An AC generator consists of a coil of 500 tur | vedclass

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(D) Given: Number of turns $N = 500$,Area $A = 2\ m^2$,Angular speed $\omega = 30\ rad\ s^{-1}$,Magnetic field $B = 0.20\ T$,Resistance $R = 1000\ \Om

Vedclass · Accepted Answer

$6$

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(D) Given: Number of turns $N = 500$,Area $A = 2\ m^2$,Angular speed $\omega = 30\ rad\ s^{-1}$,Magnetic field $B = 0.20\ T$,Resistance $R = 1000\ \Omega$. The induced electromotive force $(EMF)$ in an $AC$ generator is given by $\varepsilon = NBA\omega \sin(\omega t)$. The maximum $EMF$ is $\varepsilon_0 = NBA\omega$. The maximum current $I_0$ is given by $I_0 = \frac{\varepsilon_0}{R} = \frac{NBA\omega}{R}$. Substituting the values: $I_0 = \frac{500 	imes 0.20 	imes 2 	imes 30}{1000}$. $I_0 = \frac{6000}{1000} = 6\ A$. Therefore,the maximum current drawn from the generator is $6\ A$.

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