Electromagnetic Induction And Alternating Current Ques 36
36. A transformer consisting of 300 turns in the primary and 150 turns in the secondary gives output power of $2.2 \mathrm{~kW}$. If the current in the secondary coil is $10 \mathrm{~A}$, then the input voltage and current in the primary coil are
(Main 2019, 10 April I)
(a) $440 \mathrm{~V}$ and $5 \mathrm{~A}$
(b) $220 \mathrm{~V}$ and $20 \mathrm{~A}$
(c) $220 \mathrm{~V}$ and $10 \mathrm{~A}$
(d) $440 \mathrm{~V}$ and $20 \mathrm{~A}$
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Answer:
Correct Answer: 36.(a)
Solution:
Formula:
- Given, Number of turns in primary, $N_{1}=300$
Number of turns in secondary, $N_{2}=150$
Output power, $P_{2}=2.2 \mathrm{~kW}=2.2 \times 10^{3} \mathrm{~W}$
Current in secondary coil, $I_{2}=10 \mathrm{~A}$
Output power, $\mathrm P _ 2 =\mathrm I _ 2 \mathrm ~V _ 2 $
$$ \Rightarrow \quad V_{2}=\frac{P_{2}}{I_{2}}=\frac{2.2 \times 10^{3}}{10}=220 \mathrm{~V} $$
We know that,
$$ \begin{aligned} & \frac{N_{1}}{N_{2}}=\frac{\text { Input voltage }}{\text { Output voltage }}=\frac{V_{1}}{V_{2}} \Rightarrow V_{1}=\frac{N_{1}}{N_{2}} V_{2} \\ & \Rightarrow \quad V_{1}=\frac{300}{150} \times(220 \mathrm{~V}) \quad \text { [using Eq. (i)] } \\ & V_{1}=440 \mathrm{~V} \\ & \text { Again, } \quad \frac{V_{1}}{V_{2}}=\frac{I_{2}}{I_{1}} \\ & \Rightarrow \quad I_{1}=\frac{V_{2}}{V_{1}} \quad I_{2}=\frac{220}{440} \times 10 \\ & \Rightarrow \quad I_{1}=5 \mathrm{~A} \end{aligned} $$
BETA
