Electromagnetic Induction and Alternating Current 3 Question 1
####1. A transformer consisting of 300 turns in the primary and 150 turns in the secondary gives output power of $2.2 kW$. If the current in the secondary coil is $10 A$, then the input voltage and current in the primary coil are
(Main 2019, 10 April I)
(a) $440 V$ and $5 A$
(b) $220 V$ and $20 A$
(c) $220 V$ and $10 A$
(d) $440 V$ and $20 A$
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Answer:
Correct Answer: 1. (a)
Solution:
- Given, Number of turns in primary, $N _1=300$
Number of turns in secondary, $N _2=150$
Output power, $P _2=2.2 kW=2.2 \times 10^{3} W$
Current in secondary coil, $I _2=10 A$
Output power, $\mathrm P _ 2 =\mathrm I _ 2 \mathrm ~V _ 2 $
$$ \Rightarrow \quad V _2=\frac{P _2}{I _2}=\frac{2.2 \times 10^{3}}{10}=220 V $$
We know that,
$$ \begin{aligned} & \frac{N _1}{N _2}=\frac{\text { Input voltage }}{\text { Output voltage }}=\frac{V _1}{V _2} \Rightarrow V _1=\frac{N _1}{N _2} V _2 \\ & \Rightarrow \quad V _1=\frac{300}{150} \times(220 V) \quad \text { [using Eq. (i)] } \\ & V _1=440 V \\ & \text { Again, } \quad \frac{V _1}{V _2}=\frac{I _2}{I _1} \\ & \Rightarrow \quad I _1=\frac{V _2}{V _1} \quad I _2=\frac{220}{440} \times 10 \\ & \Rightarrow \quad I _1=5 A \end{aligned} $$
