Question

An ideal transformer has 1200 turns in the secondary coil and an output voltage of 240 V. If the primary voltage 12 V, calculate:

a. number of turns in the primary coil.

b. Power in the primary coil.

Answer 1

Explanation:

ndary terminal

The given transformer is a step down transformer as the number of turns in the secondary is less than primary. So, the voltage gets reduced.

Given,

Number of turns in the primary, N

1

= 1200

Number of turns in the secondary, N

2

= 120

primary voltage, V

1

= 240V

We have,

N

2

N

1

=

V

2

V

1

Where,

N

1

is the number of turns in the primary winding

N

2

is the number of turns in the secondary winding

V

1

is the voltage across the primary winding

V

2

is the voltage across the secondary winding

Substituting the given values in the equation we get,

120

1200

=

V

2

240

1200V

2

=28800

V

2

=

1200

28800

=24V

So the lamps will connect as secondary side as the secondary voltage is 24V only.

Answer 2

Answer:

An ideal transformer has 1200 turns in the secondary coil and an output voltage of 240 V. If the primary voltage 12 V, the number of turns in the primary coil is 60

​Explanation:

• In a transformer, there is a primary coil with a number of turns $N_{1}$ voltage, and current through it is $I_{1}$. In the secondary coil, the number of turns is $N_{2}$ voltage is $V_{2}$ and current is $I_{2}$

• The relation connecting these parameters is given by the formula

        $\frac{N_{1} }{N_{2} } =\frac{V_{1} }{V_{2} }=\frac{I_{2} }{I_{1} }$

From the question, we have

the number of turns in the secondary coil $N_{2} =1200$

the voltage in the primary coil is $V_{1} =12v$

the voltage in the secondary coil is $V_{2} =240v$

the number of turns in the primary coil $N_{1} =\frac{V_{1} }{V_{2} } N_{2}$

⇒$N_{1} =\frac{12*1200}{240}=60$