Shielding System Inductance and Resistance Calculator – IEC, IEEE

Shielding system inductance and resistance calculations are critical for ensuring electrical safety and system reliability. Accurate computation helps in designing effective grounding and shielding systems compliant with IEC and IEEE standards.

This article explores detailed methodologies, formulas, and practical examples for calculating shielding system inductance and resistance. It covers standard values, real-world applications, and provides an AI-powered calculator for precise results.

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  • Calculate inductance and resistance for a 100-meter copper shielded cable per IEC 62305.
  • Determine shielding system resistance for a 50-meter aluminum conductor using IEEE Std 80.
  • Compute inductance of a coaxial cable shield with 10 mm radius and 1 mm thickness.
  • Find resistance and inductance values for a 200-meter steel wire shield under IEC 61024 guidelines.

Common Values for Shielding System Inductance and Resistance – IEC, IEEE

MaterialResistivity (ρ) [Ω·m]Typical Shield Thickness (t) [mm]Permeability (μr)Typical Inductance Range [μH/m]Standard Reference
Copper1.68 × 10⁻⁸0.5 – 2~10.2 – 0.5IEC 62305, IEEE Std 80
Aluminum2.82 × 10⁻⁸1 – 3~10.3 – 0.7IEC 61024, IEEE Std 80
Steel (Galvanized)1.43 × 10⁻⁷1 – 5~100 – 2000.5 – 1.5IEEE Std 80, IEC 62305
Silver1.59 × 10⁻⁸0.1 – 1~10.15 – 0.4IEC 62305
Shield GeometryTypical Inductance FormulaTypical Resistance FormulaNotes
Cylindrical Shield (Wire)L = (μ₀ * μr * l / 2π) * ln(D/d)R = ρ * l / AD = outer diameter, d = wire diameter
Flat Strip ShieldL ≈ (μ₀ * μr * l / w) * kR = ρ * l / (t * w)w = width, t = thickness, k = geometry factor
Coaxial Cable ShieldL = (μ₀ * μr * l / 2π) * ln(b/a)R = ρ * l / (2π * t * r)a = inner radius, b = outer radius

Fundamental Formulas for Shielding System Inductance and Resistance

Understanding the formulas and variables involved in calculating shielding system inductance and resistance is essential for accurate design and analysis. Below are the key formulas used in IEC and IEEE standards.

Inductance Calculation

  • Inductance of a Cylindrical Wire Shield:

    L = (μ₀ × μr × l / 2π) × ln(D / d)

    • L: Inductance (Henries, H)
    • μ₀: Permeability of free space (4π × 10⁻⁷ H/m)
    • μr: Relative permeability of shield material (dimensionless)
    • l: Length of the shield (meters, m)
    • D: Outer diameter of the shield (meters, m)
    • d: Diameter of the wire or inner conductor (meters, m)
  • Inductance of a Coaxial Cable Shield:

    L = (μ₀ × μr × l / 2π) × ln(b / a)

    • b: Outer radius of the shield (meters, m)
    • a: Inner radius of the shield (meters, m)
  • Inductance of a Flat Strip Shield:

    L ≈ (μ₀ × μr × l / w) × k

    • w: Width of the strip (meters, m)
    • k: Geometry factor (dimensionless, typically 1–2)

Resistance Calculation

  • Resistance of Shielding Conductor:

    R = ρ × l / A

    • R: Resistance (Ohms, Ω)
    • ρ: Resistivity of the material (Ohm-meters, Ω·m)
    • l: Length of the conductor (meters, m)
    • A: Cross-sectional area of the conductor (square meters, m²)
  • Resistance of a Coaxial Shield:

    R = ρ × l / (2π × t × r)

    • t: Thickness of the shield (meters, m)
    • r: Mean radius of the shield (meters, m)

Detailed Real-World Examples

Example 1: Calculating Inductance and Resistance of a Copper Wire Shield

A copper wire shield is 100 meters long with an outer diameter of 10 mm and a wire diameter of 2 mm. Calculate the inductance and resistance of the shield according to IEC 62305.

  • Given:
    • Length, l = 100 m
    • Outer diameter, D = 10 mm = 0.01 m
    • Wire diameter, d = 2 mm = 0.002 m
    • Resistivity of copper, ρ = 1.68 × 10⁻⁸ Ω·m
    • Relative permeability, μr ≈ 1 (non-magnetic copper)
  • Step 1: Calculate Inductance (L)
  • Using the formula:

    L = (μ₀ × μr × l / 2π) × ln(D / d)

    Calculate ln(D/d): ln(0.01 / 0.002) = ln(5) ≈ 1.609

    Calculate μ₀ × μr × l / 2π:

    μ₀ = 4π × 10⁻⁷ H/m

    So, (4π × 10⁻⁷ × 1 × 100) / (2π) = (4π × 10⁻⁵) / (2π) = 2 × 10⁻⁵ H

    Therefore, L = 2 × 10⁻⁵ × 1.609 = 3.218 × 10⁻⁵ H = 32.18 μH

  • Step 2: Calculate Resistance (R)
  • Cross-sectional area, A = π × (d/2)² = π × (0.002/2)² = π × (0.001)² = 3.1416 × 10⁻⁶ m²

    Resistance, R = ρ × l / A = (1.68 × 10⁻⁸) × 100 / (3.1416 × 10⁻⁶) ≈ 0.535 Ω

Result: The copper wire shield has an inductance of approximately 32.18 μH and a resistance of 0.535 Ω over 100 meters.

Example 2: Shielding Resistance and Inductance of an Aluminum Coaxial Cable

An aluminum coaxial cable shield has an inner radius of 4 mm, outer radius of 5 mm, and length of 50 meters. Calculate the inductance and resistance according to IEEE Std 80.

  • Given:
    • Length, l = 50 m
    • Inner radius, a = 4 mm = 0.004 m
    • Outer radius, b = 5 mm = 0.005 m
    • Thickness, t = b – a = 0.001 m
    • Resistivity of aluminum, ρ = 2.82 × 10⁻⁸ Ω·m
    • Relative permeability, μr ≈ 1
  • Step 1: Calculate Inductance (L)
  • Using the coaxial inductance formula:

    L = (μ₀ × μr × l / 2π) × ln(b / a)

    Calculate ln(b/a): ln(0.005 / 0.004) = ln(1.25) ≈ 0.2231

    Calculate μ₀ × μr × l / 2π:

    (4π × 10⁻⁷ × 1 × 50) / (2π) = (2 × 10⁻⁵) H

    Therefore, L = 2 × 10⁻⁵ × 0.2231 = 4.462 × 10⁻⁶ H = 4.462 μH

  • Step 2: Calculate Resistance (R)
  • Mean radius, r = (a + b) / 2 = (0.004 + 0.005) / 2 = 0.0045 m

    Resistance formula for coaxial shield:

    R = ρ × l / (2π × t × r)

    Calculate denominator: 2π × 0.001 × 0.0045 = 2π × 4.5 × 10⁻⁶ ≈ 2.827 × 10⁻⁵ m²

    Calculate R: (2.82 × 10⁻⁸) × 50 / 2.827 × 10⁻⁵ ≈ 0.0498 Ω

Result: The aluminum coaxial cable shield has an inductance of approximately 4.462 μH and resistance of 0.0498 Ω over 50 meters.

Additional Technical Considerations

  • Skin Effect: At high frequencies, current tends to flow near the surface of the conductor, effectively reducing the cross-sectional area and increasing resistance. This phenomenon must be considered for shielding systems exposed to transient or high-frequency currents, as per IEEE Std 80.
  • Temperature Dependence: Resistivity varies with temperature. The temperature coefficient of resistivity (α) for copper is approximately 0.0039/°C. Resistance at temperature T can be calculated as R_T = R_20 × [1 + α × (T – 20)].
  • Magnetic Permeability: Materials like steel have high relative permeability, significantly increasing inductance. This must be accounted for in shielding design to avoid unintended inductive coupling or resonance.
  • Shield Geometry Impact: The shape and arrangement of the shield influence inductance and resistance. For example, braided shields have different effective cross-sectional areas and inductive properties compared to solid or foil shields.
  • Standard Compliance: IEC 62305 focuses on lightning protection systems, emphasizing low resistance grounding and shielding. IEEE Std 80 provides detailed guidelines for grounding and shielding in substations, including calculation methods for inductance and resistance.

References and Further Reading