What is the exact lower limit ω1?

−R/2L + √[(R/2L)^2 + ω0^2] ≈ 2119 rad/s

R/2L + √[(R/2L)^2 + ω0^2] ≈ 2719 rad/s

2073 Shrawan

Series RLC: resonant frequency, bandwidth limits, L and C

Short answer

Q = Vc/V = 400/100 = 4; L = R/BW = 100/600 = 0.1667 H; resonant frequency ω0 = Q·BW = 2400 rad/s (381.97 Hz); C = 1/(ω0²L) = 1.04 μF; half-power frequencies: ω1 ≈ 2119 rad/s, ω2 ≈ 2719 rad/s.

A voltage v(t) = 100 sin ωt is applied to a series RLC circuit. At the resonant frequency of the circuit, the maximum voltage across the capacitor is found to be 400 V. The bandwidth is known to be 600 rad/sec and the impedance at resonance is 100 Ω. Find the resonant frequency and compute the upper and lower limits of the bandwidth. Also determine the value of L and C of the circuit.

Given: Vm = 100 V, Vc,max = 400 V at resonance, BW = 600 rad/s, Z0 = R = 100 Ω.

Assumption: the 400 V is the capacitor voltage at resonance (V_C = Q·V), and the 100 V is the amplitude of the source, so both are peak values and the ratio gives Q.

Step 1: Q factor

Q = V_C / V = 400 / 100 = 4

Step 2: Inductance from bandwidth

BW = R/L  →  L = R/BW = 100/600 = 0.1667 H

Step 3: Resonant frequency

BW = ω0/Q  →  ω0 = Q × BW = 4 × 600 = 2400 rad/s
f0 = ω0/2π = 381.97 Hz

Step 4: Capacitance

C = 1/(ω0² L) = 1/(2400² × 0.1667)
  = 1/(5.76×10⁶ × 0.1667) = 1.0417 μF

Check: Q = ω0L/R = 2400 × 0.1667/100 = 4 ✓. Also I0 = V/R = 1 A and V_C = I0/(ω0C) = 400 V ✓.

Step 5: Band limits (half power frequencies)

Exact formula: ω1,2 = ∓R/2L + √[(R/2L)² + ω0²], with R/2L = 300 rad/s.

√(300² + 2400²) = 2418.68
ω2 = 300 + 2418.68 = 2718.68 rad/s   (f2 = 432.7 Hz)
ω1 = −300 + 2418.68 = 2118.68 rad/s  (f1 = 337.2 Hz)
ω2 − ω1 = 600 rad/s ✓

Approximate (high Q): ω0 ± BW/2 = 2100 and 2700 rad/s.

Answer: ω0 = 2400 rad/s (381.97 Hz); lower limit ω1 ≈ 2119 rad/s, upper limit ω2 ≈ 2719 rad/s; L = 0.1667 H; C = 1.04 μF.

✓ Answer checked by Claude Opus 5.5 (high effort)

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