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It is possible to attain maximum efficiency in a transformer when the
core losses are equal to rated full load copper losses
core losses are more than rated full load copper losses
core losses and full load copper losses are constant
copper loss also becomes constant
core losses are equal to rated full load copper losses
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.
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.
Pcu,varтАЛ=PiтАЛ тАФ Condition for maximum efficiency
╬╖=nScos╧Х+PiтАЛ+n2PcuтАЛnScos╧ХтАЛ тАФ General efficiency equation
The efficiency ╬╖ of a transformer is given by ╬╖=nScos╧Х+PiтАЛ+n2PcuтАЛnScos╧ХтАЛ, where n is the fraction of full load, S is the kVA rating, PiтАЛ is core loss, and PcuтАЛ is full load copper loss. Differentiating with respect to n and setting to zero yields n2PcuтАЛ=PiтАЛ, which implies that the load-dependent copper loss must equal the load-independent core loss.
Core losses are constant regardless of load current.
Copper losses are proportional to the square of the load current (I2R).
Maximum efficiency occurs at a specific fraction 'n' of the full load.
For most transformers, max efficiency is designed to occur near 50-75% of full load.
Minimizes heat dissipation at operating load.
Optimizes material usage for specific power ratings.
Requires precise transformer design during manufacturing.
Efficiency drops if the load deviates significantly from the designed 'n' value.
Power system planning for steady loads.
Transformer rating optimization in distribution networks.
Core loss (PiтАЛ) comprises hysteresis and eddy current losses.
Copper loss (PcuтАЛ) comprises I2R losses in primary and secondary windings.
Option B is incorrect because unequal losses lead to non-optimal energy conversion.
Option C is incorrect because copper losses are inherently variable with load, not constant.
A is correct тАФ Maximum efficiency is achieved when the variable copper losses equal the constant core losses.
Always remember that for any machine (DC motors or transformers), maximum efficiency occurs when the variable losses equal the constant losses.