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If an ideal transformer have turn ratio 5, frequency 50 Hz, root mean square value of potential difference on primary 100 volts and the resistance of the secondary winding is 500 $Ω$ , then the peak value of voltage across secondary winding will be (the efficiency of the transformer is hundred percent)

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$50 \sqrt{2} volt$

$20 \sqrt{2} volt$

$10 \sqrt{2} volt$

$500 \sqrt{2} volt$

answer is A.

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Detailed Solution

Use $\frac{v_{1}}{v_{2}} = \frac{N_{1}}{N_{2}}$

$\Rightarrow \frac{100 \sqrt{2}}{V_{2}} = \frac{1}{5}$

$\Rightarrow V_{2} = 500 \sqrt{2} volt$

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