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

Answer:

12 V, 0.5 A

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

Answer 2

Answer:

Option D) 12 V, 0.50 A

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

$\frac{V_{s} }{V_{p} } = \frac{N_{s} }{N_{p} }$

$\frac{V_{s} }{120 } = \frac{100 }{1000 }$

$V_{s} = 120*0.1\\V_{s} = 12.0 V$

Relationship between currents drawn by the secondary and primary coils

$\frac{I_{s} }{I_{p} } = \frac{N_{p} }{N_{s} }$

$\frac{I_{s} }{0.050 } = \frac{1000 }{100 }$

$I_{s} = 0.050*10\\I_{s} = 0.50 A$