Question

A transformer has 10000 turns and 240V , 0.2A in the primary coil. If the current through secondary coil is 0.4A. a)What kind of transformer is this? b)Find out the voltage and number of turns in the secondary coil. c)Calculate maximum power in the secondary coil of this transformer.

Answer 1

Given:

A transformer has 10000 turns and 240 V , 0.2 A in the primary coil. If the current through secondary coil is 0.4 A.

To Find:

a)What kind of transformer is this?

b)Find out the voltage and number of turns in the secondary coil.

c)Calculate maximum power in the secondary coil of this transformer.

Solution:

Primary Winding:

$\mathbf{Number\ of\ turn\ (N_{1})= 10000 }$

$\mathbf{Primary\ voltage\ (V_{1})= 240 \ V }$

$\mathbf{Primary\ current\ (I_{1})= 0.2 \ A }$

Secondary Winding:

$\mathbf{Secondary\ current\ (I_{2})= 0.4 \ A }$

We know the formula of transformer:

$\mathbf{ Turn\ Ratio =\dfrac{N_{1}}{N_{2}}=\dfrac{V_{1}}{V_{2}}=\dfrac{I_{2}}{I_{1}} }$

Kind of transformer:

$\mathbf{ Turn\ Ratio =\dfrac{I_{2}}{I_{1}} }$

$\mathbf{ Turn\ Ratio =\dfrac{0.4}{0.2}=2 }$

Means turn ratio is grater than 1 , so it is step down transformer.

Out put voltage and number of turns in the secondary coil:

$\mathbf{ \dfrac{N_{1}}{N_{2}}=\dfrac{V_{1}}{V_{2}}=\dfrac{I_{2}}{I_{1}} }$

On putting respective value in above equation:

$\mathbf{ \dfrac{10000}{N_{2}}=\dfrac{240}{V_{2}}=\dfrac{0.4}{0.2} }$

$\mathbf{ \dfrac{10000}{N_{2}}=\dfrac{240}{V_{2}}=2}$

On solving, we get:

$\mathbf{ Voltage\ in\ secondary\ coil\ = V_{2} =120 \ V }$

$\mathbf{ Number\ of\ turn\ in\ secondary\ coil\ = N_{2} =5000 }$

Maximum power in the secondary coil of this transformer:

Power (P) = Voltage× Current

So:

Power in the secondary coil (P) = Voltage in the secondary coil× Current in the secondary coil

$\mathbf{ Power\ in\ secondary\ coil\ (P) = V_{2}\times I_{2} }$

$\mathbf{ Power\ in\ secondary\ coil\ (P) = 120\times 0.4 }$

Power in secondary coil = 48 W

Answer 2

Answer:

step down transfarmer