7532V
For a given transformer, the ratio of the number of turns in its primary coil ([tex]N_{p}[/tex]) to the number of turns in its secondary coil ([tex]N_{s}[/tex]) is equal to the ratio of the input voltage ([tex]V_{p}[/tex]) to the output voltage ([tex]V_{s}[/tex]) of the transformer. i.e
[tex]\frac{N_p}{N_s}[/tex] = [tex]\frac{V_p}{V_s}[/tex] ----------------(i)
From the question;
[tex]N_{p}[/tex] = number of turns in the primary coil = 8 turns
[tex]N_{s}[/tex] = number of turns in the secondary coil = 515 turns
[tex]V_{p}[/tex] = input voltage = 117V
Substitute these values into equation (i) as follows;
[tex]\frac{8}{515}[/tex] = [tex]\frac{117}{V_s}[/tex]
Solve for [tex]V_{s}[/tex];
[tex]V_{s}[/tex] = 117 x 515 / 8
[tex]V_{s}[/tex] = 7532V
Therefore, the output voltage (in V) of the transformer is 7532
