Due to the losses in a transformer, the output power of a transformer is less than the input power supplied.

**.**Power output = Power input - Total losses

^{.}.**.**Power input = Power output + Total losses

^{.}. = Power output + P

_{i}_{ }+ P_{cu}_{ }_{ } The efficiency of any device is defined as the ratio of the power output to power input. So for a transformer the efficiency can be expresses as,

**.**

^{.}._{i}

_{ }+ P

_{cu}

_{ })

Now power output = V

_{2}_{ }I_{2}_{ }cos Φwhere cos Φ = Load power factor

The transformer supplies full load of current I

_{2}_{ }and with terminal voltage V_{2}. P

_{cu}_{ }= Copper losses on full load = I_{2}^{2}_{ }R_{2e }**.**η = (V

^{.}._{2}

_{ }I

_{2}

_{ }cos Φ

_{2}

_{ })/(V

_{2}

_{ }I

_{2}

_{ }cos Φ

_{2}

_{ }+ P

_{i}

_{ }+ I

_{2}

^{2}

_{ }R

_{2e})

But V

_{2}_{ }I_{2}= VA rating of a transformer This is full load percentage efficiency with,

I

_{2}= Full load secondary current But if the transformer is subjected to fractional load then using the appropriate values of various quantities, the efficiency can be obtained.

Let n =Fraction by which load is less than full load = Actual load/Full load

For example, if transformer is subjected to half load then,

n = Half load/Full load = (1/2)/2 = 0.5

when load changes, the load current changes by same proportion.

new P

In general for fractional load the efficiency is given by,

i) The total copper losses on full load.

ii) The efficiency while supplying full load at 0.9 p.f. lagging.

iii) The efficiency while supplying half load at 0.8 p.f. leading.

Assume total iron losses equal to 60 W.

V

K = 400/200 = 2

(I

(i) Total copper losses on full load,

(P

(ii) cos Φ = 0.9 lagging and full load

(iii) cos Φ = 0.8 leading, half load

As half load, n = 0.5

For example, if transformer is subjected to half load then,

n = Half load/Full load = (1/2)/2 = 0.5

when load changes, the load current changes by same proportion.

**.**new I^{.}._{2}= n (I_{2}) F.L. Similarly the output V

Similarly as copper losses are proportional to square of current then,_{2}_{ }I_{2 }cosΦ_{2}_{ }also reduces by the same fraction. Thus fraction of VA rating is available at the output.new P

_{cu}_{ }= n^{2}_{ }(P_{cu}_{ }) F.L.**Key Point**: So copper losses get reduced by n^{2}.In general for fractional load the efficiency is given by,

where n = Fraction by which load power factor lagging, leading and unity the efficiency expression does not change, and remains same.

**Example**: A 4 KVA, 200/400 V, 50 Hz, single phase transformer has equivalent resistance referred to primary as 0.15 Ω Calculate,i) The total copper losses on full load.

ii) The efficiency while supplying full load at 0.9 p.f. lagging.

iii) The efficiency while supplying half load at 0.8 p.f. leading.

Assume total iron losses equal to 60 W.

**Solution**: The given values are,V

_{1}= 200 V, V_{2}= 400 V, S = 4 KVA, R_{1e}= 0.15 Ω , P_{i }= 60 WK = 400/200 = 2

**.**R^{.}._{2e }= K^{2}R_{1e }= (2)^{2}x 0.15 = 0.6 Ω(I

_{2})F.L. = KVA/V_{2}= 4 x 10^{3}/400 = 10 A(i) Total copper losses on full load,

(P

_{cu})F.L. = {(I_{2}) F.L.}^{2}R_{2e }= (10)^{2}x 0.6 = 60 W(ii) cos Φ = 0.9 lagging and full load

(iii) cos Φ = 0.8 leading, half load

As half load, n = 0.5

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