Resonance in Series LCR Circuits: Full Explanation with Derivation & Applications

Spread the love

Ever wondered why radios can pick one station among hundreds? Or how MRI machines create detailed body images? The secret lies in resonance in series LCR circuits—one of the most powerful phenomena in physics and engineering.

In this complete, exam-ready guide, you’ll master:

The exact condition for resonance (with derivation)
Why current becomes maximum at resonance
How the Quality Factor (Q) controls selectivity
Real-world applications from radios to medical tech

Perfect for Class 12 CBSE/ISC, JEE Main & Advanced, and NEET aspirants!

Contents hide

1 ⚡ What is Resonance in Series LCR Circuits?

2 🧮 Derivation of Resonant Frequency (Critical for Exams!)

3 📊 Key Characteristics at Resonance

3.1 1. Minimum Impedance

3.2 2. Maximum Current

3.3 3. Unity Power Factor

4 🎯 Quality Factor (Q-Factor): The Selectivity Measure

5 📈 Resonance Curves: Visual Understanding

5.1 Current vs Frequency

5.2 Impedance vs Frequency

6 🔧 Practical Applications of LCR Resonance

6.1 1. Radio and TV Tuners

6.2 2. Induction Heating

6.3 3. Medical Imaging (MRI)

6.4 4. Metal Detectors

6.5 5. Wireless Charging

7 ⚠️ Common Misconceptions & Exam Traps

8 🔬 Numerical Example (JEE/NEET Level)

9 💡 Advanced Insight: Energy Perspective

10 📚 Why This Matters for Your Exams

11 🔗 Deepen Your Understanding on PhysicsQanda.com

12 ✨ Final Thought

⚡ What is Resonance in Series LCR Circuits?

Series LCR circuit diagram showing AC source, resistor, inductor and capacitor in series — Series LCR circuit configuration with common current I

Resonance occurs when the inductive reactance (X_L) equals the capacitive reactance (X_C) in a series LCR circuit. At this special frequency:

Impedance becomes minimum (equal to resistance R)
Current becomes maximum
Voltage and current come into phase (φ = 0)

This frequency is called the resonant frequency or natural frequency of the circuit.

🧮 Derivation of Resonant Frequency (Critical for Exams!)

For a series LCR circuit connected to AC source v = V₀sin(ωt):

Step 1: Impedance Expression
The total impedance Z is:

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

Where:

$X_L = \omega L$ (inductive reactance)
$X_C = \frac{1}{\omega C}$ (capacitive reactance)

Step 2: Resonance Condition
At resonance, impedance is minimum when:

$X_L = X_C$

Step 3: Solve for Resonant Frequency
Set $\omega L = \frac{1}{\omega C}$

$\omega^2 = \frac{1}{LC}$

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

Since $\omega = 2\pi f$ :

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

✅ Exam Tip: This derivation appears in NCERT Class 12 Physics (Chapter 7) and is frequently asked in JEE Main. Always write the condition $X_L = X_C$ first!

📊 Key Characteristics at Resonance

Phasor diagram at resonance showing V_L and V_C canceling each other with zero phase difference

1. Minimum Impedance

When $X_L = X_C$ :

$Z = \sqrt{R^2 + (0)^2} = R$

Impedance reduces to pure resistance!

2. Maximum Current

Current amplitude:

$I_0 = \frac{V_0}{Z} = \frac{V_0}{R}$

This is the maximum possible current for given V₀ and R.

3. Unity Power Factor

Phase angle:

$\phi = \tan^{-1}\left(\frac{X_L - X_C}{R}\right) = \tan^{-1}(0) = 0$

Voltage and current are in phase → power factor = cos(0) = 1 (most efficient power transfer)

🎯 Quality Factor (Q-Factor): The Selectivity Measure

The sharpness of resonance is quantified by the Quality Factor:

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

Q-factor comparison – High Q gives sharp resonance (selective), low Q gives broad resonance

What Q-Factor Tells You:

High Q (Q >> 1): Sharp resonance peak, narrow bandwidth, highly selective (ideal for radios)
Low Q (Q ≈ 1): Broad resonance peak, wide bandwidth, less selective

Bandwidth (Δf) is related to Q:

$\Delta f = f_2 - f_1 = \frac{f_0}{Q}$

Where f₁ and f₂ are half-power frequencies.

📈 Resonance Curves: Visual Understanding

Resonance curve showing current maximum at resonant frequency with bandwidth — Current vs frequency curve showing resonance peak and bandwidth Δf

Imagine these graphs in your mind (or see our diagrams below):

Current vs Frequency

Peak at f = f₀
Height = V₀/R
Width inversely proportional to Q

Impedance vs Frequency

Minimum at f = f₀
Value = R
Symmetric curve around f₀

💡 Pro Tip: Sketch these curves in exams—they fetch quick marks and show conceptual clarity!

🔧 Practical Applications of LCR Resonance

Radio tuner application showing resonance selecting specific frequency from multiple stations — Real-world application – Radio tuner uses LCR resonance to select specific frequencies

1. Radio and TV Tuners

Your radio uses a variable capacitor to change f₀. When f₀ matches a station’s frequency, maximum current flows for that signal while rejecting others.

2. Induction Heating

Resonant circuits generate high currents at specific frequencies to heat metals efficiently in industrial processes.

3. Medical Imaging (MRI)

MRI machines use resonant RF coils tuned to hydrogen atom frequencies to create detailed body images.

4. Metal Detectors

Resonant frequency shifts when metal objects disturb the magnetic field, triggering detection.

5. Wireless Charging

Resonant inductive coupling transfers power efficiently between coils tuned to the same frequency.

⚠️ Common Misconceptions & Exam Traps

Myth: “At resonance, voltage across L and C becomes zero.”
Truth: Individual voltages can be very high (Q times supply voltage), but they cancel each other!
Trap: “Resonant frequency depends on resistance R.”
Fact: f₀ = 1/(2π√LC) is independent of R. Only Q-factor depends on R.
Exam Trick: Questions often ask about voltages across L and C at resonance:
$V_L = I_0 X_L = \frac{V_0}{R} \cdot \omega_0 L = Q \cdot V_0$
$V_C = I_0 X_C = Q \cdot V_0$
These can be much larger than supply voltage!

🔬 Numerical Example (JEE/NEET Level)

Problem: A series LCR circuit has L = 100 mH, C = 10 μF, and R = 10 Ω connected to 220V, 50Hz supply. Find:
(a) Resonant frequency
(b) Current at resonance
(c) Q-factor

Solution:

(a) $f_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{0.1 \times 10 \times 10^{-6}}} = 159.2 \text{ Hz}$

(b) At resonance, Z = R = 10 Ω
$I_0 = \frac{V_0}{R} = \frac{220}{10} = 22 \text{ A}$

Notice: Even though supply is 50Hz, resonance occurs at 159.2Hz. At actual supply frequency, current would be much less!

💡 Advanced Insight: Energy Perspective

At resonance, energy oscillates between L and C:

When current is maximum → all energy in inductor: $\frac{1}{2}LI^2$
When voltage is maximum → all energy in capacitor: $\frac{1}{2}CV^2$

Resistance R dissipates energy, requiring continuous power input to maintain oscillations. The Q-factor represents:

$Q = 2\pi \times \frac{\text{Energy stored}}{\text{Energy dissipated per cycle}}$

📚 Why This Matters for Your Exams

NCERT Class 12: Direct derivation questions from Chapter 7 (Example 7.6, Exercise 7.7)
JEE Main: 1-2 questions yearly on resonance frequency, Q-factor calculations
JEE Advanced: Complex problems involving resonance curves and bandwidth
NEET: Conceptual questions on selectivity and applications

🎯 Exam Strategy: Memorize the resonant frequency formula cold. Practice drawing resonance curves with proper labeling of f₀, I_max, and bandwidth.

🔗 Deepen Your Understanding on PhysicsQanda.com

✨ Final Thought

Resonance in LCR circuits isn’t just textbook physics—it’s the invisible force behind wireless communication, medical diagnostics, and countless modern technologies. Master this concept, and you’ll see physics in action everywhere.

Save this guide. Practice the derivations. Sketch the curves. You’ve got this!

Follow @physics_qanda for daily physics insights, exam hacks, and quick derivations!

#LCRresonance #ResonantFrequency #Qfactor #Class12Physics #JEEMains #NEETPhysics #ACcircuits #PhysicsQanda #CBSE #CompetitiveExams

Resonance in Series LCR Circuits: Full Explanation with Derivation & Applications

⚡ What is Resonance in Series LCR Circuits?

🧮 Derivation of Resonant Frequency (Critical for Exams!)