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A transformer is used to light a 100 W and 110 V lamp from a 220 V mains. If the main current is 0.5 A , the efficiency of the transformer is approximately
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Verified Answer
The correct answer is:
$90 \%$
The efficiency of transformer
$=\frac{\text { Energy obtained from the secondary coil }}{\text { Energy given to the primary coil }}$
or
$\begin{array}{lrl}
\text { or } & \eta & =\frac{\text { Output power }}{\text { Input power }} \\
\text { or } & \eta & =\frac{V_s I_s}{V_p I_p} \\
\text { Given, } & V_s I_s & =100 \mathrm{~W}, V_p=220 \mathrm{~V}, I_p=0.5 \mathrm{~A} \\
\text { Hence, } & \eta & =\frac{100}{220 \times 0.5}=0.90=90 \%
\end{array}$
$=\frac{\text { Energy obtained from the secondary coil }}{\text { Energy given to the primary coil }}$
or
$\begin{array}{lrl}
\text { or } & \eta & =\frac{\text { Output power }}{\text { Input power }} \\
\text { or } & \eta & =\frac{V_s I_s}{V_p I_p} \\
\text { Given, } & V_s I_s & =100 \mathrm{~W}, V_p=220 \mathrm{~V}, I_p=0.5 \mathrm{~A} \\
\text { Hence, } & \eta & =\frac{100}{220 \times 0.5}=0.90=90 \%
\end{array}$
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