An LCR circuit — find resonant frequency and Q factor

Question

An LCR series circuit has L = 0.5 H, C = 20 μF, and R = 10 Ω. Find (a) the resonant frequency, and (b) the Q-factor of the circuit.

Solution — Step by Step

In an LCR series circuit, resonance occurs when the inductive reactance equals the capacitive reactance:

X_L = X_C \implies \omega L = \frac{1}{\omega C}

At resonance, impedance is minimum ( $Z = R$ ) and current is maximum.

\omega_0 = \frac{1}{\sqrt{LC}}

Resonant frequency: $f_0 = \dfrac{1}{2\pi\sqrt{LC}}$

Given: L = 0.5 H, C = 20 μF = $20 \times 10^{-6}$ F

\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.5 \times 20 \times 10^{-6}}}

= \frac{1}{\sqrt{10^{-5}}} = \frac{1}{10^{-2.5}} = \frac{1}{3.162 \times 10^{-3}}

Let’s compute step by step:

LC = 0.5 \times 20 \times 10^{-6} = 10 \times 10^{-6} = 10^{-5}

\omega_0 = \frac{1}{\sqrt{10^{-5}}} = 10^{2.5} = 10^2 \times \sqrt{10} \approx 100 \times 3.162 = 316.2 \text{ rad/s}

f_0 = \frac{\omega_0}{2\pi} = \frac{316.2}{2\pi} = \frac{316.2}{6.283} \approx 50.3 \text{ Hz}

Resonant frequency $f_0 \approx 50$ Hz (note: this is the mains AC frequency — a nicely designed example).

The quality factor (Q-factor) measures how sharp the resonance peak is — high Q means the circuit responds strongly only in a narrow frequency range.

Q = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 CR} = \frac{1}{R}\sqrt{\frac{L}{C}}

Using $Q = \frac{\omega_0 L}{R}$ :

Q = \frac{316.2 \times 0.5}{10} = \frac{158.1}{10} = 15.81

Verify using another form: $Q = \frac{1}{R}\sqrt{\frac{L}{C}} = \frac{1}{10}\sqrt{\frac{0.5}{20 \times 10^{-6}}} = \frac{1}{10}\sqrt{25000} = \frac{158.1}{10} = 15.81$ ✓

Q-factor ≈ 15.8 (dimensionless).

Why This Works

At resonance, $X_L = X_C$ exactly cancels out. The remaining impedance is just $R$ , so current $I = V/R$ — the maximum possible for a given voltage. This is why resonance is used in radio tuners: by varying C (tuning capacitor), you match the resonant frequency to the desired station’s broadcast frequency.

The Q-factor tells you the “sharpness” of resonance:

High Q (≥ 10): Sharp resonance — useful for radio tuners (you want to pick one station, not many)
Low Q (≤ 1): Broad resonance — less selective but more tolerant of frequency variation

Physically: $Q = \frac{\text{energy stored per cycle}}{\text{energy dissipated per cycle}}$ . High Q means R is small relative to $\omega_0 L$ — energy loss per cycle is small relative to energy stored.

Alternative Method — Band Width Relationship

The Q-factor is also related to the bandwidth $\Delta f$ of the resonance curve:

Q = \frac{f_0}{\Delta f}

where $\Delta f = f_2 - f_1$ is the frequency band between half-power points. So a Q of 15.8 and $f_0 = 50$ Hz means bandwidth $\approx 50/15.8 \approx 3.16$ Hz. The circuit responds strongly only in a 3.16 Hz range around 50 Hz.

Three equivalent Q-factor formulas — $\frac{\omega_0 L}{R}$ , $\frac{1}{\omega_0 CR}$ , and $\frac{1}{R}\sqrt{\frac{L}{C}}$ — all give the same result. In exams, choose the form that avoids extra calculation. If you already have $\omega_0$ , use $\frac{\omega_0 L}{R}$ . If you prefer not to compute $\omega_0$ first, use $\frac{1}{R}\sqrt{L/C}$ .

Common Mistake

Students often forget to convert capacitance to Farads before substituting. C = 20 μF = $20 \times 10^{-6}$ F. If you use C = 20 in the formula, you get an answer that is wrong by a factor of $\sqrt{10^6} = 1000$ . Unit conversion is the most common source of error in LCR problems.

Question

Solution — Step by Step

Why This Works

Alternative Method — Band Width Relationship

Common Mistake

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