---
title: "Series RLC: resonant frequency, bandwidth limits, L and C"
subject: "Electrical Circuit Theory"
chapter: 1
key: "semester_3/electrical_circuit_theory:series-resonance-q-factor-numerical"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/series-resonance-q-factor-numerical
site: BCT Engineering
---

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

Electrical Circuit Theory · Chapter 1: Network Analysis of AC circuit & Dependent Sources

- Asked in: 2073 Shrawan
- Video: [Problems #1 - Resonant Frequency, Bandwidth, Quality Factor of RLC Series Circuit - Circuit Analysis](https://www.youtube.com/watch?v=qWz0Jfsqtr4) — Padmasri Naban
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/series-resonance-q-factor-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-1

## Question

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.

## 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.

## Answer

**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

```text
Q = V_C / V = 400 / 100 = 4
```

#### Step 2: Inductance from bandwidth

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

#### Step 3: Resonant frequency

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

#### Step 4: Capacitance

```text
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.

```text
√(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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With source amplitude 100 V and capacitor voltage 400 V at resonance, what is Q?
   A. 0.25
   B. 400
   C. 2
   D. 4
   Answer: D. V_C = Q·V, so Q = 400/100 = 4. (Key point: Q factor)

2. Given R = 100 Ω and BW = 600 rad/s, what is L?
   A. R/BW = 0.1667 H
   B. BW/R = 6 H
   C. R × BW = 60 000 H
   D. 1/(R·BW)
   Answer: A. BW = R/L gives L = 100/600 = 0.1667 H. (Key point: Inductance)

3. How is ω0 found from Q and BW, and what is it?
   A. ω0 = BW/Q = 150 rad/s
   B. ω0 = Q × BW = 2400 rad/s
   C. ω0 = BW × 2π
   D. ω0 = Q/BW
   Answer: B. BW = ω0/Q so ω0 = 4 × 600 = 2400 rad/s (381.97 Hz). (Key point: Resonant frequency)

4. What is the exact lower limit ω1?
   A. 2400 − 600 = 1800 rad/s
   B. R/2L + √[(R/2L)^2 + ω0^2] ≈ 2719 rad/s
   C. −R/2L + √[(R/2L)^2 + ω0^2] ≈ 2119 rad/s
   D. 2400/4 = 600 rad/s
   Answer: C. With R/2L = 300, ω1 = −300 + √(300^2 + 2400^2) = −300 + 2418.68 ≈ 2119 rad/s. (Key point: Lower half power frequency)

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