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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
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$90 \%$
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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