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 :
- $50\, A$
- $45\, A$
- $25\, A$
- $20\, A$
The Correct Option is B
Solution and Explanation
$\eta=\frac{\text { output power}}{\text { input ower }}$
$=\frac{E_{ s } \cdot I_{ s }}{E_{ p } \cdot I_{ p }}$
Here, $\eta=90 \%, E_{ p }=2300\, V ,\, I_{ p }=5\, A$ and $E_{ s }=230\, V$
Therefore, $\frac{90}{100} =\frac{230 \times I_{ s }}{2300 \times 5}$
$I_{ s } =\frac{90 \times 2300 \times 5}{230 \times 100}=45 \,A$
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Concepts Used:
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An alternating current can be defined as a current that changes its magnitude and polarity at regular intervals of time. It can also be defined as an electrical current that repeatedly changes or reverses its direction opposite to that of Direct Current or DC which always flows in a single direction as shown below.
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