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Byju's Answer

Standard XII

Physics

Transformer

If a transfor...

Question

If a transformer have turn ratio $5$ , frequency $50 H z$ 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 in secondary winding will be (the efficiency of the transformer is hundred percent)

500 \sqrt{2}

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10 \sqrt{2}

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50 \sqrt{2}

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20 \sqrt{2}

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Solution

The correct option is D $20 \sqrt{2}$
The rms voltage in the secondary winding is given as,
$T u r n r a t i o = \frac{V_{p}}{V_{s}}$
$5 = \frac{100}{V_{s}}$
$V_{s} = 20 V$
the peak value of voltage in secondary winding $= \sqrt{2} \times V_{r m s} = 20 \sqrt{2} V$

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

The correct option is D 20√2The rms voltage in the secondary winding is given as,Turnratio=VpVs 5=100Vs Vs=20V the peak value of voltage in secondary winding =√2×Vrms=20√2V

If a transformer have turn ratio $5$ , frequency $50 H z$ 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 in secondary winding will be (the efficiency of the transformer is hundred percent)

The correct option is D $20 \sqrt{2}$
The rms voltage in the secondary winding is given as,
$T u r n r a t i o = \frac{V_{p}}{V_{s}}$
$5 = \frac{100}{V_{s}}$
$V_{s} = 20 V$
the peak value of voltage in secondary winding $= \sqrt{2} \times V_{r m s} = 20 \sqrt{2} V$