The Capacitive Reactance Formula, expressed as \( X_C = \frac{1}{2\pi f C} \), defines the opposition a capacitor offers to alternating current in an AC circuit. This formula is a core concept in Class 12 Physics, covered under Chapter 7 (Alternating Current) of the NCERT textbook. It is equally critical for JEE Main, JEE Advanced, and NEET aspirants, as AC circuit problems appear frequently in these exams. This article covers the definition, derivation, a complete formula sheet, three progressive solved examples, CBSE exam tips, common mistakes, and JEE/NEET applications of the Capacitive Reactance Formula.
Key Capacitive Reactance Formulas at a Glance
Quick reference for the most important capacitive reactance and AC circuit formulas.
- Capacitive Reactance: \( X_C = \frac{1}{2\pi f C} \)
- Angular frequency form: \( X_C = \frac{1}{\omega C} \)
- Impedance in RC circuit: \( Z = \sqrt{R^2 + X_C^2} \)
- Phase angle: \( \tan\phi = \frac{X_C}{R} \)
- Capacitive current: \( I = \frac{V}{X_C} \)
- Resonance condition: \( X_L = X_C \)
- Resonant frequency: \( f_0 = \frac{1}{2\pi\sqrt{LC}} \)
What is Capacitive Reactance?
The Capacitive Reactance Formula describes how a capacitor resists the flow of alternating current. Unlike a resistor, a capacitor does not dissipate energy. Instead, it stores and releases energy in an electric field. The opposition it offers depends on both the frequency of the AC signal and the capacitance value.
When frequency is high, the capacitor charges and discharges rapidly. This allows more current to pass, so reactance is low. When frequency is low, the capacitor takes longer to charge. Less current flows, and reactance is high. At zero frequency (DC), the capacitor blocks current entirely, making reactance theoretically infinite.
This concept is introduced in NCERT Class 12 Physics, Chapter 7 — Alternating Current. It forms the foundation for understanding RC circuits, LC oscillations, and resonance. The SI unit of capacitive reactance is the Ohm (Ω), just like resistance. This is because both quantities represent opposition to current flow, and both follow Ohm's law in their respective contexts.
Understanding this formula is essential for solving problems on impedance, phase difference, power factor, and resonance in AC circuits — all of which are high-weightage topics in CBSE board exams and competitive entrance tests.
Capacitive Reactance Formula — Expression and Variables
The standard expression for capacitive reactance is:
\[ X_C = \frac{1}{2\pi f C} \]
This can also be written using angular frequency \( \omega = 2\pi f \) as:
\[ X_C = \frac{1}{\omega C} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( X_C \) | Capacitive Reactance | Ohm (Ω) |
| \( f \) | Frequency of AC supply | Hertz (Hz) |
| \( C \) | Capacitance of the capacitor | Farad (F) |
| \( \omega \) | Angular frequency | Radian per second (rad/s) |
| \( \pi \) | Mathematical constant | Dimensionless (~3.14159) |
Derivation of the Capacitive Reactance Formula
Consider a capacitor of capacitance \( C \) connected to an AC voltage source \( v = V_0 \sin(\omega t) \). The charge on the capacitor at any instant is \( q = CV_0 \sin(\omega t) \). The instantaneous current is the rate of change of charge:
\[ i = \frac{dq}{dt} = C V_0 \omega \cos(\omega t) = \frac{V_0}{\frac{1}{\omega C}} \cos(\omega t) \]
Comparing this with Ohm's law \( i = \frac{V_0}{X_C} \cos(\omega t) \), we identify the capacitive reactance as \( X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} \). The cosine term also reveals that current leads voltage by 90° in a purely capacitive circuit. This phase relationship is a key exam point.
Complete AC Circuits Formula Sheet
The following table covers all major formulas related to AC circuits as per the NCERT Class 12 syllabus. Use this as a quick revision tool before your board exams or competitive tests.
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Capacitive Reactance | \( X_C = \frac{1}{2\pi f C} \) | f = frequency, C = capacitance | Ω | Class 12, Ch 7 |
| Inductive Reactance | \( X_L = 2\pi f L \) | f = frequency, L = inductance | Ω | Class 12, Ch 7 |
| Impedance (RLC Series) | \( Z = \sqrt{R^2 + (X_L – X_C)^2} \) | R = resistance, X_L, X_C = reactances | Ω | Class 12, Ch 7 |
| Impedance (RC Series) | \( Z = \sqrt{R^2 + X_C^2} \) | R = resistance, X_C = capacitive reactance | Ω | Class 12, Ch 7 |
| Resonant Frequency | \( f_0 = \frac{1}{2\pi\sqrt{LC}} \) | L = inductance, C = capacitance | Hz | Class 12, Ch 7 |
| Phase Angle (RC Circuit) | \( \tan\phi = \frac{X_C}{R} \) | X_C = capacitive reactance, R = resistance | Dimensionless | Class 12, Ch 7 |
| Power Factor | \( \cos\phi = \frac{R}{Z} \) | R = resistance, Z = impedance | Dimensionless | Class 12, Ch 7 |
| RMS Voltage | \( V_{rms} = \frac{V_0}{\sqrt{2}} \) | V_0 = peak voltage | Volt (V) | Class 12, Ch 7 |
| RMS Current | \( I_{rms} = \frac{I_0}{\sqrt{2}} \) | I_0 = peak current | Ampere (A) | Class 12, Ch 7 |
| Average Power (AC) | \( P = V_{rms} I_{rms} \cos\phi \) | V_rms, I_rms = RMS values, φ = phase angle | Watt (W) | Class 12, Ch 7 |
| Quality Factor | \( Q = \frac{1}{R}\sqrt{\frac{L}{C}} \) | R = resistance, L = inductance, C = capacitance | Dimensionless | Class 12, Ch 7 |
Capacitive Reactance Formula — Solved Examples
The following three examples progress from basic Class 10-level substitution to JEE-level multi-concept problems. Work through each one carefully.
Example 1 (Class 10-11 Level) — Direct Application
Problem: A capacitor of capacitance 50 μF is connected to an AC source of frequency 100 Hz. Calculate its capacitive reactance.
Given:
- Capacitance, \( C = 50\,\mu F = 50 \times 10^{-6}\,F \)
- Frequency, \( f = 100\,Hz \)
Step 1: Write the Capacitive Reactance Formula:
\[ X_C = \frac{1}{2\pi f C} \]
Step 2: Substitute the known values:
\[ X_C = \frac{1}{2 \times 3.14159 \times 100 \times 50 \times 10^{-6}} \]
Step 3: Calculate the denominator:
\[ 2 \times 3.14159 \times 100 \times 50 \times 10^{-6} = 0.031416 \]
Step 4: Compute the reactance:
\[ X_C = \frac{1}{0.031416} \approx 31.83\,\Omega \]
Answer
The capacitive reactance is approximately 31.83 Ω.
Example 2 (Class 12 / CBSE Board Level) — Nano-Farad Capacitor
Problem: Find the capacitive reactance of a 120 nF capacitor operating at a frequency of 10,000 Hz (10 kHz).
Given:
- Capacitance, \( C = 120\,nF = 120 \times 10^{-9}\,F \)
- Frequency, \( f = 10{,}000\,Hz \)
Step 1: Write the formula:
\[ X_C = \frac{1}{2\pi f C} \]
Step 2: Substitute values:
\[ X_C = \frac{1}{2 \times 3.14 \times 10000 \times 120 \times 10^{-9}} \]
Step 3: Compute the denominator:
\[ 2 \times 3.14 \times 10000 \times 120 \times 10^{-9} = 7.536 \times 10^{-3} \]
Step 4: Find the reactance:
\[ X_C = \frac{1}{7.536 \times 10^{-3}} \approx 132.7\,\Omega \]
Step 5: Verify units. Frequency is in Hz, capacitance in Farads, so the result is in Ohms (Ω). This is consistent.
Answer
The capacitive reactance of the 120 nF capacitor at 10 kHz is approximately 132.7 Ω.
Example 3 (JEE/NEET Level) — RLC Series Circuit with Impedance
Problem: A series RC circuit has a resistance of 300 Ω and a capacitor of 10 μF. It is connected to an AC source of frequency 50 Hz and peak voltage 220 V. Find: (a) the capacitive reactance, (b) the impedance, (c) the peak current, and (d) the phase angle between voltage and current.
Given:
- Resistance, \( R = 300\,\Omega \)
- Capacitance, \( C = 10\,\mu F = 10 \times 10^{-6}\,F \)
- Frequency, \( f = 50\,Hz \)
- Peak voltage, \( V_0 = 220\,V \)
Step 1: Calculate capacitive reactance using the Capacitive Reactance Formula:
\[ X_C = \frac{1}{2\pi f C} = \frac{1}{2 \times 3.14159 \times 50 \times 10 \times 10^{-6}} \]
\[ X_C = \frac{1}{3.14159 \times 10^{-3}} \approx 318.3\,\Omega \]
Step 2: Calculate impedance of the RC series circuit:
\[ Z = \sqrt{R^2 + X_C^2} = \sqrt{(300)^2 + (318.3)^2} \]
\[ Z = \sqrt{90000 + 101315} = \sqrt{191315} \approx 437.4\,\Omega \]
Step 3: Calculate peak current:
\[ I_0 = \frac{V_0}{Z} = \frac{220}{437.4} \approx 0.503\,A \]
Step 4: Calculate the phase angle:
\[ \tan\phi = \frac{X_C}{R} = \frac{318.3}{300} = 1.061 \]
\[ \phi = \arctan(1.061) \approx 46.7^\circ \]
The current leads the voltage by approximately 46.7° in this RC circuit. This is expected because capacitors cause current to lead voltage.
Answer
- (a) Capacitive Reactance: \( X_C \approx 318.3\,\Omega \)
- (b) Impedance: \( Z \approx 437.4\,\Omega \)
- (c) Peak Current: \( I_0 \approx 0.503\,A \)
- (d) Phase Angle: \( \phi \approx 46.7^\circ \) (current leads voltage)
CBSE Exam Tips 2025-26
- Always convert units first. Convert μF to F and nF to F before substituting into the Capacitive Reactance Formula. Unit errors are the most common source of lost marks in CBSE 2025-26 board exams.
- Memorise both forms. Write both \( X_C = \frac{1}{2\pi f C} \) and \( X_C = \frac{1}{\omega C} \). CBSE papers may give angular frequency \( \omega \) directly. Using the wrong form wastes time.
- State the phase relationship clearly. In a purely capacitive circuit, current leads voltage by 90°. CBSE frequently awards one mark just for stating this correctly in descriptive answers.
- We recommend drawing a phasor diagram for every RC or RLC problem. It helps you identify phase angles visually and avoids sign errors in impedance calculations.
- Know the limiting cases. At very high frequency, \( X_C \to 0 \) (capacitor acts as a short circuit). At DC (f = 0), \( X_C \to \infty \) (capacitor blocks current). These are common 1-mark theory questions in CBSE 2025-26.
- Link to resonance. At resonance in an LC circuit, \( X_L = X_C \). This condition is a 3-5 mark derivation question in board exams. Practise deriving the resonant frequency formula from this condition.
Common Mistakes to Avoid
Students frequently lose marks on capacitive reactance problems due to avoidable errors. Here are the most common ones, along with the correct approach.
- Mistake 1: Forgetting to convert capacitance units. Many students substitute 120 nF directly as 120 instead of \( 120 \times 10^{-9} \) F. Always convert to Farads before using the formula. A nano-Farad is \( 10^{-9} \) F, and a micro-Farad is \( 10^{-6} \) F.
- Mistake 2: Confusing \( X_C \) with \( X_L \). Inductive reactance \( X_L = 2\pi f L \) increases with frequency. Capacitive reactance \( X_C = \frac{1}{2\pi f C} \) decreases with frequency. Students sometimes apply the inductive formula to capacitive problems. Always check which component you are dealing with.
- Mistake 3: Ignoring the phase relationship. In RC circuits, current leads voltage. In RL circuits, voltage leads current. Mixing these up in phasor diagrams leads to incorrect phase angle calculations. Remember: “CIVIL” — in a Capacitor (C), I leads V; in an Inductor (L), V leads I.
- Mistake 4: Using diameter instead of angular frequency. Some students write \( X_C = \frac{1}{\pi f C} \) (missing the factor of 2). The correct formula always has \( 2\pi \) in the denominator when using frequency \( f \) in Hz.
- Mistake 5: Treating capacitive reactance like resistance in power calculations. Power dissipated in a pure capacitor is zero. Students sometimes calculate \( P = I^2 X_C \), which is incorrect. Power is only dissipated in resistive elements. The average power in a purely capacitive circuit is always zero.
JEE/NEET Application of Capacitive Reactance Formula
In our experience, JEE aspirants encounter the Capacitive Reactance Formula in at least 2-3 questions per year, either directly or as part of a larger RLC circuit problem. NEET questions tend to be more conceptual, testing the inverse relationship between \( X_C \) and frequency.
Application Pattern 1: Resonance Condition
JEE frequently asks for the resonant frequency of an LC or RLC circuit. At resonance, \( X_L = X_C \), which gives:
\[ 2\pi f_0 L = \frac{1}{2\pi f_0 C} \implies f_0 = \frac{1}{2\pi\sqrt{LC}} \]
Students must be able to derive this from the Capacitive Reactance Formula and the inductive reactance formula. JEE Advanced 2023 included a multi-part RLC problem where deriving \( f_0 \) was the first step.
Application Pattern 2: Frequency-Dependent Behaviour
NEET and JEE Main often present graphs of \( X_C \) vs. \( f \) or \( X_C \) vs. \( C \) and ask students to identify the correct curve. Since \( X_C \propto \frac{1}{f} \) and \( X_C \propto \frac{1}{C} \), both relationships produce rectangular hyperbolas. Knowing the shape of these graphs is a quick 1-mark win.
Application Pattern 3: Power Factor and Phase Angle in RLC Circuits
JEE Main regularly tests the power factor \( \cos\phi = \frac{R}{Z} \) in series RLC circuits. To find \( Z \), students must first calculate \( X_C \) and \( X_L \) separately. Our experts suggest always computing \( X_C \) and \( X_L \) as the very first step in any RLC impedance problem. This systematic approach prevents errors and saves time under exam conditions.
A typical JEE-style question might state: “An RLC series circuit has R = 100 Ω, L = 0.5 H, C = 100 μF, and is driven at 50 Hz. Find the power factor.” The solution requires the Capacitive Reactance Formula as its foundation. Practise at least 10 such problems before your JEE Main attempt.
FAQs on Capacitive Reactance Formula
Explore More Physics Formulas
Now that you have mastered the Capacitive Reactance Formula, strengthen your AC circuits knowledge by exploring related topics. Visit our complete Physics Formulas hub for a full list of NCERT-aligned formula articles. You may also find these resources helpful:
- Angular Velocity Formula — essential for understanding rotational motion and the relationship between angular and linear frequency in AC circuits.
- Fluid Mechanics Formula — a comprehensive guide to all fluid mechanics expressions for Class 11 and JEE preparation.
- Instantaneous Speed Formula — covers rate-of-change concepts that underpin the derivation of capacitive reactance from charge equations.
For the official NCERT Class 12 Physics syllabus and chapter details, refer to the NCERT official website.