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An ideal transformer has 50 turns in its primary and 250 turns in its secondary. 12-V ac is connected to the primary. (a) Show that 60 V ac is available at the secondary; (b) Show that 6 A of current is in a 10@Ω device con- nected to the secondary. (c) Show that the power supplied to the primary is 360 W.

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carlos2112

Answer:

(a) [tex]V_s= \frac{Ns*V_p}{N_p}[/tex]

(b) [tex]I=\frac{V}{R}[/tex]

(c) [tex]P=VI[/tex]

Explanation:

The relationship between the voltage of the primary and secondary of a transformer depends on the number of turns the primary coil and the number of turns in the secondary coil. Mathematically is given by:

[tex]\frac{V_s}{V_p} =\frac{Ns}{N_p}[/tex]

Where:

[tex]V_s=Voltage\hspace{3}in\hspace{3}secondary\hspace{3}coil[/tex]

[tex]V_p=Voltage\hspace{3}in\hspace{3}primary\hspace{3}coil[/tex]

[tex]N_s=Number\hspace{3}of\hspace{3}turns\hspace{3}on\hspace{3}secondary\hspace{3}coil[/tex]

[tex]N_p=Number\hspace{3}of\hspace{3}turns\hspace{3}on\hspace{3}primary\hspace{3}coil[/tex]

Isolating Vs and replacing the data in the previous equation:

[tex]V_s=\frac{V_p*N_s}{N_p}=\frac{250*12}{50}=60V[/tex]

(b) Using ohm's law:

[tex]V=IR[/tex]

Isolating I and replacing the data:

[tex]I=\frac{V}{R} =\frac{60}{10} =6A[/tex]

(c) The Electric power is defined by:

[tex]P=VI[/tex]

Hence:

[tex]P=(60)*(6)=360W[/tex]

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