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:

  1. 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

$$ \begin{array}{rlrl} & \eta & =\frac{\text { Output (secondary) power }}{\text { Input (primary) power }} \\ \Rightarrow \quad & \eta & =\frac{V _S \cdot I _S}{V _P \cdot I _P} \times 100 \\ \Rightarrow \quad \eta & =\frac{(230)\left(I _S\right)}{(2300)(5)} \times 100 \\ 90 & =\frac{230 I _S}{(2300) \times 5} \times 100 \\ \Rightarrow \quad & I _S & =45 A \end{array} $$



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