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(A) For an ideal transformer,the ratio of the secondary voltage $(V_s)$ to the primary voltage $(V_p)$ is equal to the ratio of the number of turns in

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$50$

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(A) For an ideal transformer,the ratio of the secondary voltage $(V_s)$ to the primary voltage $(V_p)$ is equal to the ratio of the number of turns in the secondary $(N_s)$ to the number of turns in the primary $(N_p)$: $\frac{V_s}{V_p} = \frac{N_s}{N_p}$.The given primary voltage is the peak value,$V_{p(peak)} = 28 \, V$. The r.m.s. value of the primary voltage is $V_{p(rms)} = \frac{V_{p(peak)}}{\sqrt{2}} = \frac{28}{\sqrt{2}} \, V$.Using the transformer equation: $V_{s(rms)} = V_{p(rms)} 	imes \frac{N_s}{N_p}$.Substituting the values: $V_{s(rms)} = \left( \frac{28}{\sqrt{2}} ight) 	imes \left( \frac{250}{100} ight)$.$V_{s(rms)} = \left( \frac{28}{1.414} ight) 	imes 2.5 \approx 19.8 	imes 2.5 = 49.5 \, V$.Rounding to the nearest integer,we get $V_{s(rms)} \approx 50 \, V$.

An ideal transformer has $100$ turns in the primary and $250$ turns in the secondary. The peak value of the $ac$ input voltage is $28 \, V$. The r.m.s. secondary voltage is nearest to.....$V$

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