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

Np/Ns=Is/Ip

Ns=1000/2= 500

Vs Is =VpIp

Vs=240×0.2/0.4

Vs=120

Pmax=VsIs=120×0.4=48

As Vs<Vp it is a step down transformer

I hope this helps ( ╹▽╹ )

Answer 2

Answer:

A transformer has 1000 turns and 240 voltage, 0.2 A in the primary coil. if the current through the secondary coil is 0.4 A, the voltage and number of turns in the secondary coil are 120v and 500 respectively. The transformer is step-down and maximum power in the secondary coil is 48W

​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

current in the primary coil $I_{1} =0.2A$

current in the secondary coil $I_{2} =0.4A$

the voltage in the primary coil is $V_{1} =240v$

the number of turns in the primary coil $N_{1} =1000$

the voltage in the secondary coil $V_{2} =\frac{V_{1} }{I_{2} } I_{1}$

⇒$V_{2} =240*\frac{0.2}{0.4} =120v$

since the voltage in the secondary coil is less than that of the primary the transformer is step-down

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

⇒ $N_{2} =120/240*1000=500$

maximum power in the secondary coil is $P=V_{2} I_{2}$

⇒ $P=120*0.4=48W$