An inductor of inductance 2 mH is connected to a 220 ~V, 50 ~Hz a.c. source. Let inductive reactance in the circuit is X_1 . If a 220 ~V d.c. source replaces the a.c. source in the circuit, then the inductive reactance

Correct answer: C. 0.628 , zero For AC circuits, aligned X_ L & =ω L X_ L & =2 π n L X_ L & =0.628 Hence, & X_1=0.628 aligned ( ω=2 π n) For DC circuits, ω=0 aligned X_ L & =ω L X_ L & =0 X_2 & =0 aligned Hence X_2=0

NEET 2022

Exam: NEET
Year: 2022
Subject: Physics
Chapter: Alternating Current

NEET 2022 · Physics · Alternating Current

An inductor of inductance 2 mH is connected to a 220 ~V, 50 ~Hz...

An inductor of inductance

2 \mathrm{mH}

is connected to a

220 \mathrm{~V}, 50 \mathrm{~Hz}

a.c. source. Let inductive reactance in the circuit is

X_1

. If a

220 \mathrm{~V}

d.c. source replaces the a.c. source in the circuit, then the inductive reactance in the circuit is

\mathrm{X}_2 \cdot \mathrm{X}_1

and

\mathrm{X}_2

respectively are:

Options

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

Correct answer

C. $0.628 \Omega$ , zero

Step-by-step solution

For AC circuits,

\begin{aligned} X_{\mathrm{L}} & =\omega \mathrm{L} \\ \mathrm{X}_{\mathrm{L}} & =2 \pi n \mathrm{ L} \\ \mathrm{X}_{\mathrm{L}} & =0.628 \Omega \\ \text{Hence, } & \mathrm{X}_1=0.628 \Omega\end{aligned}

\quad(\because \omega=2 \pi n)

For DC circuits,

\omega=0

\begin{aligned} X_{\mathrm{L}} & =\omega \mathrm{L} \\ \mathrm{X}_{\mathrm{L}} & =0 \\ \mathrm{X}_2 & =0 \end{aligned}

Hence

\quad X_2=0$

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