Answer:
[tex]N_s\approx41667 \hspace{3}lo ops[/tex]
Explanation:
In an ideal transformer, the ratio of the voltages is proportional to the ratio of the number of turns of the windings. In this way:
[tex]\frac{V_p}{V_s} =\frac{N_p}{N_s} \\\\Where:\\\\V_p=Primary\hspace{3} Voltage\\V_s=V_p=Secondary\hspace{3} Voltage\\N_p=Number\hspace{3} of\hspace{3} Primary\hspace{3} Windings\\N_s=Number\hspace{3} of\hspace{3} Secondary\hspace{3} Windings[/tex]
In this case:
[tex]V_p=120V\\V_s=100kV=100000V\\N_p=50[/tex]
Therefore, using the previous equation and the data provided, let's solve for [tex]N_s[/tex] :
[tex]N_s=\frac{N_p V_s}{V_p} =\frac{(50)(100000)}{120} =\frac{125000}{3} \approx41667\hspace{3}loo ps[/tex]
Hence, the number of loops in the secondary is approximately 41667.
