In an LR circuit of 3 text{ mH} inductance an | vedclass

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(A) The given $	ext{emf}$ is $E = E_{0} \cos(\omega t)$,where $E_{0} = 4 	ext{ V}$ and $\omega = 1000 	ext{ rad/s}$.The impedance $Z$ of an $LR$ ci

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$0.8 	ext{ A}$

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(A) The given $	ext{emf}$ is $E = E_{0} \cos(\omega t)$,where $E_{0} = 4 	ext{ V}$ and $\omega = 1000 	ext{ rad/s}$.The impedance $Z$ of an $LR$ circuit is given by $Z = \sqrt{R^{2} + X_{L}^{2}}$,where $X_{L} = \omega L$.Given $R = 4 	ext{ } \Omega$,$L = 3 	ext{ mH} = 3 	imes 10^{-3} 	ext{ H}$,and $\omega = 1000 	ext{ rad/s}$.First,calculate the inductive reactance $X_{L} = \omega L = 1000 	imes 3 	imes 10^{-3} = 3 	ext{ } \Omega$.Now,calculate the impedance $Z = \sqrt{4^{2} + 3^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5 	ext{ } \Omega$.The amplitude of the current $i_{0}$ is given by $i_{0} = \frac{E_{0}}{Z} = \frac{4}{5} = 0.8 	ext{ A}$.

In an $LR$ circuit of $3 \text{ mH}$ inductance and $4 \text{ } \Omega$ resistance,an $\text{emf } E = 4 \cos(1000t) \text{ V}$ is applied. The amplitude of the current is:

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