A resistance of 400,Omega is connected in ser | vedclass

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(B) Given: $R = 400\,\Omega$,$L = (3/\pi)\,H$,$e = 400 \sin(100\pi t)$.Comparing with $e = E_0 \sin(\omega t)$,we get $E_0 = 400\,V$ and $\omega = 100

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In the circuit,the potential difference across the resistance is in the same phase as the current.

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(B) Given: $R = 400\,\Omega$,$L = (3/\pi)\,H$,$e = 400 \sin(100\pi t)$.Comparing with $e = E_0 \sin(\omega t)$,we get $E_0 = 400\,V$ and $\omega = 100\pi\,rad/s$.Inductive reactance $X_L = \omega L = 100\pi 	imes (3/\pi) = 300\,\Omega$.Impedance $Z = \sqrt{R^2 + X_L^2} = \sqrt{400^2 + 300^2} = 500\,\Omega$.Peak current $I_0 = E_0 / Z = 400 / 500 = 0.8\,A = (4/5)\,A$. Thus,option $(a)$ is incorrect.In a series $RL$ circuit,the potential difference across the resistor is always in phase with the current. Thus,option $(b)$ is correct.The potential difference across an inductor always leads the current by $90^{\circ}$. Thus,options $(c)$ and $(d)$ are incorrect.

$A$ resistance of $400\,\Omega$ is connected in series with an inductor of $(3/\pi)\,H$ and this combination is connected across an $A.C.$ source as shown in the figure,where $e = 400 \sin(100\pi t)$. Which of the following statements is correct?

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