Question: 9: An inductor of inductance 2 mH is connected to a 220 \mathrm{~V}, 50 \mathrm{~Hz} a.c. source. Let the inductive reactance in the circuit is \mathrm{X}_{1}. If a 220 V dc source replaces the ac source in the circuit, then the inductive reactance in the circuit is \mathrm{X}_{2} . \mathrm{X}_{1} and \mathrm{X}_{2} respectively are :
(1) 6.28 \Omega, zero
(2) 6.28 \Omega, infinity
(3) 0.628 \Omega, zero
(4) 0.628 \Omega, infinity
Answer: Option (3)
Explanation:
The inductive reactance in an a.c. circuit is given by the formula X_L = 2\pi f L.
Here, the frequency f = 50 \mathrm{~Hz} and inductance L = 2 \times 10^{-3} \mathrm{~H}.
Substituting these values, we get X_1 = 2\pi \times 50 \times 2 \times 10^{-3}.
This gives X_1 = 0.628 \Omega.
When a d.c. source is connected, the frequency of the source is zero.
Since inductive reactance depends directly on frequency, for d.c. we have f = 0.
Therefore, the inductive reactance becomes X_2 = 2\pi \times 0 \times L = 0.
Hence, X_1 = 0.628 \Omega and X_2 = 0.