JEE Main 2023 · 31 January · evening · Single correct · +4 / −1
An alternating voltage source V=260 sin (628 t ) is connected across a pure inductor of 5 mH Inductive reactance in the circuit is :
- (a) 6.28 Ω
- (b) 0.318 Ω
- (c) 0.5 Ω
- (d) 3.14 Ω
Answer and solution
Answer (d) 3.14 Ω
Solution
Physics sir · 3 steps
The number multiplying t is already ω, so multiply it by L and you are done.
- Read off omega. Compare V = 260 sin(628t) with V = V₀ sin(ω t). So ω = 628 rad s⁻¹, which is 2π f with f = 100 Hz.
X_L = ω L
- Put in numbers. Convert L = 5 mH = 5 × 10⁻³ H. Then X_L = 628 × 5 × 10⁻³ = 3.14 Ω.
- So. The inductive reactance is 3.14 Ω, option (d). The peak voltage 260 V plays no part in it.
The idea to keep. Reactance depends only on the frequency and the component: X_L = ω L, X_C = 1/(ω C). Take ω straight from the sine argument, and change mH to H first.
Answer from the NTA final answer key. Solution by PrepWiser.