If a step-up transformer has a turn ratio of | vedclass

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(A) The turn ratio of the transformer is given as $k = \frac{N_s}{N_p} = 5$.The $RMS$ voltage of the primary coil is $V_{p(rms)} = 100 \, V$.For an id

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$500 \sqrt{2} \, V$

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(A) The turn ratio of the transformer is given as $k = \frac{N_s}{N_p} = 5$.The $RMS$ voltage of the primary coil is $V_{p(rms)} = 100 \, V$.For an ideal transformer,the relation between the secondary voltage $(V_s)$ and primary voltage $(V_p)$ is given by $\frac{V_s}{V_p} = \frac{N_s}{N_p}$.Therefore,the $RMS$ voltage of the secondary coil is $V_{s(rms)} = 5 	imes 100 \, V = 500 \, V$.The peak value of the voltage $(V_0)$ is related to the $RMS$ value by the formula $V_0 = \sqrt{2} 	imes V_{rms}$.Thus,the peak value of the voltage in the secondary winding is $V_{s(peak)} = \sqrt{2} 	imes 500 = 500 \sqrt{2} \, V$.

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