Resonant frequency, Q and bandwidth of a series RLC circuit

At resonance X_L = X_C, so the impedance is purely R.

Worked answer.

Step 1: Resonant frequency. Resonance occurs when X_L = X_C, i.e. ωL = 1/(ωC), giving
f₀ = 1 / (2π√(LC)).
LC = 10 × 10⁻³ H × 1 × 10⁻⁶ F = 1 × 10⁻⁸ s², so √(LC) = 1 × 10⁻⁴ s.
f₀ = 1 / (2π × 10⁻⁴) ≈ 1591.5 Hz (about 1.59 kHz).

Step 2: Quality factor. For a series circuit,
Q = (1/R)√(L/C) = (1/10) × √(10 × 10⁻³ / 1 × 10⁻⁶) = (1/10) × √10⁴ = (1/10) × 100 = 10.

Step 3: Bandwidth. BW = f₀ / Q = 1591.5 / 10 ≈ 159 Hz. Equivalently BW = R/(2πL) = 10/(2π × 0.01) ≈ 159.2 Hz, which confirms the result.

Step 4: Current at resonance. The reactances cancel, so Z = R = 10 Ω and I = V/R = 20/10 = 2 A rms.

Check. At f₀, X_L = 2π × 1591.5 × 0.01 ≈ 100 Ω and X_C = 1/(2π × 1591.5 × 10⁻⁶) ≈ 100 Ω. They are equal, as they must be. Note that the voltage across L (and across C) at resonance is I × X_L = 2 × 100 = 200 V, which is Q times the supply voltage. This is why a series resonant circuit is also called a voltage-magnification circuit.

Common mistake. Students often give ω₀ = 1/√(LC) = 10,000 rad/s as "the frequency" in hertz. Always divide by 2π when the question asks for f.

Likely follow-up. "What happens to the bandwidth if R is doubled?" Q halves to 5 and the bandwidth doubles to about 318 Hz, while f₀ stays the same because it does not depend on R.