How would you calculate the required wire gauge to carry 15 A over 20 meters with a maximum allowable voltage drop of 2 V in a 28 V system?

How would you calculate the required wire gauge to carry 15 A over 20 meters with a maximum allowable voltage drop of 2 V in a 28 V system?

• A. R_total = V_drop / I ≈ 0.133 Ω; with a 40 m round trip, R_per_meter ≈ 0.00333 Ω/m; select a conductor gauge with resistance per meter at or below this value and verify ampacity meets 15 A.
• B. R_total = I / V_drop = 15/2 = 7.5 Ω; R_per_meter = 7.5 / 40 = 0.1875 Ω/m.
• C. R_total = V_drop × I = 2 × 15 = 30 Ω; R_per_meter = 30 / 40 = 0.75 Ω/m.
• D. Use V = IR to compute the required conductor cross-section area directly from current, ignoring voltage drop.

Answer: (D)

To determine wire size for a given current and allowable voltage drop, start with how voltage drop relates to resistance and length. The voltage drop across the conductors is V_drop = I × R_total. Here, the allowable drop is 2 V and the current is 15 A, so the total loop resistance you can tolerate is R_total = V_drop / I = 2 / 15 ≈ 0.133 Ω. That resistance is for the entire circuit path the current travels, which includes both the forward conductor and the return conductor. If the one-way distance is 20 m, the total conductor length is 40 m. The allowable resistance per meter is then R_per_meter = R_total / total_length ≈ 0.133 Ω / 40 m ≈ 0.00333 Ω/m. Choose a conductor (material and gauge) that has a resistance per meter at or below about 0.00333 Ω/m and also has an ampacity of at least 15 A for the operating conditions. For copper, this typically points to gauges around a mid-range value (you’d compare exact resistance figures and temperature ratings from a wiring table), ensuring the insulation and temperature rating support 15 A. Remember, the 2 V drop represents about a 7% drop on a 28 V system, which should be acceptable only if the load can tolerate that loss. This approach is correct because it directly uses V_drop = I × R_total and accounts for the actual path length (round trip). The other thought processes that multiply or divide inappropriately, or that try to size the wire from current alone without considering voltage drop and length, don’t ensure the specified voltage drop is met.

To determine wire size for a given current and allowable voltage drop, start with how voltage drop relates to resistance and length. The voltage drop across the conductors is V_drop = I × R_total. Here, the allowable drop is 2 V and the current is 15 A, so the total loop resistance you can tolerate is R_total = V_drop / I = 2 / 15 ≈ 0.133 Ω.

That resistance is for the entire circuit path the current travels, which includes both the forward conductor and the return conductor. If the one-way distance is 20 m, the total conductor length is 40 m. The allowable resistance per meter is then R_per_meter = R_total / total_length ≈ 0.133 Ω / 40 m ≈ 0.00333 Ω/m.

Choose a conductor (material and gauge) that has a resistance per meter at or below about 0.00333 Ω/m and also has an ampacity of at least 15 A for the operating conditions. For copper, this typically points to gauges around a mid-range value (you’d compare exact resistance figures and temperature ratings from a wiring table), ensuring the insulation and temperature rating support 15 A. Remember, the 2 V drop represents about a 7% drop on a 28 V system, which should be acceptable only if the load can tolerate that loss.

This approach is correct because it directly uses V_drop = I × R_total and accounts for the actual path length (round trip). The other thought processes that multiply or divide inappropriately, or that try to size the wire from current alone without considering voltage drop and length, don’t ensure the specified voltage drop is met.