########################################################################
# owuitools prompt test result
# prompt dir : medium/m03-dc-wire-gauge
# chat id    : bf37b531-70d2-4c26-82df-d9de4fb173d1
# chat title : Copper Wire Gauge for 12V Load
# user       : the operator
# model      : qwen3.8:latest
# turns      : 2 messages on selected branch
# tool calls : 2  wiring_calc x2
# dumped at  : 2026-09-11T04:54:24Z
########################################################################

=== USER  (2026-09-11T04:53:31Z) ===
I need to run 12 V DC to a 20 A load that’s 30 feet from the battery. What copper wire gauge keeps the voltage drop under 3%? Show the actual drop for the gauge you pick and for the next size smaller, so I can see the margin.

=== ASSISTANT  (2026-09-11T04:53:31Z, model qwen3.8:latest) ===
usage: prompt_eval_count=12528 eval_count=677 prompt_tokens=12528 completion_tokens=677 total=20.1s prefill=3.7s decode=16.3s 41.6 tok/s

=== [1] THINKING ===
Let me think about this. 12 V DC, 20 A load, 30 feet one-way. So total circuit length = 60 feet (out and back).

Voltage drop = I × R_total, where R_total = R_per_foot × 60 feet.

3% of 12 V = 0.36 V. So max allowed drop = 0.36 V.

Required total resistance = 0.36 V / 20 A = 0.018 ohms over 60 feet = 0.0003 ohm/foot.

Let me compute wire gauges. AWG resistances (copper, at 20°C):
- 10 AWG: 1.0 Ω/1000 ft = 0.001 Ω/ft
- 8 AWG: 0.65 Ω/1000 ft = 0.00065 Ω/ft
- 6 AWG: 0.408 Ω/1000 ft

Let me use the wiring_calc tool for actual numbers.

10 AWG: R = 1.028 Ω/1000 ft (copper). Over 60 ft: 0.0617 Ω. Drop = 20 × 0.0617 = 1.23 V. That's 10.3% — too high.

8 AWG: 0.65 Ω/1000 ft. Over 60 ft: 0.039 Ω. Drop = 20 × 0.039 = 0.78 V = 6.5%. Still too high.

6 AWG: 0.408 Ω/1000 ft. Over 60 ft: 0.0245 Ω. Drop = 20 × 0.0245 = 0.49 V = 4.1%. Still over 3%.

4 AWG: 0.253 Ω/1000 ft. Over 60 ft: 0.0152 Ω. Drop = 20 × 0.0152 = 0.304 V = 2.53%. Under 3%! 

So 4 AWG is the answer. Let me verify with the tool.

Wait, let me double-check. 3% of 12V = 0.36V. 

4 AWG: 0.253 Ω/1000ft × 60ft = 0.01518 Ω. × 20A = 0.304V = 2.53%. Good, under 3%.

6 AWG: 0.408 × 60 = 0.02448 Ω × 20 = 0.4896V = 4.08%. Over 3%.

So the answer is 4 AWG, and the next size smaller (6 AWG) shows the margin — it fails.

Let me use the wiring_calc tool to get proper numbers rather than relying on memory.

=== [2] TOOL CALL: wiring_calc  (call_0jfdhq9b) ===
{
  "calculation": "voltage_drop",
  "awg": 4,
  "length": "9.144",
  "current": "20",
  "voltage": "12"
}

=== [3] TOOL CALL: wiring_calc  (call_9bbvffif) ===
{
  "calculation": "voltage_drop",
  "awg": 6,
  "length": "9.144",
  "current": "20",
  "voltage": "12"
}

=== [4] TOOL RESULT: wiring_calc  (call_0jfdhq9b) ===
# Voltage drop over 9.144 m of 4 AWG

**Drop: 298.1 mV**  (round trip, 18.288 m of conductor)

Conductor: 4 AWG = 21.15 mm² (5.19 mm diameter), 0.815 Ω/km
Loop resistance: 14.91 mΩ
Power lost as heat: 5.963 W

That is 2.48% of a 12 V supply — within the usual 3% guideline.
Voltage at the load: 11.7 V

## Nearby gauges, same run

-   1 AWG ( 42.41 mm²): 148.7 mV = 1.24%
-   2 AWG ( 33.63 mm²): 187.5 mV = 1.56%
-   3 AWG ( 26.67 mm²): 236.4 mV = 1.97%
-   5 AWG ( 16.77 mm²): 375.9 mV = 3.13%
-   6 AWG ( 13.30 mm²): 474 mV = 3.95%
-   7 AWG ( 10.55 mm²): 597.8 mV = 4.98%

Lower AWG numbers mean thicker wire and less drop.


**For mains or building wiring, follow your local electrical code (NEC 310.16 in the US, BS 7671 in the UK) rather than these figures.** They are free-air rules of thumb; real installations derate for bundling, insulation rating, ambient temperature and conduit fill.

=== [5] TOOL RESULT: wiring_calc  (call_9bbvffif) ===
# Voltage drop over 9.144 m of 6 AWG

**Drop: 474 mV**  (round trip, 18.288 m of conductor)

Conductor: 6 AWG = 13.30 mm² (4.12 mm diameter), 1.296 Ω/km
Loop resistance: 23.7 mΩ
Power lost as heat: 9.481 W

That is 3.95% of a 12 V supply — above the usual 3% guideline — consider thicker wire or a shorter run.
Voltage at the load: 11.53 V

## Nearby gauges, same run

-   3 AWG ( 26.67 mm²): 236.4 mV = 1.97%
-   4 AWG ( 21.15 mm²): 298.1 mV = 2.48%
-   5 AWG ( 16.77 mm²): 375.9 mV = 3.13%
-   7 AWG ( 10.55 mm²): 597.8 mV = 4.98%
-   8 AWG (  8.37 mm²): 753.8 mV = 6.28%
-   9 AWG (  6.63 mm²): 950.5 mV = 7.92%

Lower AWG numbers mean thicker wire and less drop.


**For mains or building wiring, follow your local electrical code (NEC 310.16 in the US, BS 7671 in the UK) rather than these figures.** They are free-air rules of thumb; real installations derate for bundling, insulation rating, ambient temperature and conduit fill.

=== [6] THINKING (2s) ===
The tool used the length as one-way in meters (9.144 m = 30 ft) and computed round trip. Good.

So the answer: 4 AWG keeps it under 3% (2.48%, 298 mV). The next size smaller, 6 AWG, is 3.95% (474 mV) — over the limit. Note 5 AWG is 3.13%, also over. So 4 AWG is the smallest gauge that meets the requirement.

Let me present clearly.

=== [7] ASSISTANT TEXT ===
## 4 AWG is your wire

For a 30 ft one‑way run (60 ft / 18.3 m of conductor total), 20 A, 12 V:

| Gauge | Drop | % of 12 V | At the load |
|-------|------|-----------|-------------|
| **4 AWG (pick)** | **298 mV** | **2.48%** | 11.70 V |
| 6 AWG (next smaller) | 474 mV | 3.95% | 11.53 V |

**Margin:** 4 AWG lands at 2.48%, so you're 0.52 percentage points under the 3% ceiling — about 23% headroom on the drop itself (0.36 V allowed vs. 0.298 V used). The load still sees 11.70 V.

The next size down, 6 AWG, drops to 3.95% — nearly a full point over your limit, and the load only gets 11.53 V. (5 AWG sits at 3.13%, also over.) So 4 AWG is the *smallest* gauge that clears 3%.

A couple of things worth flagging:

- **This is a DC, low‑voltage run, so the 3% drop rule is the right constraint** — not ampacity. But 4 AWG also comfortably carries 20 A (its free‑air rating is well above that), so you're fine on both counts.
- **Heat:** ~6 W is dissipated in the wire at 4 AWG (9.5 W at 6 AWG). At 20 A that's a real load, so keep the wire well‑ventilated and away from anything heat‑sensitive.
- **If you'd rather not run thick 4 AWG**, the usual alternative is to bump the system voltage (e.g. 24 V halves the drop for the same wire) — but that changes the load side, so I'd only suggest it if 4 AWG is genuinely impractical to route.

One caveat: these are free‑air copper figures at room temperature. If the wire is bundled, in conduit, or running hot, the resistance rises a few percent — 4 AWG still clears 3% with room to spare, but 6 AWG would not, so don't drop down to it on the assumption of a cooler run.

=== STATUS HISTORY (live progress indicator) ===
- Processing prompt
- Thinking · prompt processed in 7.0 s
- Thought: Let me think about this. 12 V DC, 20 A load, 30 feet one-way. So total circuit length = 60 feet (out and back). Voltage drop = I × R_total, where R_total =…
- Calling wiring_calc
- Calling wiring_calc
- Thinking
- Thought: The tool used the length as one-way in meters (9.144 m = 30 ft) and computed round trip. Good. So the answer: 4 AWG keeps it under 3% (2.48%, 298 mV). The next…
- Writing response
- Done · 2 tool calls
