An alternating emf E = 440 sin(100 pi t) is a | vedclass

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(C) Given: $E = 440 \sin(100 \pi t)$ and $L = \frac{\sqrt{2}}{\pi} 	ext{ H}$.Comparing with $E = E_0 \sin(\omega t)$,we get $E_0 = 440 	ext{ V}$ and

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$2.2$

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(C) Given: $E = 440 \sin(100 \pi t)$ and $L = \frac{\sqrt{2}}{\pi} 	ext{ H}$.Comparing with $E = E_0 \sin(\omega t)$,we get $E_0 = 440 	ext{ V}$ and $\omega = 100 \pi 	ext{ rad/s}$.The inductive reactance $X_L$ is given by $X_L = \omega L = (100 \pi) 	imes \left( \frac{\sqrt{2}}{\pi} ight) = 100 \sqrt{2} \, \Omega$.The peak current $I_0$ is $I_0 = \frac{E_0}{X_L} = \frac{440}{100 \sqrt{2}} = \frac{4.4}{\sqrt{2}} 	ext{ A}$.An $AC$ ammeter measures the $RMS$ value of the current.$I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{4.4 / \sqrt{2}}{\sqrt{2}} = \frac{4.4}{2} = 2.2 	ext{ A}$.Thus,the reading of the ammeter is $2.2 	ext{ A}$.

An alternating $emf$ $E = 440 \sin(100 \pi t)$ is applied to a circuit containing an inductance of $\frac{\sqrt{2}}{\pi} \text{ H}$. If an $AC$ ammeter is connected in the circuit,its reading will be $....... \text{ A}$.

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