A transformer, designed to convert the voltage from 240 V a.c. mains to 12 V, has 4000 turns on the primary coil. How many turns should be on the secondary coil? If the transformer were 100% efficient, what current would flow through the primary coil when the current in the secondary coil was 0.4 A? Ans. (200, 0.02A)



Data Given:


Supplied Voltage to the primary coil of transformer  = Vp = 240 V 


Output Voltage of the secondary coil of transformer  = Vs = 12 V

Number of turns on the Primary coil = Np = 4000



To Find:


Number of turns on the secondary coil = Ns = ?

Current in the primary coil = Ip



Solution:

The relationship between voltages and the number of turns of Primary and secondary coils are as follows

`\frac{Ns}{Np}` = `\frac{Vs}{Vp}`

or 

Ns = `\frac{Vs}{Vp}` ｘ Np

by putting values

Ns = `\frac{12 V}{240 V}` ï½˜ 4000

Ns = 200 --------------Ans.1
Thus, the number of turn of the secondary coil is 200



To find Ip (Current in the primary coil) by using the following formula for an ideal transformer that is 

Power of primary coils = power of the secondary coil

Pp = Ps   

or

Vp Ip = Vs Is    (∴ P= VI )

or 

Ip = `\frac{Vs}{Vp}` ï½˜ Is  

By putting values

Ip = `\frac{12}{240}` ｘ0.4 A

by simplifying we get

Is = 0.02 A------ Ans.2
Thus the current in the primary coil is 0.02 A

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