KCET · Physics · Current Electricity
A transformer is used to light \( 100 \mathrm{~W}-110 \mathrm{~V} \) lamp from \( 220 \mathrm{~V} \) mains. If the main current is \( 0.5 \) A, the efficiency of the transformer is
- A \( 90 \% \)
- B \(95\%\)
- C \( 96 \% \)
- D \( 99 \% \)
Answer & Solution
Correct Answer
(A) \( 90 \% \)
Step-by-step Solution
Detailed explanation
Given, transformer lights \(100 \mathrm{~W}-100 \mathrm{~V}\) lamp; input voltage \(=220 \mathrm{~V}\); input current \(=0.5 \mathrm{~A}\)
Therefore, efficiency \(\eta=\frac{\text { Output power }}{\text { Input power }} \times 100\)
Here, output power \(=100 \mathrm{~W}\)
Input power \(=\) Input voltage \(\times\) Input current \(=220 \times 0.5=110 \mathrm{~W}\)
Thus, \(\eta=\frac{100}{110} \times 100=90 \%\)
Therefore, efficiency \(\eta=\frac{\text { Output power }}{\text { Input power }} \times 100\)
Here, output power \(=100 \mathrm{~W}\)
Input power \(=\) Input voltage \(\times\) Input current \(=220 \times 0.5=110 \mathrm{~W}\)
Thus, \(\eta=\frac{100}{110} \times 100=90 \%\)
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