Question

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

• A. $50\, A$
• B. $45\, A$
• C. $25\, A$
• D. $20\, A$

Answer 1

Answer: (B)

$45\, A$

Answer 2

We know that, efficiency is given by
$\eta=\frac{\text { output power}}{\text { input ower }}$
$=\frac{E_{ s } \cdot I_{ s }}{E_{ p } \cdot I_{ p }}$
Here, $\eta=90 \%, E_{ p }=2300\, V ,\, I_{ p }=5\, A$ and $E_{ s }=230\, V$
Therefore, $\frac{90}{100} =\frac{230 \times I_{ s }}{2300 \times 5}$
$I_{ s } =\frac{90 \times 2300 \times 5}{230 \times 100}=45 \,A$