Electromagnetic Induction and Alternating Current 7 Question 2
2. 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 $5 A$ and its efficiency is $90 %$, the output current would be
(Main 2019, 9 Jan II)
(a) $45 A$
(b) $50 A$
(c) $25 A$
(d) $35 A$
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Answer:
Correct Answer: 2. (a)
Solution:
- For a transformer, there are two circuits which have $N _p$ and $N _s$ (number of coil turns), $I _p$ and $I _S$ (currents) respectively as shown below.
Here, input voltage, $V _p=2300 V$
Number of turns in primary coil, $N _P=4000$
Output voltage, $V _S=230$ volt
Output power, $P _S=V _S \cdot I _S$
Input power, $P _P=V _P I _P$ $\therefore$ The efficiency of the transformer is
$$ \begin{array}{rlrl} & \eta & =\frac{\text { Output (secondary) power }}{\text { Input (primary) power }} \\ \Rightarrow \quad & \eta & =\frac{V _S \cdot I _S}{V _P \cdot I _P} \times 100 \\ \Rightarrow \quad \eta & =\frac{(230)\left(I _S\right)}{(2300)(5)} \times 100 \\ 90 & =\frac{230 I _S}{(2300) \times 5} \times 100 \\ \Rightarrow \quad & I _S & =45 A \end{array} $$
