In an L-R circuit,the value of L is (frac{0.4 | vedclass

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(D) The impedance $Z$ of an $L-R$ circuit is given by $Z = \sqrt{R^2 + X_L^2}$,where $X_L = 2\pi fL$.Given $R = 30 \, \Omega$,$f = 50 \, Hz$,and $L =

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$50 \, \Omega, 4 \, A$

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(D) The impedance $Z$ of an $L-R$ circuit is given by $Z = \sqrt{R^2 + X_L^2}$,where $X_L = 2\pi fL$.Given $R = 30 \, \Omega$,$f = 50 \, Hz$,and $L = \frac{0.4}{\pi} \, H$.First,calculate the inductive reactance $X_L = 2\pi 	imes 50 	imes \frac{0.4}{\pi} = 100 	imes 0.4 = 40 \, \Omega$.Now,calculate the impedance $Z = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \, \Omega$.The current $I$ in the circuit is given by $I = \frac{V}{Z} = \frac{200}{50} = 4 \, A$.Thus,the impedance is $50 \, \Omega$ and the current is $4 \, A$.

In an $L-R$ circuit,the value of $L$ is $(\frac{0.4}{\pi}) \, H$ and the value of $R$ is $30 \, \Omega$. If an alternating e.m.f. of $200 \, V$ at $50 \, Hz$ is connected to the circuit,what are the impedance and the current in the circuit?

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