A transformer has a primary coil with 108 turns and a secondary coil of 340 turns. The AC voltage across the primary coil has a maximum of 122 V and the AC current through the primary coil has a maximum of 4 A.
What are the maximum values of the voltage and current for the secondary coil?

Respuesta :

Answer:

[tex]V_s=384.074V\\I_s=1.108A[/tex]

Explanation:

In an ideal transformer, due to the electromagnetic induction, the voltages in the windings are proportional to the variation of the magnetic flux that crosses them and the number of turns of the winding. it follows that the relationship between tensions is proportional to the relationship between the number of turns of the windings. In this way:

[tex]\frac{V_p}{V_s} =\frac{N_p}{N_s} =\frac{I_s}{I_p}[/tex]

Where:

[tex]N_p=Number\hspace{3}of\hspace{3}turns\hspace{3}on\hspace{3}primary\hspace{3}coil\\N_s=Number\hspace{3}of\hspace{3}turns\hspace{3}on\hspace{3}secondary\hspace{3}coil\\V_p=Voltage\hspace{3}on\hspace{3}primary\hspace{3}coil\\V_s=Voltage\hspace{3}on\hspace{3}secondary\hspace{3}coil\\I_p=Current\hspace{3}on\hspace{3}primary\hspace{3}coil\\I_s=Current\hspace{3}on\hspace{3}secondary\hspace{3}coil[/tex]

So:

[tex]V_s=\frac{V_p*N_s}{N_p}=\frac{122*340}{108}  =384.074V[/tex]

[tex]I_s=\frac{I_p*N_p}{N_s}=\frac{4*108}{340} =1.108A[/tex]