JEE Main Alternating Current PYQ

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11. (AIEEE 2002)

In a transformer, number of turns in the primary coil are $140$ and that in the secondary coil are $280.$ If current in primary coil is $4 A,$ then that in the secondary coil is

A. 4A

B. 2A

C. 6A

D. 10A

Correct Option is (B)

$N_{p} = 140, N_{s} = 280, I_{p} = 4 A, I_{s} = ?$

For a transformer $\frac{I_{s}}{I_{p}} = \frac{N_{p}}{N_{s}}$

$\Rightarrow \frac{I_{s}}{4} = \frac{140}{280} \Rightarrow I_{s} = 2 A$