Answer

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**Hint**First calculate the time period of the AC current from the frequency given. Then using electromagnetic theory of induction derive the formula for voltage induced in the secondary. Substituting suitable values in this equation determines the answer.

Formulas used

$t = \dfrac{1}{f}$ where $t$ is the time period and $f$ is the frequency of the waveform.

${V_2} = M\dfrac{{d{i_1}}}{{dt}}$ where $M$is the mutual inductance, $d{i_1}$ is the change in current and $dt$ is the time difference.

**Complete step by step answer**

A transformer is an electrical device which can transfer electrical energy from one circuit to another. It usually consists of two windings with mutual inductance. One of the windings is called the primary while the other is called secondary. AC source is connected to the primary and a load to the secondary. The transformer can step up or step down the generator voltage.

The frequency of the given current is $50Hz$ so its time period is given by,

$t = \dfrac{1}{f} = \dfrac{1}{{50}}$$s$

In a transistor, the source can deliver power to the load via the magnetic coupling between the windings.

So, by formula,

${V_2} = M\dfrac{{d{i_1}}}{{dt}}$ where $M$is the mutual inductance, $d{i_1}$ is the change in current and $dt$ is the time difference.

We use this formula because a voltage is induced in the secondary coils of the transformer due to the change in current in the primary.

Now, the crest current is given as $1A$. In a complete waveform there is a crest and a trough having values $1A$ and $ - 1A$ respectively. So the change in current is $\left\{ {1 - \left( { - 1} \right)} \right\} = 2A$

The time difference between the crest and the trough of the current waveform is half of its total time period.

Substituting the values we get,

${V_2} = 0.5 \times \dfrac{2}{{\dfrac{1}{{2 \times 50}}}} = 100V$

Therefore, the crest voltage induced in the secondary is $100V$

**Hence, the correct answer is C.**

**Note**In a transformer, for low frequency operation laminated iron core is used but for radio frequency operation, to minimize eddy current while maintaining high flux density ferrite core is used.

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