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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 $230 \,V$ by the transformer. If the current in the primary of the transformer is $5A$ and its efficiency is $90\%$, the output current (in ampere) would be:

Alternating Current

Solution:

Efficiency, $\eta = \frac{P_{out}}{P_{in}} = \frac{V_s I_s}{V_p I_p}$
$ \Rightarrow 0.9 = \frac{230 \times I_s}{2300 \times 5}$
$\Rightarrow I_s = 0.9 \times 50 = 45\,A$
Output current $= 45\, A$