Axiom Physics | Class 12 - Series LCR Circuit & Resonance

⚡ 01. The Spark (Analogy)

Socho ek child ke swing (jhule) ko: inductor us jhule ka bhari mass (inertia) hai aur capacitor uski elastic rope/spring tension hai. Inductor voltage ko $90^\circ$ aage dhakelta hai, jabki capacitor current ko $90^\circ$ aage khinchta hai—yani dono ek dusre ke theek $180^\circ$ opposite (anti-parallel) ladte hain! Jab applied AC frequency natural frequency se match kar jaati hai, toh inductor aur capacitor ka reactance ek dusre ko 100% cancel kar deta hai ($X_L = X_C$). Poore circuit ka opposition drop hokar sirf pure resistance $R$ bachta hai, aur current achanak explode hokar peak par pahunch jata hai—isi ko Electrical Resonance kehte hain! Hamare radio receiver ka knob ghumana asal mein capacitor $C$ ko badal kar isi resonance frequency ko target station ke frequency se match karna hota hai!

📐 02. The Core Law & Formulae

1. Impedance & Phase Angle

Total effective opposition $Z$ and phase lag/lead $\phi$:

$$Z = \sqrt{R^2 + (X_L - X_C)^2}$$ $$\tan\phi = \frac{X_L - X_C}{R} = \frac{\omega L - \frac{1}{\omega C}}{R}$$

If $X_L > X_C$: Inductive ($\phi > 0$, $v$ leads $i$). If $X_C > X_L$: Capacitive ($\phi < 0$, $i$ leads $v$). If $X_L = X_C$: Purely resistive ($\phi = 0$).

2. Resonance & Quality Factor (Q)

Condition for maximum current amplitude $I_{\max} = V_0 / R$:

$$\omega_0 = \frac{1}{\sqrt{LC}} \implies f_0 = \frac{1}{2\pi\sqrt{LC}}$$ $$Q = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 C R} = \frac{1}{R}\sqrt{\frac{L}{C}} = \frac{\omega_0}{2\Delta\omega}$$

Bandwidth $2\Delta\omega = \frac{R}{L}$. Lower $R$ gives a sharper resonance peak and higher selectivity for tuning circuits.

03. Interactive Visualizer: Series LCR Impedance Triangle & Resonance Response

Frequency ($f$) 50.0 Hz

Inductance ($L$) 200 mH

Capacitance ($C$) 50.0 µF

Resistance ($R$) 25.0 Ω

Total Impedance ($Z$):

25.00 Ω

Resonant Frequency ($f_0$):

50.33 Hz

Phase Angle ($\phi$):

0.0° (Resistive)

Quality Factor ($Q$):

2.53 (Sharpness)

📝 04. Stepwise Derivations Blueprint

Primary Board Derivation

Impedance & Phase Angle of Series LCR Circuit via Phasor Analysis

Fig 7.2A: Series LCR Circuit and corresponding phasor diagram; $V_L$ and $V_C$ oppose each other along the vertical axis.

1. In a series circuit, current $i = I_0\sin\omega t$ is identical through all three components at every instant.

2. Individual voltage phasors with respect to current $i$:

• Voltage across resistor $\vec{V}_R$ is in phase with current: $V_R = I R$.

• Voltage across inductor $\vec{V}_L$ leads current by $\pi/2$: $V_L = I X_L = I(\omega L)$.

• Voltage across capacitor $\vec{V}_C$ lags current by $\pi/2$: $V_C = I X_C = I(1/\omega C)$.

3. Since $\vec{V}_L$ and $\vec{V}_C$ point along the exact opposite directions ($180^\circ$ apart), their net vector resultant is $(\vec{V}_L - \vec{V}_C)$ perpendicular to $\vec{V}_R$.

4. Applying the Pythagorean theorem on the right-angled phasor triangle:

$$V_0^2 = V_R^2 + (V_L - V_C)^2 = (I_0 R)^2 + [I_0(X_L - X_C)]^2 = I_0^2 \left[ R^2 + (X_L - X_C)^2 \right]$$

5. Total circuit Impedance $Z = \frac{V_0}{I_0}$:

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

6. The phase angle $\phi$ by which applied voltage leads current:

$$\tan\phi = \frac{V_L - V_C}{V_R} = \frac{X_L - X_C}{R} = \frac{\omega L - \frac{1}{\omega C}}{R}$$

Derivation B

Series Resonance Condition, Bandwidth & Sharpness (Quality Factor Q)

1. Current amplitude is $I_0 = \frac{V_0}{\sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}}$. Current reaches its absolute maximum when impedance $Z$ is minimum.

2. $Z$ is minimum when the net reactance term vanishes identically:

$$X_L - X_C = 0 \implies \omega_0 L = \frac{1}{\omega_0 C} \implies \omega_0^2 = \frac{1}{LC}$$

$$\omega_0 = \frac{1}{\sqrt{LC}} \implies f_0 = \frac{1}{2\pi\sqrt{LC}}$$

3. At resonance ($\omega = \omega_0$):

• Minimum Impedance: $Z_{\min} = R$.

• Maximum Current: $I_{\max} = \frac{V_0}{R}$.

• Power Factor: $\cos\phi = \frac{R}{Z} = 1$ (Phase angle $\phi = 0^\circ$).

4. Bandwidth & Quality Factor ($Q$): At half-power frequencies ($\omega = \omega_0 \pm \Delta\omega$), current drops to $I_{\max}/\sqrt{2}$. Bandwidth is $2\Delta\omega = \frac{R}{L}$. Quality factor $Q$ measures the sharpness of resonance:

$$Q = \frac{\omega_0}{2\Delta\omega} = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 C R} = \frac{1}{R}\sqrt{\frac{L}{C}}$$

Tuning Selectivity: A higher $Q$ factor (smaller $R$, larger $L/C$ ratio) produces a narrower, taller resonance curve, enabling a radio receiver to reject adjacent channel noise and tune sharply into one target broadcasting frequency.

⚠️ 05. Red Alert Exam Trap Zone

Trap 1: The "Algebraic Sum of Voltages" Fallacy

*Board Question:* In a series LCR circuit, voltmeters read $V_R = 80\text{ V}$, $V_L = 100\text{ V}$, and $V_C = 40\text{ V}$. What is the total applied supply voltage?
*Trap Answer:* $80 + 100 + 40 = 220\text{ V}$ (Scalar addition is completely wrong!).
*Correct Vector Addition:* $$V = \sqrt{V_R^2 + (V_L - V_C)^2} = \sqrt{80^2 + (100 - 40)^2} = \sqrt{6400 + 3600} = \mathbf{100\text{ Volts}}!$$ AC voltages add as orthogonal vectors (phasors), never as scalar numbers.

Trap 2: Voltage Magnification at Resonance ($V_L > V_{\text{source}}$)

*Counter-Intuitive Reality:* At resonance, total supply voltage drops purely across resistor ($V_R = V$). However, the individual voltage across inductor or capacitor is: $$V_L = I_0 X_L = \left(\frac{V}{R}\right)(\omega_0 L) = \left(\frac{\omega_0 L}{R}\right) V = \mathbf{Q \times V}!$$ If $Q = 20$ and supply is $10\text{ V}$, $V_L$ and $V_C$ can reach 200 Volts each! (They cancel out across the pair, but individual insulation can break down).

Trap 3: Can a Series LC Circuit Resonate with DC ($f = 0$)?

*Exam Trap:* Resonance requires $X_L = X_C$. At $f = 0$, $X_L = 0$ while $X_C \to \infty$. Thus, a series LCR circuit CANNOT resonate at DC. The capacitor permanently blocks current flow ($I = 0$).

Trap 4: Effect of Frequency Variation on Circuit Nature

• At low frequencies ($f < f_0$): $X_C > X_L \implies$ Circuit is predominantly Capacitive ($i$ leads $v$).
• At resonant frequency ($f = f_0$): $X_L = X_C \implies$ Circuit is purely Resistive ($i$ and $v$ in phase).
• At high frequencies ($f > f_0$): $X_L > X_C \implies$ Circuit is predominantly Inductive ($v$ leads $i$).

💡 06. Handcrafted Exemplars

Board Classic Level

Problem: A series LCR circuit with $R = 20\text{ }\Omega$, $L = 1.5\text{ H}$, and $C = 35\text{ }\mu\text{F}$ is connected to a variable-frequency $200\text{ V}$ AC supply. (i) What is the resonant angular frequency $\omega_0$ of the circuit? (ii) What is the current amplitude at resonance? (iii) Find the Quality factor $Q$ of the circuit.

Solution:

1. Resonant angular frequency ($\omega_0$):

$$\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{1.5 \times (35 \times 10^{-6})}} = \frac{1}{\sqrt{5.25 \times 10^{-5}}} = \frac{1}{7.245 \times 10^{-3}} \approx 138.0\text{ rad/s}$$

2. At resonance, total impedance is minimum: $Z = R = 20\text{ }\Omega$.

$$V_0 = \sqrt{2} \times V_{\text{rms}} = 1.414 \times 200 \approx 282.8\text{ V}$$

Peak current amplitude at resonance:

$$I_0 = \frac{V_0}{R} = \frac{282.8\text{ V}}{20\text{ }\Omega} \approx 14.14\text{ A} \quad (I_{\text{rms}} = 10\text{ A})$$

3. Quality factor $Q$:

$$Q = \frac{\omega_0 L}{R} = \frac{138.0 \times 1.5}{20} = \frac{207.0}{20} = 10.35$$

JEE/NEET Competitive Trap

Problem: A series LCR circuit has $L = 8.0\text{ H}, C = 0.5\text{ }\mu\text{F}$, and $R = 100\text{ }\Omega$. Calculate the resonant frequency, the bandwidth of the circuit, and the two half-power frequencies at which the power dissipated drops to half of its maximum value.

Solution:

1. Resonant angular frequency:

$$\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{8.0 \times (0.5 \times 10^{-6})}} = \frac{1}{\sqrt{4.0 \times 10^{-6}}} = \frac{1}{2.0 \times 10^{-3}} = 500\text{ rad/s}$$ $$f_0 = \frac{\omega_0}{2\pi} = \frac{500}{2\pi} \approx 79.58\text{ Hz}$$

2. Bandwidth ($2\Delta\omega$):

$$2\Delta\omega = \frac{R}{L} = \frac{100\text{ }\Omega}{8.0\text{ H}} = 12.5\text{ rad/s} \implies \Delta\omega = 6.25\text{ rad/s}$$

3. Half-power angular frequencies ($\omega_1$ and $\omega_2$):

$$\omega_1 = \omega_0 - \Delta\omega = 500 - 6.25 = 493.75\text{ rad/s}$$ $$\omega_2 = \omega_0 + \Delta\omega = 500 + 6.25 = 506.25\text{ rad/s}$$

Quality factor for reference: $Q = \frac{\omega_0}{2\Delta\omega} = \frac{500}{12.5} = 40.0$.

🗂️ 07. Quick Revision Flashcard Core Formulae

Total Impedance

$Z = \sqrt{R^2 + (X_L - X_C)^2}$

Resonant Angular Freq

$\omega_0 = \frac{1}{\sqrt{LC}}$ ($Z_{\min} = R$)

Quality Factor (Q)

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

Bandwidth

$2\Delta\omega = \frac{R}{L}$

Series LCR Circuit, Resonance & Quality Factor