Cable Sizing

AC and DC Cable Performance Worked Examples

Textbook-style AC and DC cable performance worked examples with equations for temperature, resistance, voltage drop, power loss and indicative life.

Updated June 1, 2026

These worked examples show how cable performance calculations can combine current rating, operating temperature, resistance, voltage drop, power loss and an indicative life estimate. The calculations are presented as engineering steps rather than software code.

A.C. cable performance example

Given data

InputValue
System voltage, U400 V three-phase
Current rating, Iz222 A
Design current, Ib180 A
Length, L56 m
Ambient temperature, Ta20 °C
Maximum conductor temperature, Tc90 °C
Resistance at 20 °C, R200.443 mΩ/m
Reactance, X0.099 mΩ/m
Temperature coefficient, α204.03 × 10-3 per °C

Step 1: estimate conductor operating temperature

t = (Ib / Iz)^2 × (Tc – Ta) + Tat = (180 / 222)^2 × (90 – 20) + 20t = 66.0 °C

Step 2: adjust resistance for operating temperature

R = R20 × [1 + α20 × (t – 20)]R = 0.443e-3 × [1 + 4.03e-3 × (66.0 – 20)]R = 5.252e-4 Ω/m

Step 3: calculate voltage drop

For this three-phase example, b = 1. A power factor of 0.8 is used, so cos φ = 0.8 and sin φ = 0.6.

dU% = 100 × b × (R cos φ + X sin φ) × Ib × L / UdU% = 100 × 1 × [(5.252e-4 × 0.8) + (0.099e-3 × 0.6)] × 180 × 56 / 400dU = 1.21%

Step 4: calculate power loss

P = Ib^2 × R × n × LP = 180^2 × 5.252e-4 × 3 × 56P = 2,859 W

Step 5: indicative life estimate

The original example used an Arrhenius relationship for XLPE with slope factor m = -0.0693 and intercept b = 18.5678.

Life hours = e^(m × t + b)Life hours = e^(-0.0693 × 66.0 + 18.5678)Life hours = 1,193,838 hours, approximately 136 years

D.C. cable performance example

Given data

InputValue
System voltage, U720 V d.c.
Current rating, Iz152 A
Design current, Ib132 A
Length, L56 m
Ambient temperature, Ta30 °C
Maximum conductor temperature, Tc90 °C
Resistance at 20 °C, R201.24 mΩ/m
Reactance, X0 Ω/m
Temperature coefficient, α203.93 × 10-3 per °C

Step 1: estimate conductor operating temperature

The original d.c. example used a linear load-to-rating ratio for this step.

t = (Ib / Iz) × (Tc – Ta) + Tat = (132 / 152) × (90 – 30) + 30t = 82.1 °C

Design note: many thermal calculations use an I2 relationship for heat generation, so this temperature assumption should be reviewed before using the example as a design method.

Step 2: adjust resistance for operating temperature

R = R20 × [1 + α20 × (t – 20)]R = 1.24e-3 × [1 + 3.93e-3 × (82.1 – 20)]R = 1.543e-3 Ω/m

Step 3: calculate voltage drop

For a two-wire d.c. circuit, b = 2.

dU% = 100 × b × R × Ib × L / UdU% = 100 × 2 × 1.543e-3 × 132 × 56 / 720dU = 3.17%

Step 4: calculate power loss

P = Ib^2 × R × n × LP = 132^2 × 1.543e-3 × 2 × 56P = 3,010 W

Step 5: indicative life estimate

Life hours = e^(m × t + b)Life hours = e^(-0.0693 × 82.1 + 18.5678)Life hours = 391,564 hours, approximately 45 years

These examples connect to the articles on conductor resistance, reactance, voltage drop, power loss and estimating cable life.

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