Q2 (a) — [8 marks]

Resonance, Half-Power Frequencies & Bandwidth in Series R-L-C Circuit

Problem. What do you mean by resonance in an R-L-C series circuit? Define half-power frequencies and bandwidth in an R-L-C series circuit and obtain an expression for them.

1 · Resonance in a Series R-L-C Circuit

Fig. 1 — Series R-L-C circuit driven by an ac source of variable frequency.

The impedance of the series circuit is

$$Z = R + j\!\left(\omega L - \frac{1}{\omega C}\right) = R + j(X_L - X_C)$$

Magnitude:

$$|Z| = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^{\!2}}$$

Resonance: A series R-L-C circuit is said to be in resonance when the inductive reactance equals the capacitive reactance, i.e. the net reactance is zero and the circuit behaves as a pure resistance.

At resonance:

$$X_L = X_C \;\Rightarrow\; \omega_0 L = \frac{1}{\omega_0 C}$$ $$\boxed{\;\omega_0 = \frac{1}{\sqrt{LC}} \quad\text{or}\quad f_0 = \frac{1}{2\pi\sqrt{LC}}\;}$$

Properties at Resonance

Impedance is minimum: $\;Z_{\min} = R$.
Current is maximum: $\;I_0 = V/R$.
Power factor is unity ($\cos\varphi = 1$); voltage and current are in phase.
$V_L$ and $V_C$ are equal in magnitude and 180° out of phase, so they cancel.
The circuit absorbs maximum power: $P_{\max} = V^2/R$.

2 · Frequency Response & Half-Power Frequencies

Fig. 2 — Current vs. frequency. At $\omega_0$ current is maximum; at $\omega_1$ and $\omega_2$ the current is $I_{\max}/\sqrt{2}$ (half-power points).

Half-power frequencies ($\omega_1, \omega_2$): The two frequencies on either side of $\omega_0$ at which the current falls to $\dfrac{I_{\max}}{\sqrt{2}}$ (i.e., $0.707\,I_{\max}$) and consequently the power dissipated drops to half of its maximum value $P_{\max}$.
$\omega_1$ = lower half-power (cut-off) frequency, $\omega_2$ = upper half-power (cut-off) frequency.

Bandwidth (BW): The frequency range between the two half-power points: $$\text{BW} = \omega_2 - \omega_1 \quad (\text{rad/s}) \quad\text{or}\quad f_2 - f_1 \;(\text{Hz}).$$

3 · Derivation of $\omega_1$, $\omega_2$ and Bandwidth

Power at resonance vs. at half-power point

Maximum power dissipated occurs at $\omega_0$:

$$P_{\max} = I_{\max}^2\,R = \frac{V^2}{R}$$

At half-power frequencies, by definition,

$$P = \frac{P_{\max}}{2} = I^2 R \;\Rightarrow\; I = \frac{I_{\max}}{\sqrt{2}}$$

Condition on impedance

Since $I = V/|Z|$ and $I_{\max} = V/R$, the half-power condition becomes

$$\frac{V}{|Z|} = \frac{1}{\sqrt{2}}\cdot\frac{V}{R} \;\Rightarrow\; |Z| = \sqrt{2}\,R$$

Substituting $|Z|^2 = R^2 + (\omega L - 1/\omega C)^2$:

$$R^2 + \left(\omega L - \frac{1}{\omega C}\right)^{\!2} = 2R^2$$ $$\left(\omega L - \frac{1}{\omega C}\right)^{\!2} = R^2$$ $$\boxed{\;\omega L - \frac{1}{\omega C} = \pm R\;}$$

Upper half-power frequency $\omega_2$ (take $+R$, since $\omega_2 > \omega_0$ ⇒ $X_L > X_C$)

$$\omega_2 L - \frac{1}{\omega_2 C} = +R$$

Multiplying by $\omega_2$:

$$L\,\omega_2^2 - R\,\omega_2 - \frac{1}{C} = 0$$

Solving the quadratic and taking the positive root:

$$\omega_2 = \frac{R}{2L} + \sqrt{\left(\frac{R}{2L}\right)^{\!2} + \frac{1}{LC}}$$

Lower half-power frequency $\omega_1$ (take $-R$)

$$\omega_1 L - \frac{1}{\omega_1 C} = -R \;\Rightarrow\; L\,\omega_1^2 + R\,\omega_1 - \frac{1}{C} = 0$$ $$\omega_1 = -\frac{R}{2L} + \sqrt{\left(\frac{R}{2L}\right)^{\!2} + \frac{1}{LC}}$$

Bandwidth — subtract $\omega_1$ from $\omega_2$

$$\omega_2 - \omega_1 = \left[\frac{R}{2L} + \sqrt{\cdots}\right] - \left[-\frac{R}{2L} + \sqrt{\cdots}\right] = \frac{R}{L}$$

Bandwidth

$$\boxed{\;\text{BW} = \omega_2 - \omega_1 = \dfrac{R}{L}\ \text{rad/s}\;}\qquad \text{or}\qquad f_2 - f_1 = \dfrac{R}{2\pi L}\ \text{Hz}$$

Useful relation: $\omega_0$ is the geometric mean of $\omega_1, \omega_2$

Multiplying $\omega_1$ and $\omega_2$:

$$\omega_1\,\omega_2 = \left[\sqrt{\left(\tfrac{R}{2L}\right)^{\!2} + \tfrac{1}{LC}}\right]^{\!2} - \left(\tfrac{R}{2L}\right)^{\!2} = \frac{1}{LC} = \omega_0^2$$ $$\boxed{\;\omega_0 = \sqrt{\omega_1\,\omega_2}\;}$$

4 · Quality Factor (Q) and its Link with Bandwidth

The selectivity of the circuit is measured by the quality factor:

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

Combining with bandwidth:

$$\text{BW} = \frac{R}{L} = \frac{\omega_0}{Q} \;\Rightarrow\; \boxed{\;Q = \dfrac{\omega_0}{\text{BW}}\;}$$

Interpretation: A higher Q-factor means a narrower bandwidth, sharper resonance peak, and higher selectivity — desirable in tuning circuits (radio, filters).

Summary of Key Expressions

Quantity	Expression
Resonant (angular) frequency	$\omega_0 = \dfrac{1}{\sqrt{LC}}, \quad f_0 = \dfrac{1}{2\pi\sqrt{LC}}$
Maximum current	$I_{\max} = V/R$
Lower half-power freq.	$\omega_1 = -\dfrac{R}{2L} + \sqrt{\left(\dfrac{R}{2L}\right)^{\!2}+\dfrac{1}{LC}}$
Upper half-power freq.	$\omega_2 = +\dfrac{R}{2L} + \sqrt{\left(\dfrac{R}{2L}\right)^{\!2}+\dfrac{1}{LC}}$
Bandwidth	$\text{BW} = \omega_2 - \omega_1 = \dfrac{R}{L}$
Geometric mean property	$\omega_0 = \sqrt{\omega_1 \omega_2}$
Quality factor	$Q = \dfrac{\omega_0 L}{R} = \dfrac{\omega_0}{\text{BW}}$

Resonance condition

X_L = X_C

Resonant frequency

ω₀ = 1/√(LC)

Half-power current

I_max / √2

Bandwidth

BW = R / L