Question

A transformer has 1000 turns in the primary coil and 100 turns in the secondary coil. If the primary coil is connected to a 120 v outlet and draws 0.050 a, what are the voltage and current of the secondary coil?1200v,0.0050 a1200v, 0.50a12v,00050a12v,0.50a

Answer 1

Option D) 12 V, 0.50 A

Step-by-step explanation:

Number of turns in the primary coil,
`N_(p) = 1000`

Number of turns in the secondary coil,
`N_(s) = 100`

Voltage in primary coil,
`V_(p) = 120 V`

Current drawn by primary coil,
`I_(p) = 0.050 A`

Voltage in secondary coil,
`V_(s) = ?`

Current in secondary coil,
`I_(s) = ?`

Relationship between voltages applied in the secondary and primary coils

`(V_(s) )/(V_(p) ) = (N_(s) )/(N_(p) )`

`(V_(s) )/(120 ) = (100 )/(1000 )`

`V_(s) = 120*0.1\\V_(s) = 12.0 V`

Relationship between currents drawn by the secondary and primary coils

`(I_(s) )/(I_(p) ) = (N_(p) )/(N_(s) )`

`(I_(s) )/(0.050 ) = (1000 )/(100 )`

`I_(s) = 0.050*10\\I_(s) = 0.50 A`

Answer 2

12 V, 0.5 A

Step-by-step explanation:

For the Voltage,

Using,

Vs/Vp = Ns/Np.......................... Equation 1

Where Vs = Voltage in the secondary coil, Vp = Voltage in the primary coil, Ns = number of turn in the secondary coil, Np = number of turns in the primary coil.

Make Vs the subject of the equation

Vs = Vp(Ns/Np)........................ Equation 2

Given: Vp = 120 v, Np = 1000 turns, Ns = 100 turns

Substitute into equation 2

Vs = 120(100/1000)

Vs = 120×0.1

Vs = 12 v

For the current,

Using

Ns/Np = Ip/Is....................... Equation 3

Where Ip = current in the primary coil, Is = current in the secondary coil

make Is the subject of the equation

Is = Ip(Np/Ns).................. Equation 4

Given: Np = 1000 turns, Ns = 100 turns, Ip = 0.05 A

Substitute into equation 4

Is = 0.05(1000/100)

Is = 0.05×10

Is = 0.5 A.

Hence the voltage and the current in the secondary coil is 12 V, 0.5 A