in nature. So, the power will be lagging due to the inductive property of the circuit. So LPF (Low Power Factor) Wattmeter is used in open circuit test of transformer.

It is used to find

  • The core loss of the transformer
  • The no-load current
  • Equivalent resistance referred to the metering side

 

Short circuit test:

SSC JE Electrical 52 17Q Jan 27 Second Shift Machines Hindi images Q4

  • It is done by shorting one of the winding terminals (usually low voltage terminal) and applying a small voltage across the other winding terminals (high voltage terminal because the current in HV terminal will be less and easy to handle) and using a wattmeter to measure the power dissipated in the LV terminal. The wattmeter will indicate the full load copper loss.
  • In the short circuit test, the secondary winding of the transformer is short-circuited. As the secondary side is short-circuited the entire coil will be purely resistive in nature. So, the power factor will be High or unity.

 

A short circuit test is done to find

  • The full load copper loss
  • Short circuit current
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