FISEKON GmbH – Fischer Elektrokonstruktion

Discuss a project

Contact
+49 (0) 5247 4070-66info@fisekon.comAuf Deutsch

07 — Calculators

Calculate parallel cables

Beyond a certain size a single conductor is no longer workable — 240 mm² cannot sensibly be bent any more. Cables are then run in parallel. The arithmetic is simple; the trap lies elsewhere: the current only divides evenly if the cables really are identical.

Input

Already including grouping — parallel cables necessarily lie close together.

With unequal length or routing the current does not divide evenly. 10 % is a common allowance.

Result

Total cross-section per line conductor
—mm²
Current per cable if evenly divided
—A
Current in the most heavily loaded cableThis cable governs, not the average.
—A
Margin on the most loaded cable
—%

A guide, not a design to standard. We give no warranty for the correctness of the results, the applicable standards and case-by-case verification govern.

Worked example

The default case the calculator starts with, worked through once. Change the values above and it recalculates immediately.

Inputs

Number of parallel cables per line conductor
2
Cross-section per cable
95 mm²
Total current
400 A
Current-carrying capacity per cable
230 A
Assumed current imbalance
10 %

Result

Total cross-section per line conductor
190 mm²
Current per cable if evenly divided
200 A
Current in the most heavily loaded cable
220 A
Margin on the most loaded cable
4.35 %

Formula

  • A_total = n · A
  • I per cable = I_total / n
  • I_worst = I per cable · (1 + imbalance)
  • Margin = (I_z − I_worst) / I_z

Assumptions and standards

  • The calculation assumes equal cross-section and equal material for all parallel cables.
  • The capacity per cable is entered and must already include grouping: parallel cables lie close together and heat one another.
  • The imbalance is an allowance, not a calculation. It grows with differences in length, installation method and position within the bundle.

Frequently asked

Why does the current not divide evenly?

Because the impedances differ. A few metres of length difference, a different position in the bundle or one extra bend shifts the division. The cable with the lowest impedance takes the most current — and that one governs the heating.

What matters for the protection?

A common protective device only sees the total current. If one of the parallel cables fails, the others take over its share and become overloaded without the device operating. That is why execution matters here: equal length, equal cross-section, equal installation method, no branches.

How many cables may I run in parallel?

There is no fixed technical limit; in practice it becomes unwieldy beyond three or four. Check first whether busbar trunking is the better answer — it solves the symmetry question by construction.

Calculating is the easy part.

A formula gives you a number. Designing a plant also demands installation method, grouping, discrimination, the standards in force and a look at the installed base. That is what we take on.

Discuss a project

More tools

Contact

Tell us what it is about.

A phone call or three lines is enough. From the very start you talk to the people who will later work on your project, not to a distribution list.

Direct

Locations

Registered office and postal addressStellmacherstr. 7
33428 Marienfeld (Harsewinkel)
Gütersloh officeWagenfeldstr. 2
33332 Gütersloh

Or write to us

We usually reply within one working day.

Write email

Opens your email program with the message already written. Read it, change whatever you like, and send it yourself — from your mailbox to ours.

Go to the contact form