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Dc Cable Loss Calculator

Power Loss Formula:

\[ Power\ Loss = I² × R × Length × 2 / 1000 \]

A
Ω/km
km

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1. What is DC Cable Power Loss?

DC cable power loss refers to the energy dissipated as heat in electrical cables due to resistance. This calculator estimates power loss in DC cables based on current, resistance per kilometer, and cable length, accounting for round-trip current flow.

2. How Does the Calculator Work?

The calculator uses the power loss formula:

\[ Power\ Loss = I² × R × Length × 2 / 1000 \]

Where:

Explanation: The formula calculates power loss using Joule's law (P = I²R), multiplied by length and doubled for round-trip current, then divided by 1000 for proper unit conversion.

3. Importance of Power Loss Calculation

Details: Calculating power loss is crucial for designing efficient electrical systems, selecting appropriate cable sizes, minimizing energy waste, and ensuring system reliability and safety.

4. Using the Calculator

Tips: Enter current in amperes, resistance per kilometer in Ω/km, and cable length in kilometers. All values must be positive numbers. The result shows power loss in watts.

5. Frequently Asked Questions (FAQ)

Q1: Why is the formula multiplied by 2?
A: The multiplication by 2 accounts for round-trip current flow - current flows to the load and returns through the cable.

Q2: What is typical resistance for copper cables?
A: Copper cable resistance varies by size: approximately 0.5-5 Ω/km for common DC power cables. Check manufacturer specifications for exact values.

Q3: How does cable length affect power loss?
A: Power loss increases linearly with cable length. Doubling the cable length doubles the power loss, assuming constant current and resistance.

Q4: Why is current squared in the formula?
A: Power loss is proportional to the square of current (I²R), meaning small increases in current cause significant increases in power loss.

Q5: How can I reduce power loss in DC cables?
A: Use larger diameter cables (lower resistance), reduce cable length, or lower the current. Higher voltage systems with lower current are more efficient.

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