Question

the current in the secondary coil of a transformer is 1A and that is the primary is 0.5 if two hundred volt is available in secondary coil of the transformer what is the voltage of primary​

Answer 1

Solution :-

We know :

$\boxed{\bf \dfrac{I_p}{I_s} = \dfrac{V_s}{V_p}}$

• $\sf I_p$ = Current flow in primary coil .

• $\sf I_s$ = Current flow in secondary coil.

• $\sf V_p$ = Voltage across primary coil.

• $\sf V_s$ = Voltage across secondary coil.

Here :-

• $\sf I_p$ = 0.5 A

• $\sf I_s$ = 1A

• $\sf V_s$ = 200V

Substitute the given values :-

$\longrightarrow \sf \dfrac{0.5}{1} = \dfrac{200}{V_p} \\\\\longrightarrow 0.5 \times V_p = 200 \\\\\longrightarrow V_p = \dfrac{200}{0.5} \\\\\longrightarrow\bf V_p = 400$

Voltage across primary coil = 400V

Extra information :-

• Transformer cannot work on DC. It changes voltages/currents.

• A transformer is essentially an AC device.

• Transformer does not affect the frequency of AC.

Answer 2

Answer:

The current in the secondary coil of a transformer is 1A and that is the primary is 0.5 if 200v is available in the secondary coil of the transformer, the voltage of the primary​ coil is 400v

​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 $I_{1} =0.5A$

current in the secondary coil $I_{2} =1A$

the voltage in the secondary coil is $V_{2} =2000v$

the ratio of currents  $I_{1} /I_{2} =2$

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

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

⇒$V_{1} =200/0.5=400v$