Page 566 - Basic Electrical Engineering

at one-third of full load copper loss

2
Thus, at x load, copper loss = x W cu

satisfying the condition for maximum efficiency

2
W = x W cu
i

or,

where W is the full-load copper loss and x is the fraction of the full-load at
cu
which efficiency will be maximum.
If we want to know the kVA of the transformer at maximum efficiency, we

would determine it as follows:

kVA at η max = x × Full-load kVA

Therefore, kVA at maximum efficiency = × Full-load kVA

Efficiency of a transformer is often expressed in terms of energy output in
24 hours in a day to the energy input. Such a calculated efficiency is known
as all-day efficiency which is explained as follows.

6.15 ALL-DAY EFFICIENCY

A transformer when connected to the load has to remain energized all the

time ready to supply the load connected to it. Even when all the loads are
switched off, i.e., no one is utilizing any electricity, the transformer has to

remain on. Thus, irrespective of the load on the transformer, the core loss will
occur for all the 24 hours of the day. However, the copper loss will depend on