Reactance Calculator — XL and XC

Calculate inductive reactance (XL = 2πfL) and capacitive reactance (XC = 1/2πfC) for any frequency. Also computes the LC resonant frequency where XL = XC. Essential for filter and impedance matching design.

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Formula

X_L = 2\pi f L, \quad X_C = \frac{1}{2\pi f C}, \quad f_{res} = \frac{1}{2\pi\sqrt{LC}}

Reference: Pozar, Microwave Engineering, 4th ed.

X_L— Inductive reactance (Ω)

X_C— Capacitive reactance (Ω)

f— Frequency (Hz)

L— Inductance (H)

C— Capacitance (F)

f_res— Resonant frequency where XL = XC (Hz)

How It Works

Reactance is the opposition to alternating current caused by inductors and capacitors, expressed in ohms (Ω). Inductive reactance XL = 2πfL increases linearly with frequency, while capacitive reactance XC = 1/(2πfC) decreases inversely with frequency. At the resonant frequency f_res = 1/(2π√LC), XL equals XC and the two reactances cancel — a fundamental principle underlying every LC filter, oscillator, and impedance matching network. Unlike resistance, reactance stores energy (magnetic field in inductors, electric field in capacitors) rather than dissipating it, making it a purely imaginary component of complex impedance. Understanding XL and XC is prerequisite knowledge for Pozar's microwave filter design methods and any RF impedance matching task.

Worked Example

Design a 50 Ω matching network at 100 MHz. Step 1: Compute XL for L = 80 nH — XL = 2π × 100×10⁶ × 80×10⁻⁹ = 50.27 Ω. This inductor presents 50 Ω reactance at the target frequency. Step 2: Compute XC for C = 31.8 pF — XC = 1/(2π × 100×10⁶ × 31.8×10⁻¹²) = 50.03 Ω. Both components present ~50 Ω at 100 MHz. Step 3: Resonant frequency of the LC pair — f_res = 1/(2π√(80e-9 × 31.8e-12)) = 99.97 MHz ≈ 100 MHz. The components resonate at the design frequency, confirming the matching network design. Key insight: XL doubles every octave (2× frequency), XC halves every octave — making frequency scaling straightforward in filter design.

Practical Tips

✓Per Pozar's 'Microwave Engineering', use quarter-wave transmission lines as lossless reactance elements above 500 MHz — a λ/4 open-circuited stub presents a short circuit (zero reactance), while a short-circuited stub presents an open circuit
✓For bypass capacitors, target XC < 1/10 of the supply impedance at the noise frequency; a 100 nF capacitor has XC = 1.6 Ω at 1 MHz but only 0.16 Ω at 10 MHz, making frequency choice critical for power supply filtering
✓To hit a specific reactance value, use standard E96 component values and apply reactance scaling: for XL = 100 Ω at 10 MHz, L = XL/(2πf) = 100/(2π×10⁷) = 1.59 µH — the nearest E24 value is 1.5 µH (XL = 94.2 Ω, 5.8% low)

Common Mistakes

✗Confusing the frequency dependence directions — XL increases with frequency (inductive impedance rises) while XC decreases (capacitor becomes a short at high frequency), leading to swapped component selections in filter designs
✗Ignoring parasitic reactance — a real 100 nH inductor has 5–10 pF winding capacitance, creating a self-resonant frequency (SRF); above the SRF the component behaves capacitively, invalidating XL calculations
✗Using the wrong unit multiplier — entering 100 MHz as 100 instead of 1e8 Hz, or mixing µH with nH, produces reactance errors of 3 orders of magnitude

Frequently Asked Questions

What is the difference between reactance and impedance?

Reactance (X) is the imaginary part of impedance (Z). For a pure inductor, Z = jXL; for a pure capacitor, Z = -jXC. When combined with resistance R, total impedance is Z = R + j(XL - XC). Magnitude |Z| = √(R² + (XL-XC)²). Reactance stores and returns energy each cycle, while resistance dissipates it as heat.

Why does an inductor block high frequencies?

XL = 2πfL increases linearly with frequency. A 10 µH inductor presents 63 Ω at 1 MHz but 628 Ω at 10 MHz. At high frequencies the inductor opposes rapid current changes (Faraday's law: V = L·dI/dt), making it an effective RF choke. This is why ferrite beads and inductors are used to block switching noise above 10 MHz in power supply designs.

How do I choose L and C values to resonate at a specific frequency?

Start with f_res = 1/(2π√LC) and choose a convenient impedance level. For 50 Ω at 100 MHz: set Z₀ = √(L/C) = 50 Ω, giving L/C = 2500. Combined with f_res: L = Z₀/(2πf) = 50/(6.28×10⁸) = 79.6 nH, C = 1/(Z₀×2πf) = 31.8 pF. Alternatively, use Murata's online SimSurfing tool to find standard value pairs within 1% of target.

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