The frequencies at which the current amplitude in an LCR series circuit becomes 1/√2 times its maximum value, are 212 rad s⁻¹ and 232 rad s⁻¹. The value of resistance in the circuit is R=5 Ω. The self inductance in the circuit is __________ mH.

Solution

Physics sir · 4 steps

The gap between the two half-power frequencies is the bandwidth, and that bandwidth is just R/L.

1. Half-power points. The current drops to 1/√2 of its peak when the net reactance equals R, that is, when |ω L - 1/(ω C)| = R. This happens at ω = ω₀ ± R/2L (very nearly).
2. Bandwidth. Subtract the lower frequency from the upper one. The ω₀ cancels and the two R/2L terms add.

   Δω = ω₂ - ω₁ = R/L

3. Put in numbers. Δω = 232 - 212 = 20 rad s⁻¹, so L = R/Δω = 5/20 = 0.25 H.
4. So. L = 0.25 H = 250 mH.

The common slip: 125. Taking the bandwidth as R/2L instead of R/L gives L = 5/40 H = 125 mH. Each half-power point sits R/2L from ω₀, so the full gap is R/L.

The idea to keep. In a series LCR circuit the half-power bandwidth is Δω = R/L. Given the two half-power frequencies and R, the inductance follows in one line, with no need for C.

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