JEE Main 2022 · 30 June · morning · Single correct · +4 / −1

In series RLC resonator, if the self inductance and capacitance become double, the new resonant frequency (f₂) and new quality factor (Q₂) will be :
(f₁ = original resonant frequency, Q₁ = original quality factor)

  1. (a) f₂ = f₁/2 and Q₂ = Q₁
  2. (b) f₂ = f₁ and Q₂ = Q₁/Q₂
  3. (c) f₂ = 2f₁ and Q₂ = Q₁
  4. (d) f₂ = f₁ and Q₂ = 2Q₁
Answer and solution

Answer (a) f₂ = f₁/2 and Q₂ = Q₁

Solution

Physics sir · 3 steps

Frequency depends on the product LC, Q on the ratio L/C. Doubling both changes only one.

  1. Resonant frequency. At resonance f = 1/2π√LC. When L and C both double, the product becomes 4LC, so √LC becomes twice as large.

    f₂ = 1/2π√(2L)(2C) = 1/2·1/2π√LC = f₁/2

  2. Quality factor. For a series RLC circuit Q = (ω₀ L)/R = 1/R√(L/C). The ratio 2L/2C = L/C does not change, and R is the same, so Q₂ = Q₁.
  3. So. The resonant frequency halves and the quality factor stays the same: f₂ = f₁/2 and Q₂ = Q₁, which is option (a).

The idea to keep. In series RLC, f₀ depends on LC and Q depends on √(L/C)/R. Before calculating, check whether the change alters the product, the ratio, or both.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Alternating Current

Asked about 140 times in JEE Main since 2020. This question is from 2022.