Hi friends in this video we are going to
discuss about efficiency of a
transformer by using that we will check
what should be the load on a transformer
in order to get maximum efficiency
efficiency is nothing but a ratio of
output power to input power
now if I am giving the input power
maximum part of input power is go to
output as it is with some losses
happening so what I can say input power
has to supply some of the losses so I
can write efficiencies output power upon
input power I will write like this
output power plus losses now transformer
rating is an apparent power to get a
actual power I need to multiply it
through the power factor so for full
load I can say efficiency equal to
output power so at the second aside I am
applying the load having a power factor
cos Phi 2 with set under side voltage V
2 and secondary side full load current I
2 I will get output power V 2 I 2 cos
Phi 2 upon output power V 2 I 2 cos Phi
2 and there are two losses occurring in
a transformer those are core losses
which also called as iron losses and
second is copper losses so I can write
over here WI plus w CU w CU is a copper
loss depends upon current so I can write
efficiency equal to V 2 I 2 cos Phi 2
upon V 2 I 2
five two plus iron losses copper losses
I will write like this by 2 square R 2 e
r 2 is nothing but resistance referred
to secondary side because load is
connected at the secondary side but this
is a formula for efficiency when full
load is applied but that isn't the case
all the time sometimes we apply
fractional load so for fractional load
we need to define efficiency but before
that let us see what is the fractional
load sometimes it may happen that we
just apply half load or 60% of load so
that fraction is nothing but ratio of
two currents so X is nothing but actual
load upon full load and that is related
to the currents so it is like this
secondary current at actual load upon
secondary side full load current so from
this what I can get I 2 is nothing but X
multiplied by I 2 fl so the formula of
efficiency will become instead of i2 if
I put X into I 2 fl it will be like this
X multiplied by v2 i2 FL cos phi 2 upon
X multiplied by v2 i2 FL
cos Phi 2 plus iron losses WI and
instead of I 2 if I substitute this I
will get X square I to FL square
multiplied by r2 e now full or Quran we
are getting with rated voltage so I can
say this is nothing but X multiplied by
this thing is nothing but V rating of a
transformer cos Phi 2 is power factor wi
iron losses plus X square since full
load current and passing
so hence losses at the secondary will be
full load copper losses so this is the
formula of efficiency which we use for
solving numericals
and any fractional load efficiency you
can calculate with the help of this
formula next we will go to maximum
efficiency
we may say full or means high current is
passing
hence obviously at full load I should
get maximum efficiency but hold on
that is not the case we need to find out
a fraction at which I am getting a full
load and most of the time it is never a
full load then what we will check
shortly first of all we will derive
condition for maximum efficiency for
that I will write equation of efficiency
so efficiency is nothing but V 2 I 2 cos
Phi 2 upon V 2 I 2 cos Phi 2 plus iron
loss plus I 2 square R 2 e copper loss I
am NOT talking about any load whether it
is a full load half load that we need to
determine to get a maximum efficiency
I will differentiate efficiency with
respect to variable what will be the
variable in this equation obviously I -
so differentiate efficiency with respect
to I 2 so D efficiency upon di 2 equal
to u by B rule we have to used it is a V
2 I 2 cos Phi 2 plus WI plus I 2 square
R 2 e multiplied by derivative of this
term which is V 2 cos Phi 2 minus V 2 I
2 cos Phi 2
derivative of this term which is V 2 cos
Phi 2 WI constant losses
hence derivative is zero whole divided
by square of this
I want efficiency maximum so to get
maximum efficiency
I will equate this derivative to zero so
if I do that I will get V 2 I 2 cos Phi
2 plus WI plus I 2 square R 2 e
multiplied by V 2 cos Phi 2 minus V 2 I
2 cos Phi 2 in bracket V 2 cos Phi 2
plus 2 times i - r - e equal to 0 I can
take V 2 cos 5 to common in bracket I
will get V 2 I 2 cos Phi 2 plus WI plus
I 2 square R - E - if V 2 cos Phi 2 I
will take common I 2 I will multiply
with this which will give me V 2 I 2 cos
Phi 2 - 2 I 2 square R 2 e equal to 0
this will get over here cancel out V 2 I
2 cos Phi 2
- Vito 2 cos Phi 2 will also get
cancelled out ultimately I will get WI
and this will be minus I 2 square R 2 e
equal to 0 so I can say WI equal to I 2
square R 2 e
what will WI WI is nothing but ion
losses I to square R to e copper losses
so if I am getting copper losses same as
iron losses then I can say I will get
maximum efficiency so this is the
condition for maximum efficiency so to
get a maximum efficiency from a
transformer I need to make sure that
copper losses which are variable should
be equal to constant losses those are
nothing but core losses now we will
check what should be a load to get
maximum efficiency now let us check what
is the load current at maximum
efficiency
we got a condition for maximum
efficiency I to square R 2 e should be
equal to WI we can nothing but core
losses I 2 is nothing but X multiplied
by I 2 FL so if I substitute I will get
X square I to FL square multiplied by r2
e equal to WI I 2 FL square r2 e is
nothing but full lower copper losses so
X square WC u FL equal to WI so X equal
to root of WI upon WC u FL so this
should be the fraction of full lower
current pass through resistance in order
to get maximum efficiency so while
calculating maximum efficiency first
thing we have to get what is the
factional load we need to connect what
does that mean that is X now what should
be kv a supplied at maximum efficiency
it is very simple KVA at maximum
efficiency is nothing but fraction x x
kv a rating of a transformer and finally
value of maximum efficiency it is given
as we are going to calculate x for
maximum efficiency so x multiplied by v
rating of a transformer x power factor
upon x v rating of a transformer x power
factor plus 2 times iron losses because
we got a condition for maximum
efficiency when copper losses are equal
to iron losses so instead of copper
losses I can put WI hence total losses
will be 2 times W I so this is the value
of maximum efficiency so we have seen
losses based on those losses what is the
condition to get a maximum efficiency
and how to get a efficiency at different
type of fractional loads thank you
