Q. A power transmission line feeds input power at 2300 V to a step down transformer with its primary windings having 4000 turns. The output power is delivered at 230 V bv the transformer. If the current in the primary of the transformer is 5A and its efficiency is 90 %, the output current would be :

Solution:

$\eta = \frac{P_{out }}{P_{in}} = \frac{V_{S}I_{S}}{V_{P}I_{P}} $
$ \Rightarrow 0.9 = \frac{23 \times I_{S}}{230 \times5} $
$ \Rightarrow I_{S} = 45A $

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