(C) For an $AC$ source,the inductive reactance is given by $X_L = \omega L = 2 \pi f L$.Given $L = 2 \, mH = 2 \times 10^{-3} \, H$ and $f = 50 \, Hz$

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(C) For an $AC$ source,the inductive reactance is given by $X_L = \omega L = 2 \pi f L$.Given $L = 2 \, mH = 2 \times 10^{-3} \, H$ and $f = 50 \, Hz$.$X_1 = 2 \times \pi \times 50 \times 2 \times 10^{-3} = 100 \pi \times 2 \times 10^{-3} = 0.2 \pi \, \Omega$.Using $\pi \approx 3.14$,we get $X_1 = 0.2 \times 3.14 = 0.628 \, \Omega$.For a $DC$ source,the frequency $f = 0$,so the angular frequency $\omega = 2 \pi f = 0$.The inductive reactance $X_2 = \omega L = 0 \times L = 0 \, \Omega$.Thus,$X_1 = 0.628 \, \Omega$ and $X_2 = 0$.

Class 12 Physics Alternating Current Only Inductor, Only Capacitor and Only Resistor Circuit

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An inductor of inductance $2 \, mH$ is connected to a $220 \, V, 50 \, Hz$ $AC$ source. Let the inductive reactance in the circuit be $X_1$. If a $220 \, V$ $DC$ source replaces the $AC$ source in the circuit,then the inductive reactance in the circuit is $X_2$. $X_1$ and $X_2$ respectively are:

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A
$6.28 \, \Omega$,zero
B
$6.28 \, \Omega$,infinity
C
$0.628 \, \Omega$,zero
D
$0.628 \, \Omega$,infinity

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Only Inductor, Only Capacitor and Only Resistor Circuit

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An inductor of inductance $2 \, mH$ is connected to a $220 \, V, 50 \, Hz$ $AC$ source. Let the inductive reactance in the circuit be $X_1$. If a $220 \, V$ $DC$ source replaces the $AC$ source in the circuit,then the inductive reactance in the circuit is $X_2$. $X_1$ and $X_2$ respectively are:

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An alternating voltage $E = 200\sqrt{2} \sin(100t)$ is connected to a $1 \mu F$ capacitor through an $AC$ ammeter. The reading of the ammeter shall be $mA$.

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$A$ $10 \mu F$ capacitor is connected to a $210 \text{ V}, 50 \text{ Hz}$ source as shown in the figure. The peak current in the circuit is nearly $(\pi = 3.14)$ : (in $\text{ A}$)

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