Calculate a low-voltage DC distribution example

What you should be able to do

Calculate conductor resistance and voltage drop in a simple two-wire DC circuit while keeping assumptions visible.

Worked through

Source voltage at its output terminals under the stated load: 24.0 V DC. Load current at the evaluated operating point: 0.50 A. One-way cable route: 100 ft. Two equal conductors, each 100 ft long. Assumed resistance per conductor: 0.010 ohm per foot at the evaluated condition. Additional connection, protection-device and switching losses: omitted for this simplified exercise.

The resistance is a fictional teaching value, not a wire-gauge specification. No actual cable product is selected. The load is assumed to draw the stated current at this operating point; a constant-power or resistive load can behave differently as voltage changes.

Step 1 - Find each conductor's resistance

R = length x resistance per unit length. Outgoing: 100 ft x 0.010 ohm/ft = 1.00 ohm. Return: 100 ft x 0.010 ohm/ft = 1.00 ohm.

Step 2 - Add the wire resistances

R loop = 1.00 + 1.00 = 2.00 ohms. There are 200 ft of conductor, although the source and device are 100 ft apart along the route. Speco's paired-wire guidance expressly includes both feed and return. If a table already gives loop resistance or paired-run voltage drop, do not double it again.

Step 3 - Calculate the wire drop

V drop = I x R loop. 0.50 A x 2.00 ohms = 1.00 V. Each equal conductor contributes 0.50 V of that drop. The same 0.50 A circulates through both; the source current is not 1.00 A merely because two conductors are present.

Step 4 - Calculate voltage across the load

V load = V source - V drop. 24.0 - 1.0 = 23.0 V. This is the voltage between the device's two supply terminals in the simplified model. The return conductor is part of the circuit, not an ideal zero-resistance reference.

As an optional arithmetic check, wire loss is I squared x R loop = 0.50 squared x 2.00 = 0.50 W. Source power is 24.0 x 0.50 = 12.0 W; load power is 23.0 x 0.50 = 11.5 W. The difference is the modeled conductor loss.

Read real cable data carefully

Belden's 9578 product information illustrates that conductor DC resistance is a specified cable property. Check the actual product, units, temperature basis and whether the value is nominal or maximum. Do not use a nominal room-temperature value as a guaranteed worst-case design limit without further evaluation. A conductor's resistance and its allowable current are different properties.

Evaluate the real operating conditions

Include relevant connection and device drops. Evaluate the minimum source voltage, including applicable battery operation, and the actual load behavior. Repeat for relevant peak or startup conditions. Compare the resulting device-terminal voltage with that device's permitted range. Do not compensate by raising a supply voltage without checking every connected device and the approved design.

This model describes one load and one dedicated outgoing/return pair. Shared feeders or shared returns require segment-by-segment currents. A simple doubling shortcut is not automatically valid for every topology.

Practice

1. If the one-way route is 150 ft with the same conductors, what is loop resistance?
2. At 0.50 A on that route, what are the drop and load voltage?
3. In the original 100 ft example, what is the drop at an assumed 0.75 A?

Answers

Calculation record

Source voltage and operating condition: One-way length and topology: Conductor resistance source, units and temperature: Loop/segment resistance: Evaluated current: Connection and equipment drops: Calculated device voltage: Device limits and unresolved assumptions:

Where beginners go wrong

Sources