In a transformer the number of turns of primary coil and secondary coil are 2.5 and 2 respectively and 140 V is applied on primary coil, then find the ratio of current in primary and secondary coils.

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Solution

In transformer
$\frac{V_{1}}{V_{2}} = \frac{N_{1}}{N_{2}} = \frac{I_{2}}{I_{1}}$
$\frac{I_{1}}{I_{2}} = \frac{N_{2}}{N_{1}} = \frac{2}{2.5} = 4 : 5$

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In a transformer the number of turns of primary coil and secondary coil are 2.5 and 2 respectively and 140 V is applied on primary coil, then find the ratio of current in primary and secondary coils.

In transformer V1V2=N1N2=I2I1I1I2=N2N1=22.5=4:5

In transformer
$\frac{V_{1}}{V_{2}} = \frac{N_{1}}{N_{2}} = \frac{I_{2}}{I_{1}}$
$\frac{I_{1}}{I_{2}} = \frac{N_{2}}{N_{1}} = \frac{2}{2.5} = 4 : 5$