Question

in a primary coil of 500 turns of a transformer has a power of 2kw and current is 0.5A then the voltage in the secondary coil is obtained 200 volt. no. of turns in secondary coil is ?

Answer 1

25 turns
Ns/Np=vs/vp
np =500
ns=x
vs=200
p=i×v 200watts=0.5×v
v=4000 volts

Answer 2

Answer:

In a primary coil of 500 turns of a transformer has a power of 2kw and current is 0.5A then the voltage in the secondary coil is obtained 200 volts. the no. of turns in the secondary coil is 5

​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 primary coil $N_{1} =500$

current in the primary coil $I_{1} =0.5A$

power in the primary coil $P=2KW=2000W$

the voltage in the primary coil is $V_{1} =P/I_{1}$

⇒ $V_{1} =2000/0.5=4000v$

voltage in the secondary coil $V_{2} =200v$

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

⇒$N_{2} =\frac{200}{4000} *500=5$