An inductor of self-inductance 1 H is connec | vedclass

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(A) Given: Self-inductance $L = 1 \ H$,Resistance $R = 100 \pi \ \Omega$,Voltage $V_{rms} = 100 \pi \ V$,Frequency $f = 50 \ Hz$.First,calculate the i

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(A) Given: Self-inductance $L = 1 \ H$,Resistance $R = 100 \pi \ \Omega$,Voltage $V_{rms} = 100 \pi \ V$,Frequency $f = 50 \ Hz$.First,calculate the inductive reactance $X_L = \omega L = 2 \pi f L = 2 \pi 	imes 50 	imes 1 = 100 \pi \ \Omega$.The impedance of the $LR$ series circuit is $Z = \sqrt{R^2 + X_L^2} = \sqrt{(100 \pi)^2 + (100 \pi)^2} = \sqrt{2 	imes (100 \pi)^2} = 100 \pi \sqrt{2} \ \Omega$.The $RMS$ current is $I_{rms} = \frac{V_{rms}}{Z} = \frac{100 \pi}{100 \pi \sqrt{2}} = \frac{1}{\sqrt{2}} \ A$.The maximum current $I_{max}$ is given by $I_{max} = I_{rms} \sqrt{2} = \frac{1}{\sqrt{2}} 	imes \sqrt{2} = 1 \ A$.

An inductor of self-inductance $1 \ H$ is connected in series with a resistor of $100 \pi \ \Omega$ and an $AC$ supply of $100 \pi \ V$,$50 \ Hz$. The maximum current flowing in the circuit is . . . . . . $A$.

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