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

4 A

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2 A

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6 A

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8 A

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Solution

The correct option is B $2 A$
For an ideal transformer,

$\frac{N_{P}}{N_{S}} = \frac{E_{P}}{E_{S}} = \frac{I_{S}}{I_{P}}$

$\Rightarrow I_{S} = \frac{N_{P} I_{P}}{N_{S}} = \frac{140 \times 4}{280} = 2 A$

Hence, option $(B)$ is the correct answer.

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In an ideal transformer, number of turns in the primary coil are 140 and those in the secondary coil are 280. If current in the primary coil is 4 A, then that in the secondary coil is -

The correct option is B 2 AFor an ideal transformer, NPNS=EPES=ISIP ⇒IS=NPIPNS=140×4280=2 A Hence, option (B) is the correct answer.

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

The correct option is B $2 A$
For an ideal transformer,

$\frac{N_{P}}{N_{S}} = \frac{E_{P}}{E_{S}} = \frac{I_{S}}{I_{P}}$

$\Rightarrow I_{S} = \frac{N_{P} I_{P}}{N_{S}} = \frac{140 \times 4}{280} = 2 A$

Hence, option $(B)$ is the correct answer.