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# owuitools prompt test result
# prompt dir : easy/e04-resistor-power (round v3)
# chat id    : 9d8504aa-f3e2-4925-a127-18acf9fd3b4c
# chat title : Resistor Current, Power, and Rating
# user       : the operator
# model      : qwen3.8:latest
# turns      : 2 messages on selected branch
# tool calls : 1  ohms_law_power x1
# dumped at  : 2026-09-11T23:10:06Z
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=== USER  (2026-09-11T22:57:30Z) ===
I'm putting a 47 Ω resistor straight across a 12 V supply. What current will flow, how much power will the resistor dissipate, and will a standard 1/4 W resistor survive? If not, what rating should I use?

=== ASSISTANT  (2026-09-11T22:57:30Z, model qwen3.8:latest) ===
usage: prompt_eval_count=9664 eval_count=558 prompt_tokens=9664 completion_tokens=558 total=18.6s prefill=2.1s decode=16.4s 33.9 tok/s

=== [1] THINKING ===
I'll try calculating using ohms_law_power.

=== [2] TOOL CALL: ohms_law_power  (call_6jozxlyi) ===
{
  "voltage": "12",
  "resistance": "47"
}

=== [3] TOOL RESULT: ohms_law_power  (call_6jozxlyi) ===
# Ohm's law

Voltage:    12 V
Current:    255.3 mA
Resistance: 47 Ω
Power:      3.064 W

Given voltage and resistance; used I = V / R, P = V² / R.

Note: 3.064 W dissipation needs a resistor rated well above that. A common rule is to pick at least twice the calculated power.

=== [4] THINKING (6s) ===
Answer: I = 255.3 mA, P = 3.06 W. A quarter-watt resistor can't withstand it. Rule of thumb: use about 2x or more → 6.3 W or more, so a 5 W or 7 W resistor. In practice, the standard ratings are: 1/8, 1/4, 1/2, 1 W, 2 W, 5 W. Using 2x margin, 6.1 W → 7 W or more, but many people would say a 5 W wirewound is the minimum practical option. However, by the 2x rule, you'd want a 7 W or 10 W one. Should I recommend a 5 W as the minimum? Better yet: with a safety margin (typically around 2x), choose a 7 W or 10 W wirewound resistor. If you must go lower than that, at least 5 W.

=== [5] ASSISTANT TEXT ===
Here's the math for 12 V across 47 Ω:

- **Current:** I = V/R = 12 / 47 ≈ **255 mA**
- **Power:** P = V²/R = 144 / 47 ≈ **3.06 W**

**Will a ¼ W resistor survive?** No — not even close. It's being asked to dissipate about **12× its rating**. It would run white-hot, likely burn out almost immediately, and could let go in a way that shorts the supply.

**What to use instead.** The usual practice is to pick a rating of at least ~2× the calculated dissipation for reliability and thermal margin:

- Minimum workable: a **5 W** resistor (e.g., a wirewound type) — it would run quite hot, so keep it clear of anything flammable and away from other components.
- Better choice: **7 W or 10 W** wirewound, giving you comfortable headroom.

A few practical notes:

- If this is meant to be a permanent load (e.g., a dummy load), a wirewound or metal-oxide power resistor is the right part — standard small film resistors don't come above ~2 W.
- Check your supply can actually deliver 255 mA without current-limiting or shutting down.
- If the goal was just to "use up" the supply or test something, make sure the 12 V source itself can handle ~3.7 W of output (it has to supply the resistor's dissipation plus its own losses).

=== STATUS HISTORY (live progress indicator) ===
- Processing prompt
- Thinking · first token after 2.4 s
- Thought: I'll try calculating using ohms_law_power.
- Calling ohms_law_power
- Thinking
- Thought: Answer: I = 255.3 mA, P = 3.06 W. A quarter-watt resistor can't withstand it. Rule of … resistor. If you must go lower than that, at least 5 W.
- Writing response
- Done · 1 tool call · prefill 2.3 s for 12,862 tokens · 616 generated
