Maximum efficiency of a transformer depends on :

the working conditions of technicians

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whether copper loss

=

1/2 x iron loss

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whether copper loss

=

iron loss

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whether copper loss

=

2 x iron loss

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Solution

The correct option is C whether copper loss $=$ iron loss
The efficiency of the transformer is defined as ratio of useful power output to the input power. The two being measured is the same unit. Its unit is either watts or kw. Transformer efficiency is denoted by $η$

$η = \frac{output power}{input power} == \frac{output power}{input power + losses}$

$η = \frac{output power}{input power + iron losses of copper + losses}$

$η = \frac{V_{2} I_{2} cos ϕ_{2}}{V_{2} I_{2} cos ϕ_{2} + P_{i} + P_{c}}$

Where

$V_{2} =$ secondary terminal voltage

$I_{2} =$ Full load secondary current.

$cos ϕ_{2} =$ Power factor of the load

$P_{i} =$ Iron losses = constant loss = exactly current loss + hysteresis loss

$P_{c} =$ full load copper loss

The efficiency is a function of load i.e. lead current $I_{2}$ assuming $cos ϕ$ constant. The secondary terminal voltage $V_{2}$ is also assumed constant.

$\frac{d η}{d I_{2}} = 0$

Now $η = \frac{V_{2} I_{2} cos ϕ_{2}}{V_{2} I_{2} cos ϕ_{2} + P_{i} + I_{2}^{2} R_{2 e}}$

$\frac{d η}{d I_{2}} = \frac{d}{d I_{2}} (\frac{V_{2} I_{2} cos ϕ_{2}}{V_{2} I_{2} cos ϕ_{2} + P_{i} + I_{2}^{2} R_{2 e}}) = 0$

$(V_{2} I_{2} cos ϕ_{2} + P_{i} + I_{2}^{2} R_{2 e}) \frac{d}{d I_{2}} (V_{2} I_{2} cos ϕ_{2}) - (V_{2} I_{2} cos ϕ_{2}) \frac{d}{d I_{2}} (V_{2} I_{2} cos ϕ_{2} + P_{i} + I_{2}^{2} R_{2 e}) = 0$

$(V_{2} I_{2} cos ϕ_{2} + P_{i} + I_{2}^{2} R_{2 e}) (V_{2} cos ϕ_{2}) - (V_{2} I_{2} cos ϕ_{2}) (V_{2} cos ϕ_{2} + 2 I_{2} R_{2 e}) = 0$

Cancelling $V_{2} cos ϕ_{2}$ from both the terms we get

$V_{2} I_{2} cos ϕ_{2} + P_{i} + I_{2}^{2} R_{2 e} - V_{2} I_{2} cos ϕ_{2} - 2 I^{2} R_{2 e} = 0$

$P_{i} - I^{2} R_{2 e} = 0$

$P_{i} = I^{2} R_{2 e} = P_{C u}$

$P_{i} = P_{C u}$

So, condition to achieve maximum efficiency is that

Copper Loss = Iron Loss

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