ry coil can be given by:

$$ \mathcal{E} = -M \frac{dI}{dt} $$

where:

$ M = 1 \, \text{H} $ is the mutual inductance,

$ \frac{dI}{dt} $ is the rate of change of current in the primary coil.

The current in the primary coil is an alternating current, expressed as:

$$ I(t) = I_0 \sin(\omega t) $$

where:

$ I_0 = \frac{2}{\pi} \, \text{A} $ is the peak current,

$ \omega = 2\pi f $ is the angular frequency, with $ f = 50 \, \text{Hz} $.

Calculate $ \omega $:

$$ \omega = 2\pi \times 50 = 100\pi \, \text{rad/s} $$

The derivative of the current with respect to time is:

$$ \frac{dI}{dt} = I_0 \omega \cos(\omega t) $$

Substitute $ I_0 $ and $ \omega $:

$$ \frac{dI}{dt} = \frac{2}{\pi} \times 100\pi \cos(\omega t) = 200 \cos(\omega t) $$

The peak value of $ \cos(\omega t) $ is 1, so the peak rate of change of current is:

$$ \left( \frac{dI}{dt} \right)_{\text{peak}} = 200 \, \text{A/s} $$

Thus, the peak e.m.f. in the secondary coil is:

$$ \mathcal{E}_{\text{peak}} = M \times \left( \frac{dI}{dt} \right)_{\text{peak}} = 1 \times 200 = 200 \, \text{V} $$

Therefore, the peak e.m.f. induced in the secondary coil is 200 V.

The correct option is D: 200 V.

Alternating Current

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