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