Alternating current of peak value left(frac{2 | vedclass

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(B) The peak value of the current is given as $I_0 = \frac{2}{\pi} 	ext{ A}$.The coefficient of mutual inductance is $M = 1 	ext{ H}$.The frequency

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(B) The peak value of the current is given as $I_0 = \frac{2}{\pi} 	ext{ A}$.The coefficient of mutual inductance is $M = 1 	ext{ H}$.The frequency of the alternating current is $f = 50 	ext{ Hz}$.The angular frequency is $\omega = 2\pi f = 2 	imes \pi 	imes 50 = 100\pi 	ext{ rad/s}$.The instantaneous current in the primary coil is $i = I_0 \sin(\omega t)$.The induced $EMF$ in the secondary coil is given by $e_2 = M \frac{di}{dt}$.Substituting the values: $e_2 = M \frac{d}{dt} (I_0 \sin(\omega t)) = M I_0 \omega \cos(\omega t)$.The peak value of the induced $EMF$ is $e_{2, 	ext{peak}} = M I_0 \omega$.Substituting the known values: $e_{2, 	ext{peak}} = 1 	imes \left(\frac{2}{\pi}ight) 	imes (100\pi) = 200 	ext{ V}$.

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