MHT CET · Physics · Alternating Current
A transformer is used to light a \(100 \mathrm{~W}\) and \(110 \mathrm{~V}\) lamp from a \(220 \mathrm{~V}\) mains. If the main current is \(0.5 \mathrm{~A}\), the efficiency of the transformer is approximately
- A \(30 \%\)
- B \(50 \%\)
- C \(90 \%\)
- D \(10 \%\)
Answer & Solution
Correct Answer
(C) \(90 \%\)
Step-by-step Solution
Detailed explanation
The efficiency of transformer \(=\frac{\text { Energy obtained from the secondary coil }}{\text { Energy given to the primary coil }}\)
Or \(\eta=\frac{\text { Output power }}{\text { Input power }}\)
or
\(
\eta=\frac{V_{s} I_{s}}{V_{p} I_{p}}
\)
Given, \(\quad V_{s} I_{s}=100 \mathrm{~W}, V_{p}=220 \mathrm{~V}, I_{p}=0.5 \mathrm{~A}\)
\(
\text { Hence, } \quad \eta=\frac{100}{220 \times 0.5}=0.90=90 \%
\)
Note The efficiency of an ideal transformer is 1 (or \(100 \%\) ). But in practice due to loss in energy, the efficiency of transformer is always less than 1 (or less than \(100 \%\) ).
Or \(\eta=\frac{\text { Output power }}{\text { Input power }}\)
or
\(
\eta=\frac{V_{s} I_{s}}{V_{p} I_{p}}
\)
Given, \(\quad V_{s} I_{s}=100 \mathrm{~W}, V_{p}=220 \mathrm{~V}, I_{p}=0.5 \mathrm{~A}\)
\(
\text { Hence, } \quad \eta=\frac{100}{220 \times 0.5}=0.90=90 \%
\)
Note The efficiency of an ideal transformer is 1 (or \(100 \%\) ). But in practice due to loss in energy, the efficiency of transformer is always less than 1 (or less than \(100 \%\) ).
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