In a series LCR resonant circuit, the quality factor is measured as 100. If the inductance is increased by two fold and resistance is decreased by two fold, then the quality factor after this change will be __________.

Solution

Physics sir · 4 steps

Write Q in terms of L, C and R only, then see how each change scales it.

1. Rewrite Q. At resonance Q = (ω₀ L)/R with ω₀ = 1/√LC. Substitute ω₀, and Q depends on L only through a square root.

   Q = 1/R√(L/C)

2. Apply the changes. L becomes 2L, so √L grows by √2. R becomes R/2, so 1/R grows by 2. C is unchanged.
3. Put in numbers. Q' = 2√2 × Q = 2 × 1.414 × 100 = 282.8.
4. So. The new quality factor is about 283.

The common slip: 400. Using Q = ω₀ L/R with ω₀ held fixed doubles Q for the inductance as well and gives 400, but ω₀ drops by √2 when L doubles.

The idea to keep. Changing L also shifts the resonant frequency, so never hold ω₀ fixed. Use Q = 1/R√(L/C): Q goes as √L, as 1/R, and as 1/√C.

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