Question

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.

Answer 1

(a)
`V_s= (Ns*V_p)/(N_p)`

(b)
`I=(V)/(R)`

(c)
`P=VI`

Step-by-step 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:

`(V_s)/(V_p) =(Ns)/(N_p)`

Where:

`V_s=Voltage\hspace{3}in\hspace{3}secondary\hspace{3}coil`

`V_p=Voltage\hspace{3}in\hspace{3}primary\hspace{3}coil`

`N_s=Number\hspace{3}of\hspace{3}turns\hspace{3}on\hspace{3}secondary\hspace{3}coil`

`N_p=Number\hspace{3}of\hspace{3}turns\hspace{3}on\hspace{3}primary\hspace{3}coil`

Isolating Vs and replacing the data in the previous equation:

`V_s=(V_p*N_s)/(N_p)=(250*12)/(50)=60V`

(b) Using ohm's law:

`V=IR`

Isolating I and replacing the data:

`I=(V)/(R) =(60)/(10) =6A`

(c) The Electric power is defined by:

`P=VI`

Hence:

`P=(60)*(6)=360W`