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(C) The transformer equation relating the voltage and the number of turns is given by:$\frac{V_s}{V_p} = \frac{N_s}{N_p}$Given:Number of turns in prim

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$3000$

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(C) The transformer equation relating the voltage and the number of turns is given by:$\frac{V_s}{V_p} = \frac{N_s}{N_p}$Given:Number of turns in primary coil,$N_p = 20$Number of turns in secondary coil,$N_s = 100$Maximum primary voltage,$V_{p, 	ext{max}} = 600 	ext{ V}$To find the maximum secondary voltage $(V_{s, 	ext{max}})$,we use the ratio:$V_{s, 	ext{max}} = \frac{N_s}{N_p} 	imes V_{p, 	ext{max}}$$V_{s, 	ext{max}} = \frac{100}{20} 	imes 600$$V_{s, 	ext{max}} = 5 	imes 600 = 3000 	ext{ V}$Thus,the maximum value of the secondary output voltage is $3000 	ext{ V}$.

$A$ transformer has $20$ turns in the primary and $100$ turns in the secondary coil. An $AC$ voltage of $V_{\text{in}} = 600 \sin 314t$ is applied to the primary terminal of the transformer. The maximum value of the secondary output voltage obtained in volts is:

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