---
title: "Parallel resonant circuit: f0, Q and bandwidth"
subject: "Electrical Circuit Theory"
chapter: 1
key: "semester_3/electrical_circuit_theory:parallel-resonant-coil-capacitor-numerical"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/parallel-resonant-coil-capacitor-numerical
site: BCT Engineering
---

# Parallel resonant circuit: f0, Q and bandwidth

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

- Asked in: 2068 Chaitra, 8 marks
- Video: [Numerical on Parallel resonant circuit](https://www.youtube.com/watch?v=KAPFTEHGy_I) — Engg-Course-Made-Easy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/parallel-resonant-coil-capacitor-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-1

## Question

In the parallel resonant circuit shown in the figure below, find resonance frequency, Q factor and band width.
[Figure: a coil of 2 Ω resistance in series with 1 mH inductance, connected in parallel with a 10 μF capacitor]

## Short answer

Resonance angular frequency ω0 = √(1/LC − R²/L²) = 9798 rad/s (f0 ≈ 1.56 kHz). Quality factor Q = ω0L/R = 4.9. Bandwidth BW = ω0/Q = 2000 rad/s ≈ 318 Hz.

## Answer

**Given:** coil with R = 2 Ω and L = 1 mH in parallel with C = 10 μF.

#### Resonance frequency

For a coil in parallel with a capacitor:

```text
ω0 = √(1/LC − R²/L²)
1/LC = 1/(1×10⁻³ × 10×10⁻⁶) = 1×10⁸
R²/L² = (2/10⁻³)² = 4×10⁶
ω0 = √(1×10⁸ − 4×10⁶) = √(9.6×10⁷) = 9798 rad/s
f0 = ω0/2π = 1559.4 Hz ≈ 1.56 kHz
```

(Approximation neglecting R: ω0 = 1/√LC = 10⁴ rad/s, f0 = 1591.5 Hz.)

#### Quality factor

```text
Q0 = ω0L/R = 9798 × 10⁻³/2 = 4.90
```

(Using ω = 10⁴ gives Q = 5.)

#### Bandwidth

```text
BW = ω0/Q0 = R/L = 2/10⁻³ = 2000 rad/s
BW (Hz) = f0/Q0 = 318.3 Hz
```

#### Dynamic impedance (extra)

```text
Z0 = L/(CR) = 10⁻³/(10×10⁻⁶ × 2) = 50 Ω
```

**Answer: f0 ≈ 1.56 kHz (ω0 = 9798 rad/s), Q ≈ 4.9, bandwidth ≈ 2000 rad/s ≈ 318 Hz.**

## Check yourself

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

1. Which formula applies to a coil (R, L) in parallel with C?
   A. ω0 = 1/√(RC)
   B. ω0 = R/L
   C. ω0 = √(1/LC + R^2/L^2)
   D. ω0 = √(1/LC − R^2/L^2)
   Answer: D. The coil resistance modifies the frequency from 1/√LC. (Key point: Resonance frequency formula)

2. What is ω0 for R = 2 Ω, L = 1 mH, C = 10 μF?
   A. √(10^8 − 4×10^6) = 9798 rad/s
   B. 10^4 rad/s exactly
   C. √(10^8 + 4×10^6) = 10 198 rad/s
   D. 2000 rad/s
   Answer: A. 1/LC = 10^8 and R^2/L^2 = 4×10^6. (Key point: Resonance value)

3. What is Q0 of the circuit?
   A. R/(ω0L) ≈ 0.2
   B. ω0L/R ≈ 4.9
   C. ω0RC ≈ 0.2
   D. ω0L·R ≈ 19.6
   Answer: B. Q0 = 9798 × 10^−3/2 = 4.90. (Key point: Q factor)

4. What is the bandwidth?
   A. L/R = 0.5 ms
   B. 1/(RC) = 50 000 rad/s
   C. R/L = 2000 rad/s
   D. ω0·Q
   Answer: C. BW = ω0/Q0 = R/L = 2/10^−3 = 2000 rad/s. (Key point: Bandwidth)

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