--- title: "Single-phase Cu cable for three loads" subject: "Industrial Power Distribution and Illumination" chapter: 5 key: "semester_6/industrial_power_distribution_and_illumination:single-phase-cable-three-loads" source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/single-phase-cable-three-loads site: BCT Engineering --- # Single-phase Cu cable for three loads Industrial Power Distribution and Illumination · Chapter 5: Electrification of Industrial Building - Asked in: 2077 Chaitra, 4 marks - Video: [Everything you need to know to solve Voltage Drop Calculations!!](https://www.youtube.com/watch?v=6t3wUX5NIqM) — Craig Michaud- Electrical Instructor - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/single-phase-cable-three-loads - Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-5 ## Question Calculate the appropriate size of Cu cable for a single phase circuit of 230V connecting three loads. The loads 10kW and 12kW heaters at a distance of 75m and 35m from the supply end and one 10HP motor at a pf of 0.8 lagging at a distance of 70m from the supply end. The permissible voltage drop is 2% of the rated voltage. Necessary table is given below. | Conductor area in sq. mm | 1 | 1.5 | 2.5 | 4 | 6 | 10 | 16 | 25 | 50 | |---|---|---|---|---|---|---|---|---|---| | Current rating in Amp | 10 | 13 | 18 | 20 | 30 | 40 | 50 | 70 | 90 | ## Short answer Using the current-moment method, total current is 130 A, required area for 2% drop is 56.94 mm². The given table’s largest cable is 50 mm² (90 A) insufficient alone, so use two 50 mm² Cu cables in parallel per conductor, giving 2.62 V drop (1.14%). ## Answer The cable must carry the total current from the supply end and keep the voltage drop at the farthest point within 2 %. Since the loads are distributed along the run, the drop is found by the current-moment method. **Assumptions**: 10 HP is the motor output with η taken as 1 (not given); heaters are at unity pf; ρ (Cu) = 0.0178 Ω·mm²/m; drop in both go and return conductors (2-wire circuit), reactance neglected. #### Load currents ```text 10 kW heater (75 m): I = 10000 / 230 = 43.48 A 12 kW heater (35 m): I = 12000 / 230 = 52.17 A 10 HP motor (70 m): P = 10 × 746 = 7460 W I = 7460 / (230 × 0.8) = 40.54 A ``` #### Total current at supply end ```text I = 43.48 + 52.17 + 40.54 (0.8 − j0.6) = 128.08 − j24.33 |I| = 130.38 A (arithmetic sum 136.20 A) ``` The largest cable in the table (50 mm², 90 A) cannot carry 130.38 A, so one cable from the table is not enough by current rating. #### Voltage drop condition ```text Allowed drop = 2 % of 230 = 4.60 V Σ I·cosφ·L = 52.17×35 + 40.54×0.8×70 + 43.48×75 = 1826.09 + 2270.43 + 3260.87 = 7357.39 A·m ΔV = 2 ρ Σ(I cosφ L) / A ≤ 4.60 A ≥ 2 × 0.0178 × 7357.39 / 4.60 = 56.94 mm² ``` #### Selection | Option | Rating (A) | ΔV (V) | OK? | |---|---|---|---| | 1 × 50 mm² | 90 | 5.24 | No (current and drop) | | 2 × 50 mm² in parallel | 180 | 2.62 | Yes | | 1 × 70 mm² (standard, not in table) | about 110–200 by type | 3.74 | Drop OK; check rating | **Answer: required area ≥ 56.94 mm² and current ≈ 130 A; using the given table, use two 50 mm² Cu cables in parallel per conductor (100 mm² total, 180 A, drop 2.62 V = 1.14 %).** A single 70 mm² or 95 mm² Cu cable could be used if its rating (from a fuller table) exceeds 130 A. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is voltage drop found for loads spread along the run? A. Using only the largest load's current B. By the current-moment method, ΔV = 2ρ Σ(I cosφ L) / A C. By dividing total current by the number of loads D. By adding the three drops at 2 % Answer: B. Each load's I cosφ x distance is summed, with the factor 2 for go and return. (Key point: Method) 2. How is the 10 HP motor current found (η = 1, pf 0.8)? A. 7460 / 230 = 32.4 A B. 10 / (230 x 0.8) = 0.05 A C. 7460 / (230 x 0.8) = 40.54 A D. 7460 x 0.8 / 230 = 25.9 A Answer: C. 10 HP = 7460 W; I = P/(V cosφ) = 40.54 A. (Key point: Motor current) 3. Total current is 130.38 A and the minimum area is 56.94 mm². What is chosen from the given table? A. One 50 mm² cable B. Two 50 mm² cables in parallel (180 A, drop 2.62 V) C. One 25 mm² cable D. One 16 mm² cable Answer: B. A single 50 mm² gives only 90 A and a 5.24 V drop, both failing; two in parallel pass. 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