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If a step-up transformer have turn ratio 5, frequency $50 H z$ root mean square value of potential difference of primary winding is 100 volts and the resistance of the secondary winging is $500 Ω$ then the peak value of voltage in secondary winding will be (the efficiency of the transformer is hundred percent)

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$50 \sqrt{2}$

$10 \sqrt{2}$

$500 \sqrt{2}$

$20 \sqrt{2}$

answer is A.

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Detailed Solution

$v_{0} = \sqrt{2} v_{r m s} = 100 \sqrt{2} \frac{V_{S}}{V_{P}} = \frac{N_{s}}{N_{P}} \Rightarrow V_{S} = \frac{N_{S}}{N_{P}} \times V_{P} = 5 \times 100 \sqrt{2} = 500 \sqrt{2}$

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