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ElectricalBasic Electrical
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It is possible to attain maximum efficiency in a transformer when the

A

core losses are equal to rated full load copper losses

B

core losses are more than rated full load copper losses

C

core losses and full load copper losses are constant

D

copper loss also becomes constant

Correct Answer

тЪЩя╕П TE тАв Technical Concept & PrincipleElectricalBasic Electrical
Option A

core losses are equal to rated full load copper losses

Quick Summary:

Maximum efficiency in a transformer occurs when the variable losses (copper losses) are equal to the constant losses (core or iron losses). This is a consequence of setting the derivative of the efficiency equation with respect to load current to zero.

тЪЩя╕ПTETechnical SolutionConcept & Principle
ЁЯТб Explanation

Maximum efficiency in a transformer occurs when the variable losses (copper losses) are equal to the constant losses (core or iron losses). This is a consequence of setting the derivative of the efficiency equation with respect to load current to zero.

ЁЯФв Key Formulas

Pcu,var=PiP_{cu,var} = P_iPcu,varтАЛ=PiтАЛ тАФ Condition for maximum efficiency

╬╖=nScosтБб╧ХnScosтБб╧Х+Pi+n2Pcu\eta = \frac{n S \cos \phi}{n S \cos \phi + P_i + n┬▓ P_{cu}}╬╖=nScos╧Х+PiтАЛ+n2PcuтАЛnScos╧ХтАЛ тАФ General efficiency equation

тЪЩя╕П Working Principle

The efficiency ╬╖\eta╬╖ of a transformer is given by ╬╖=nScosтБб╧ХnScosтБб╧Х+Pi+n2Pcu\eta = \frac{n S \cos \phi}{n S \cos \phi + P_i + n┬▓ P_{cu}}╬╖=nScos╧Х+PiтАЛ+n2PcuтАЛnScos╧ХтАЛ, where nnn is the fraction of full load, SSS is the kVA rating, PiP_iPiтАЛ is core loss, and PcuP_{cu}PcuтАЛ is full load copper loss. Differentiating with respect to nnn and setting to zero yields n2Pcu=Pin┬▓ P_{cu} = P_in2PcuтАЛ=PiтАЛ, which implies that the load-dependent copper loss must equal the load-independent core loss.

ЁЯУМ Key Points
  • тЦ╕

    Core losses are constant regardless of load current.

  • тЦ╕

    Copper losses are proportional to the square of the load current (I2RI^2RI2R).

  • тЦ╕

    Maximum efficiency occurs at a specific fraction 'n' of the full load.

  • тЦ╕

    For most transformers, max efficiency is designed to occur near 50-75% of full load.

тЬЕ Advantages
  • тЦ╕

    Minimizes heat dissipation at operating load.

  • тЦ╕

    Optimizes material usage for specific power ratings.

тЭМ Disadvantages / Limitations
  • тЦ╕

    Requires precise transformer design during manufacturing.

  • тЦ╕

    Efficiency drops if the load deviates significantly from the designed 'n' value.

ЁЯЫая╕П Applications / Uses
  • тЦ╕

    Power system planning for steady loads.

  • тЦ╕

    Transformer rating optimization in distribution networks.

ЁЯУД Additional Information
  • тЦ╕

    Core loss (PiP_iPiтАЛ) comprises hysteresis and eddy current losses.

  • тЦ╕

    Copper loss (PcuP_{cu}PcuтАЛ) comprises I2RI^2RI2R losses in primary and secondary windings.

  • тЦ╕

    Option B is incorrect because unequal losses lead to non-optimal energy conversion.

  • тЦ╕

    Option C is incorrect because copper losses are inherently variable with load, not constant.

ЁЯУК Diagram / Illustration
Condition for Max Efficiency
Variable Copper Loss (Pcu,varP_{cu,var}Pcu,varтАЛ)
Constant Core Loss (PiP_iPiтАЛ)
Pcu,var=PiP_{cu,var} = P_iPcu,varтАЛ=PiтАЛ
тЬЕ

A is correct тАФ Maximum efficiency is achieved when the variable copper losses equal the constant core losses.

Core Concepts Used
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Transformer Efficiency Iron Loss (Core Loss) Copper Loss (I^2R Loss)
ЁЯТб EXAM TIP

Always remember that for any machine (DC motors or transformers), maximum efficiency occurs when the variable losses equal the constant losses.

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