In a noiseless transformer an alternating current of $2 A$ is flowing in the primary coil. The number of turns on the primary and secondary coils are $100$ and $20$ respectively. The value of the current in the secondary coil is

0.08 A

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0.4 A

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5 A

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10 A

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Solution

The correct option is D $10 A$
Given : $I_{p} = 2 A$ $N_{p} = 100$ turns $N_{s} = 20$ turns
Using $\frac{I_{s}}{I_{p}} = \frac{N_{p}}{N_{s}}$

$∴$ $\frac{I_{s}}{2} = \frac{100}{20}$ $⟹ I_{s} = 10 A$

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In a noiseless transformer an alternating current of 2A is flowing in the primary coil. The number of turns on the primary and secondary coils are 100 and 20 respectively. The value of the current in the secondary coil is

The correct option is D 10AGiven : Ip=2A Np=100 turns Ns=20 turns Using IsIp=NpNs∴ Is2=10020 ⟹Is=10A

In a noiseless transformer an alternating current of $2 A$ is flowing in the primary coil. The number of turns on the primary and secondary coils are $100$ and $20$ respectively. The value of the current in the secondary coil is

The correct option is D $10 A$
Given : $I_{p} = 2 A$ $N_{p} = 100$ turns $N_{s} = 20$ turns
Using $\frac{I_{s}}{I_{p}} = \frac{N_{p}}{N_{s}}$

$∴$ $\frac{I_{s}}{2} = \frac{100}{20}$ $⟹ I_{s} = 10 A$