As we know Cu loss depends on current & iron loss depends
on voltage and not affected by power fector hence always
transformer ratting defined in KVA not in Kw.

actually rating of the machine (kva or kw) depends upon the
power factor since the load power factor to which
transformer is supplying power is not known , it may be
capacitive ,inductive , or resistive that is why its rating
is in kva and not in kw

Since Transformer is a Static device and it is having two
losses namely core loss and Cu loss. The core loss depends
on the Voltage (V) only, where Cu loss depends on Current
(I) only.So for this purpose the transformer always called
its rating in apparent power only(VA).

bcoz ..cu loss depends the current.iron loss depends the
voltage..so that total loss of transformer is depends
voltage,current...so the transformer ratings in KVA..

Transformer is a static device having two type of losses
such as ironlossed and cupper losess. since iron losses
depends upon voltage(V) and cupper lossed depends upon on
the curret(I) so rating of Transformer is in KVA

The leading/lagging power factor has
magnetizing/demagnetizing effect on armature reaction in a
rotating electrical machine. Hence the losses depends on
the load power factor. Unity PF has neither magnetizing nor
de-magnetizing effect.

Whereas in a transformer, the power factor doesn't play any
role in machine losses i.e. Cu & core losses are
independent of power factor.

Hence the rating of transformer is always given in KVA and
the rating of generators is given in kW at certain pf i.e.
usually at 0.8 lag.

The copper loss of a Transformer depends on current i.e.
I^2 R and the iron loss depends on voltage. Hence the total
losses depends on volt-ampere (VA) and not on phase angle
between voltage & current.

Therefor the losses in a transformer are independent of
power factor which is cosine of angle between voltage &
current.

That's why the rating of transformers is in KVA and not in
kW.

Now coming to Efficiency of a transformer: Yes it depends
on power factor of load. Because the
Efficiency = Output/Input
= (Input - Losses) / Input
= 1 - (Losses / Input)
= 1 - [Losses / (Vs * Is * Cos phi + Losses)]

Efficiency is inverted V curve (Y-scale) and the peak of
curve appears between 50-60% load (X-scale)for various load
power factors. Efficiency increases as the power factor
increases.

kVA is the unit for apparent power. Apparent power consists
of active and reactive power. Active power is the share of
the apparent power which transmits energy from the source
(generator) to the user. Reactive power is the share of the
apparent power which represents a useless oscillation of
energy from the source to the user and back again. It
occurs when on account of some »inertia« in the system
there is a phase shift between voltage and current. This
means that the current does not change polarity synchronous
with the voltage. But the heat generated in a winding as
well as the eddy current losses generated in a transformer
core depend on the current only, regardless of whether it
aligns with the voltage or not. Therefore the heat is
always proportional to the square of the current amplitude,
irrespective of the phase angle (the shift between voltage
and current). So a transformer has to be rated (and
selected) by apparent power.

The X-mer o/p is limited by heating and hence by the losses
in the X-mer. There are two kinds of losses in the X-mer:
copper losses (ohmic losses) and iron losses. ohmic losses
(I*I*R) depend upon current while core losses depend upon X-
mer voltage and are almost unaffected by X-mer P.F.. Hence
the rated o/p is expressed in VA (V*I) or in KVA and not in
KW.

Transformer is the statics device.actually rating of the
machine (kva or kw) depends upon the power factor since the
load power factor to which transformer is supplying power
is not known , it may be capacitive ,inductive , or
resistive that is why its rating is in kva and not in kw.
P=VIcosų
cosų is the depen upon the load & load not cosider the
manufacturing compony of transformer.then only VI
consider.so its rating is in kva.

a wire in which bulb is connected
there is two terminal
when one terminal is connected to phase than why the value of
current shock in nutral terminal is max. or large??