A) 30%
B) 50%
C) 90%
D) 10%
Correct Answer: C
Solution :
The efficiency of transformer \[\text{=}\,\frac{\text{Energy}\,\text{obtained}\,\text{from}\,\text{the}\,\text{secondary}\,\text{coil}}{\text{Energy}\,\text{given}\,\text{to}\,\text{the}\,\text{primary}\,\text{coil}}\] or \[\eta =\frac{\text{Output}\,\text{power}}{\text{Input}\,\text{power}}\] or \[\eta =\frac{{{V}_{s}}{{I}_{s}}}{{{V}_{p}}{{I}_{p}}}\] Given, \[{{V}_{s}}{{I}_{s}}=100\,W,{{V}_{p}}=220\,V,{{I}_{p}}=0.5A\,\] Hence, \[\eta =\frac{100}{220\times 0.5}=0.90=90%\] Note The efficiency of an ideal transformer is 1 (or 100%). But in practice due to loss in energy, theefficiency of transformer is always less than 1 (or less than 100%).
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