The Inductive Reactance Formula gives the opposition offered by an inductor to alternating current, expressed as \( X_L = 2\pi f L \), where \( f \) is the frequency and \( L \) is the inductance. This concept is a core part of NCERT Class 12 Physics, Chapter 7 (Alternating Current). It is also heavily tested in JEE Main, JEE Advanced, and NEET. This article covers the formula, its derivation, a complete AC formula sheet, three solved examples, CBSE exam tips, common mistakes, and JEE/NEET applications.
Key Inductive Reactance Formulas at a Glance
Quick reference for the most important inductive reactance and AC circuit formulas.
- Inductive Reactance: \( X_L = 2\pi f L = \omega L \)
- Angular frequency: \( \omega = 2\pi f \)
- Impedance (RL circuit): \( Z = \sqrt{R^2 + X_L^2} \)
- Voltage across inductor: \( V_L = I \cdot X_L \)
- Phase angle: \( \tan\phi = \dfrac{X_L}{R} \)
- Resonant frequency (LC): \( f_0 = \dfrac{1}{2\pi\sqrt{LC}} \)
- Quality factor: \( Q = \dfrac{\omega_0 L}{R} \)
What is Inductive Reactance?
Inductive reactance is the opposition that an inductor offers to the flow of alternating current (AC). Unlike resistance, inductive reactance does not dissipate energy as heat. Instead, it stores energy in a magnetic field and returns it to the circuit. The Inductive Reactance Formula shows that this opposition increases with both frequency and inductance.
An inductor is essentially a coil of wire. When AC flows through it, the changing current creates a changing magnetic flux. By Faraday's law, this changing flux induces a back-EMF (electromotive force) that opposes the change in current. This opposition is what we call inductive reactance, denoted by \( X_L \).
In NCERT Class 12 Physics, Chapter 7 (Alternating Current), inductive reactance is introduced as a key quantity in analysing AC circuits. The concept is also connected to Lenz's law and electromagnetic induction from Chapter 6. At DC (frequency = 0), inductive reactance is zero, so an ideal inductor behaves as a short circuit for DC. At very high frequencies, \( X_L \) becomes very large, effectively blocking AC signals. This frequency-dependent behaviour makes inductors essential in filters, transformers, and oscillators.
Inductive Reactance Formula — Expression and Variables
The standard expression for the Inductive Reactance Formula is:
\[ X_L = 2\pi f L = \omega L \]
Here, \( X_L \) is the inductive reactance in ohms (Ω), \( f \) is the frequency of the AC supply in hertz (Hz), \( L \) is the self-inductance of the inductor in henries (H), and \( \omega = 2\pi f \) is the angular frequency in radians per second (rad/s).
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( X_L \) | Inductive Reactance | Ohm (Ω) |
| \( f \) | Frequency of AC supply | Hertz (Hz) |
| \( L \) | Self-Inductance | Henry (H) |
| \( \omega \) | Angular Frequency | rad/s |
| \( V_L \) | Voltage across inductor | Volt (V) |
| \( I \) | Current through inductor | Ampere (A) |
Derivation of the Inductive Reactance Formula
Consider an inductor of inductance \( L \) connected to an AC source: \( V = V_0 \sin(\omega t) \).
Step 1: Apply Kirchhoff's voltage law. The voltage across the inductor equals \( L \dfrac{dI}{dt} \).
Step 2: Set up the equation: \( V_0 \sin(\omega t) = L \dfrac{dI}{dt} \).
Step 3: Integrate both sides with respect to time:
\[ I = \frac{V_0}{\omega L} \sin\left(\omega t – \frac{\pi}{2}\right) = I_0 \sin\left(\omega t – \frac{\pi}{2}\right) \]
Step 4: Compare with Ohm's law. The peak current is \( I_0 = \dfrac{V_0}{\omega L} \). Therefore, the effective resistance (reactance) is:
\[ X_L = \omega L = 2\pi f L \]
The negative sign in the phase indicates that current lags voltage by 90° in a pure inductor. This phase relationship is a critical exam point.
Complete AC Circuit Formula Sheet
The following table provides all important AC circuit formulas as listed in the NCERT Class 12 Physics syllabus. Use this as a quick revision sheet before your board exams and JEE/NEET preparation.
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Inductive Reactance | \( X_L = 2\pi f L \) | f = frequency, L = inductance | Ω | Class 12, Ch 7 |
| Capacitive Reactance | \( X_C = \dfrac{1}{2\pi f C} \) | f = frequency, C = capacitance | Ω | Class 12, Ch 7 |
| Impedance (RL Circuit) | \( Z = \sqrt{R^2 + X_L^2} \) | R = resistance, \( X_L \) = inductive reactance | Ω | Class 12, Ch 7 |
| Impedance (RC Circuit) | \( Z = \sqrt{R^2 + X_C^2} \) | R = resistance, \( X_C \) = capacitive reactance | Ω | Class 12, Ch 7 |
| Impedance (RLC Circuit) | \( Z = \sqrt{R^2 + (X_L – X_C)^2} \) | R, \( X_L \), \( X_C \) | Ω | Class 12, Ch 7 |
| Phase Angle (RL) | \( \tan\phi = \dfrac{X_L}{R} \) | \( X_L \) = inductive reactance, R = resistance | Dimensionless | Class 12, Ch 7 |
| Resonant Frequency | \( f_0 = \dfrac{1}{2\pi\sqrt{LC}} \) | L = inductance, C = capacitance | Hz | Class 12, Ch 7 |
| Quality Factor | \( Q = \dfrac{\omega_0 L}{R} = \dfrac{1}{\omega_0 CR} \) | \( \omega_0 \) = resonant angular frequency | Dimensionless | Class 12, Ch 7 |
| RMS Voltage | \( V_{rms} = \dfrac{V_0}{\sqrt{2}} \) | \( V_0 \) = peak voltage | V | Class 12, Ch 7 |
| Average Power (AC) | \( P = V_{rms} I_{rms} \cos\phi \) | \( \phi \) = phase angle | Watt (W) | Class 12, Ch 7 |
| Power Factor | \( \cos\phi = \dfrac{R}{Z} \) | R = resistance, Z = impedance | Dimensionless | Class 12, Ch 7 |
Inductive Reactance Formula — Solved Examples
Example 1 (Class 10-11 Level — Direct Application)
Problem: An inductor has a self-inductance of 0.5 H. It is connected to an AC source of frequency 50 Hz. Calculate the inductive reactance of the inductor.
Given: L = 0.5 H, f = 50 Hz
Step 1: Write the Inductive Reactance Formula: \( X_L = 2\pi f L \)
Step 2: Substitute the given values:
\[ X_L = 2 \times \pi \times 50 \times 0.5 \]
Step 3: Simplify: \( X_L = 2 \times 3.1416 \times 50 \times 0.5 = 157.08 \) Ω
Answer
Inductive Reactance \( X_L \approx 157.1 \) Ω
Example 2 (Class 12 Level — RL Circuit Impedance)
Problem: A series RL circuit has a resistance of 30 Ω and an inductance of 0.1 H. The AC supply frequency is 100 Hz. Find (a) the inductive reactance, (b) the impedance of the circuit, and (c) the phase angle between voltage and current.
Given: R = 30 Ω, L = 0.1 H, f = 100 Hz
Step 1: Calculate inductive reactance using \( X_L = 2\pi f L \):
\[ X_L = 2 \times \pi \times 100 \times 0.1 = 20\pi \approx 62.83 \text{ Ω} \]
Step 2: Calculate impedance using \( Z = \sqrt{R^2 + X_L^2} \):
\[ Z = \sqrt{(30)^2 + (62.83)^2} = \sqrt{900 + 3947.6} = \sqrt{4847.6} \approx 69.6 \text{ Ω} \]
Step 3: Calculate phase angle using \( \tan\phi = \dfrac{X_L}{R} \):
\[ \tan\phi = \frac{62.83}{30} = 2.094 \implies \phi = \arctan(2.094) \approx 64.5^\circ \]
Answer
(a) \( X_L \approx 62.83 \) Ω (b) \( Z \approx 69.6 \) Ω (c) \( \phi \approx 64.5^\circ \) (current lags voltage)
Example 3 (JEE/NEET Level — Resonance and Reactance)
Problem: In a series RLC circuit, R = 10 Ω, L = 100 mH, and C = 25 μF. The circuit is connected to an AC source of variable frequency. Find (a) the resonant frequency, (b) the inductive reactance at resonance, and (c) the quality factor of the circuit.
Given: R = 10 Ω, L = 100 mH = 0.1 H, C = 25 μF = 25 × 10&sup6; F
Step 1: Calculate resonant frequency \( f_0 = \dfrac{1}{2\pi\sqrt{LC}} \):
\[ f_0 = \frac{1}{2\pi\sqrt{0.1 \times 25 \times 10^{-6}}} = \frac{1}{2\pi\sqrt{2.5 \times 10^{-6}}} \]
\[ f_0 = \frac{1}{2\pi \times 1.581 \times 10^{-3}} = \frac{1}{9.934 \times 10^{-3}} \approx 100.7 \text{ Hz} \]
Step 2: At resonance, \( \omega_0 = 2\pi f_0 \approx 2\pi \times 100.7 \approx 632.5 \) rad/s. Calculate inductive reactance at resonance:
\[ X_L = \omega_0 L = 632.5 \times 0.1 = 63.25 \text{ Ω} \]
Step 3: At resonance, \( X_L = X_C \), confirming the calculation. Calculate quality factor:
\[ Q = \frac{\omega_0 L}{R} = \frac{63.25}{10} = 6.325 \]
Step 4: Verify using alternate formula: \( Q = \dfrac{1}{R}\sqrt{\dfrac{L}{C}} = \dfrac{1}{10}\sqrt{\dfrac{0.1}{25 \times 10^{-6}}} = \dfrac{1}{10} \times 63.25 = 6.325 \) ✓
Answer
(a) \( f_0 \approx 100.7 \) Hz (b) \( X_L \approx 63.25 \) Ω (c) Q ≈ 6.325
CBSE Exam Tips 2025-26
- Memorise the phase relationship: In a pure inductor, current lags voltage by exactly 90°. CBSE frequently asks this as a one-mark question. We recommend writing this as a mnemonic: “ELI the ICE man” — in an inductor (L), E leads I.
- Units matter: Always express \( X_L \) in ohms (Ω). Many students lose marks by omitting units. The SI unit of inductive reactance is the same as resistance.
- Distinguish reactance from resistance: Reactance does not dissipate energy. Resistance does. CBSE 2025-26 marking schemes specifically reward this distinction in 3-mark answers.
- Draw phasor diagrams: For RL and RLC circuits, always draw the phasor diagram. CBSE awards 1 mark specifically for a correct phasor diagram in 3-mark and 5-mark questions.
- Frequency dependence: Be ready to sketch the graph of \( X_L \) vs. \( f \). It is a straight line through the origin with slope \( 2\pi L \). Our experts suggest practising this graph for the 2025-26 boards.
- Formula variants: Know both forms \( X_L = 2\pi f L \) and \( X_L = \omega L \). CBSE sometimes gives \( \omega \) directly instead of \( f \).
Common Mistakes to Avoid
Students frequently lose marks due to these avoidable errors. Review each one carefully before your exam.
- Mistake 1: Confusing reactance with resistance. Inductive reactance \( X_L \) and resistance \( R \) both have units of ohms. However, they cannot be added directly. In an RL circuit, use \( Z = \sqrt{R^2 + X_L^2} \), not \( Z = R + X_L \). Adding them directly is the most common error in CBSE board exams.
- Mistake 2: Using the wrong frequency. The formula \( X_L = 2\pi f L \) uses the ordinary frequency \( f \) in Hz. If the problem gives angular frequency \( \omega \), use \( X_L = \omega L \) directly. Do not substitute \( \omega \) for \( f \) in the formula with \( 2\pi \).
- Mistake 3: Ignoring the phase angle direction. In an inductor, current lags voltage. In a capacitor, current leads voltage. Mixing these up in phasor diagrams costs marks. Remember: “ELI the ICE man.”
- Mistake 4: Applying inductive reactance to DC circuits. The formula \( X_L = 2\pi f L \) gives \( X_L = 0 \) for DC (f = 0). An ideal inductor offers zero opposition to steady DC. Students sometimes incorrectly apply the formula to DC problems.
- Mistake 5: Incorrect unit conversion. Always convert inductance to henries (H) and frequency to hertz (Hz) before substituting. If L is given in mH, divide by 1000 first. Skipping this step leads to answers that are off by factors of 1000.
JEE/NEET Application of Inductive Reactance Formula
In our experience, JEE aspirants encounter the Inductive Reactance Formula in multiple question types. Understanding these patterns helps you solve problems faster under exam conditions.
Pattern 1: Reactance and Impedance Calculation
JEE Main frequently asks you to find the impedance of a series RLC circuit at a given frequency. The approach is always the same. First, calculate \( X_L = \omega L \). Then calculate \( X_C = \dfrac{1}{\omega C} \). Finally, find \( Z = \sqrt{R^2 + (X_L – X_C)^2} \). JEE 2023 and 2024 both had direct questions on this pattern. Practise this three-step approach until it is automatic.
Pattern 2: Resonance Condition
At resonance in a series RLC circuit, \( X_L = X_C \). This gives \( \omega_0 = \dfrac{1}{\sqrt{LC}} \). JEE Advanced tests this concept in multi-concept problems combining resonance with power dissipation. At resonance, impedance is minimum (\( Z = R \)) and current is maximum. NEET also tests resonance in the context of radio tuning circuits.
Pattern 3: Power and Power Factor
A pure inductor has a power factor of zero (\( \cos\phi = 0 \) since \( \phi = 90^\circ \)). This means a pure inductor consumes zero average power. JEE frequently combines this fact with energy calculations. The average power in an RL circuit is \( P = I_{rms}^2 R \), not \( I_{rms}^2 Z \). In our experience, this distinction between R and Z in power calculations is a top source of errors in JEE Main. Always remember: only the resistive component dissipates power. The reactive component (inductor or capacitor) does not.
NEET-Specific Tip
NEET focuses more on conceptual understanding than numerical calculation. Expect questions like: “How does inductive reactance change when frequency is doubled?” Since \( X_L \propto f \), doubling the frequency doubles the reactance. Similarly, \( X_L \propto L \). These proportionality-based questions are straightforward if you understand the formula deeply.
FAQs on Inductive Reactance Formula
Explore More Physics Formulas
Strengthen your AC circuit concepts by exploring these related formula articles on ncertbooks.net. Each article follows the same structured approach with solved examples and exam tips.
- Learn how charge flows in capacitors with the Electric Flux Formula — essential for understanding capacitive reactance.
- Revise rotational motion concepts with the Angular Speed Formula, which connects to angular frequency \( \omega \) in AC circuits.
- Build your foundation in kinematics with the Average Acceleration Formula for a complete Class 11-12 physics revision.
- Browse our full Physics Formulas Hub for a complete list of NCERT and competitive exam formulas.
For the official NCERT Class 12 Physics syllabus and textbook, visit the NCERT official website.