Distribution Theory and Wiring Design

In a single-phase three-wire system, the voltage between an outer wire and the neutral is 100 V, and between the two outer wires it is 200 V. 141 V is 100×√2 and 173 V is 100×√3; neither is the outer-to-outer voltage here. 100 V is the outer-to-neutral voltage.

A 200 V three-phase three-wire system uses three wires, and the voltage between any two of them is 200 V. There are three wires and no neutral as in the single-phase three-wire system. 200÷√3 (about 115 V) is not a line-to-line voltage. It is mainly used for motors (power loads) in factories; household 100 V outlets are fed from single-phase circuits.

Article 38 of the Ordinance requires a value "not deviating more than 6 V above or below 101 V" for a standard voltage of 100 V. 101−6=95 and 101+6=107, so the range is 95 V to 107 V. 90–110 V is ±10 V around 100 V and 97–103 V is ±3 V around 100 V; neither is the standard. 101–107 V looks only at the upper side.

Article 38 requires a value "not deviating more than 20 V above or below 202 V" for 200 V. 202−20=182 and 202+20=222. 180–220 V wrongly centres on 200 V, and 190–210 V and 196–208 V use the wrong width.

The neutral current is the difference between the upper and lower currents: 10−10=0 A. When the loads are balanced, no current flows in the neutral. 20 A adds the currents, 10 A just repeats one side, and 5 A halves it; all are wrong.

With resistive loads, the neutral current is the difference: 15−9=6 A. 24 A is the sum, 12 A is the average, and 0 A applies only when balanced.

Resistance R=V²/P: the 200 W load is 100²÷200=50 Ω and the 600 W load is 100²÷600≈16.7 Ω. With the neutral broken, the two are in series across 200 V. Current = 200÷(50+16.7)=3 A, so the 200 W load gets 3×50=150 V. 50 V is the voltage on the 600 W load, 100 V is the voltage before the break, and 200 V is the total. The danger is that the lower-power load receives the higher voltage.

In two-wire with current I, the loss is 2I²r. In balanced three-wire, the outer-wire current is I/2 and the neutral carries 0, so the loss is 2×(I/2)²×r=I²r/2. I²r/2 ÷ 2I²r = 1/4. Loss is proportional to the square of current, so halving the current gives 1/4, not 1/2. "2" and "No change" miss the reduced current.

In two-wire, current flows in both the outgoing and return wires, so v=2Ir=2×20×0.1=4 V. 2 V counts only one wire, 8 V counts four wires, and 0.2 V forgets to multiply by the current (2×0.1).

The drop is 2Ir=2×10×0.15=3 V, so the load voltage is 104−3=101 V. 102.5 V subtracts only one wire (1.5 V), 104 V ignores the drop, and 98 V counts the drop twice (104−6).

Resistance of one wire: r=8×25÷1000=0.2 Ω. For two-wire, v=2Ir=2×15×0.2=6 V. 3 V counts one wire only, 12 V doubles the round trip again, and 0.6 V comes from a unit-conversion slip (taking r=0.02 Ω).

Loss P=2I²r=2×10²×0.2=40 W. 20 W counts one wire, 80 W counts four, and 4 W forgets to square the current (2×10×0.2).

For three-phase three-wire, v=√3·Ir=1.73×20×0.15≈5.2 V. 3.0 V is Ir (one wire), 6.0 V uses the two-wire formula 2Ir, and 9.0 V uses 3Ir.

Each of the three wires loses I²r, so P=3I²r=3×10²×0.1=30 W. 17 W is √3·I²r, 20 W is the two-wire 2I²r, and 10 W is one wire only. Count losses by the number of wires; √3 is not used here.

Since the loads are balanced, no current flows in the neutral, and the drop is one outer wire: Ir=20×0.1=2 V. The load voltage is 105−2=103 V. 101 V subtracts 2Ir as in two-wire, 105 V ignores the drop, and 104 V halves the current.

The neutral carries the difference, 20−10=10 A. Load 1 voltage = 105 − (outer-wire drop 20×0.1) − (neutral drop 10×0.1) = 105−2−1 = 102 V. 103 V forgets the neutral drop, 101 V assumes 20 A in the neutral too, and 104 V forgets the outer-wire drop. (Load 2 gets 105−1+1=105 V.)

The motors' total (40 A) is larger than the heaters' total (15 A) and is 50 A or less, so multiply the motor total by 1.25 and add the heater total: 15 + 1.25 × 40 = 15 + 50 = 65 A. 55 A simply adds them (ignoring motor starting current), 59 A uses the factor 1.1 that applies only above 50 A (15 + 44), and 68.75 A wrongly applies 1.25 to the heaters as well (55 × 1.25).

As a rule, a branch-circuit breaker must be within 3 m of the branch point. It may be farther if the wire is 8 m or shorter and its allowable current is at least 35% of the rating of the breaker protecting the feeder: 100 × 0.35 = 35 A. 55 A (55%) is required only when there is no length limit, not for 6 m. 100 A is just the breaker rating and is not a rule value. 25 A is below 35%.

Current makes a wire heat up; if it gets too hot, the insulation is damaged and short circuits or fires can result. The allowable current is the limit that prevents this. Voltage drop depends on length and current and is not held constant by the allowable current. Wire weight and meter range are not the reason.

When many wires share a conduit, heat builds up, so the original allowable current is multiplied by a current reduction factor (電流減少係数) smaller than 1. The more wires, the smaller the factor. Increasing it or leaving it unchanged ignores the heat build-up.

The allowable current of 2.0 mm copper wire is 35 A (Table 146-1). At 30 °C or lower the temperature correction factor is 1, and the current reduction factor for three or fewer wires is 0.70 (Table 146-4), so 35 × 1 × 0.70 = 24.5 A. 18.9 A is the value for 1.6 mm wire (27 A × 0.70), 33.6 A is for 2.6 mm wire (48 A × 0.70), and 35 A forgets the reduction factor.

At a higher ambient temperature, heat escapes less easily and the insulation heats up more, so the allowable current becomes smaller. Under Article 146 of the Interpretation, the allowable current of PVC-insulated wire is multiplied by the correction factor √((60 − θ) ÷ 30) (θ = ambient temperature, taken as 30 at 30 °C or below), which falls below 1 above 30 °C. "Larger" and "no relation" ignore heat dissipation. The current reduction factor depends on the number of wires and is never greater than 1.

Area = πr², and the radius is half the diameter, 1.0 mm: π×1.0²≈3.14 mm². 12.6 mm² uses the diameter as the radius (π×2.0²), 6.28 mm² uses the circumference formula (2πr), and 2.0 mm² simply reuses the diameter.

With three wires in the conduit, the current reduction factor is 0.70. 1.6 mm: 27 × 0.70 = 18.9 A; 2.0 mm: 35 × 0.70 = 24.5 A; 2.6 mm: 48 × 0.70 = 33.6 A; 3.2 mm: 62 × 0.70 = 43.4 A. The thinnest wire with at least 25 A is 2.6 mm. 2.0 mm gives 24.5 A, just short (comparing 35 A without the factor would wrongly select it). 1.6 mm is also too small. 3.2 mm would work but is not the thinnest.

Resistance is inversely proportional to area, so a thicker wire has lower resistance. At the same current, both voltage drop (IR) and power loss (I²R) fall, and the allowable current rises. The other options state the wrong direction or no change in resistance.

A branch circuit branches off the feeder (幹線) and runs to electrical equipment such as lights and outlets. Article 63 of the Ministerial Ordinance on Technical Standards (電気設備に関する技術基準を定める省令) describes it as a low-voltage circuit branching from the low-voltage feeder to equipment. The pole-to-entrance wire is the service drop (引込線), the entrance-to-board line is the feeder, and a transformer winding is not a wiring circuit.

I=P÷V: 15 A + 10 A + 5 A = 30 A. All load currents add up in the feeder. 15 A counts only the largest load, 20 A adds only the 1,500 W and 500 W heaters, and 45 A counts the 1,500 W heater twice.

For a breaker rated over 20 A up to 30 A, the wire must be at least 2.6 mm in diameter (Table 149-1) and the outlet must be rated 20 A to 30 A, excluding outlets that accept plugs rated below 20 A (Table 149-3). So "2.6 mm, 30 A" is acceptable. 2.0 mm and 1.6 mm wires are too thin for a 30 A breaker, and a 15 A outlet is rated too low.

An earth-leakage breaker detects a ground fault—current leaving its proper path and flowing to the ground—and opens the circuit. Article 15 of the Ministerial Ordinance requires measures such as ground-fault breakers to prevent damage, electric shock and fire from ground faults. Opening on overcurrent is the job of an overcurrent breaker. It neither raises voltage nor acts as a surge arrester (避雷器).

The breaker exists to protect the wire. A 40 A breaker on wire rated 20 A may not trip when the current exceeds 20 A and the wire overheats, which can cause a fire. Frequent tripping happens when the breaker is too small. The breaker rating does not double the voltage or eliminate loss. Table 149-1 of the Interpretation requires copper wire of at least 8 mm² for a branch circuit with a 40 A breaker.