Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

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:

6.28 $\Omega$ , zero

6.28 $\Omega$ , infinity

0.628 $\Omega$ , zero

0.628 $\Omega$ , infinity

Step-by-Step Solution

According to the sources, inductive reactance ( $X_L$ ) is given by the formula $X_L = \omega L = 2\pi f L$ . For the AC source, with $f = 50$ Hz and $L = 2$ mH ( $2 \times 10^{-3}$ H), the reactance $X_1 = 2 \times \pi \times 50 \times 2 \times 10^{-3} = 0.2\pi \approx 0.628 \ \Omega$ . For a DC source, the frequency ( $f$ ) is zero, resulting in an inductive reactance $X_2 = 2\pi(0)L = 0$ (zero) .

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