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# owuitools prompt test result
# prompt dir : easy/e04-resistor-power (round v4)
# chat id    : 1c4b6e00-2fd9-465f-ae6a-3478966a36e8
# chat title : Resistor Current and Power Calculation
# user       : the operator
# model      : qwen3.8:latest
# turns      : 2 messages on selected branch
# tool calls : 1  ohms_law_power x1
# dumped at  : 2026-09-12T04:07:22Z
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=== USER  (2026-09-12T04:06:43Z) ===
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  (2026-09-12T04:06:43Z, model qwen3.8:latest) ===
usage: prompt_eval_count=9829 eval_count=554 prompt_tokens=9829 completion_tokens=554 total=17.9s prefill=1.3s decode=16.5s 33.6 tok/s

=== [1] THINKING ===
Simple ohm's law problem: I = 12/47 = 0.255 A, P = V²/R = 144/47 = 3.06 W. A 1/4 W resistor would fail. Use at least 5 W (with 2x derating margin, 6.1 W → 7 W or 10 W). Let's run the calculation tool.

=== [2] TOOL CALL: ohms_law_power  (call_66a996tm) ===
{
  "voltage": "12",
  "resistance": "47"
}

=== [3] TOOL RESULT: ohms_law_power  (call_66a996tm) ===
# 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 (2s) ===
Answer: 255 mA, 3.06 W. No good for a 1/4 W resistor. With a 2x derating (margin), use a 6 W or higher rating → the standard is 7 W or 10 W (axial leaded wirewound type). A 5 W rating would be borderline (only 1.6x margin).

=== [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 heat up almost immediately, likely burn out, discolor the board, and possibly let go of something smoky.

**What to use instead:**

Standard practice is to pick a rating of at least **2× the actual dissipation**, so you want ≥ ~6 W. That rules out the common carbon-film sizes (¼ W, ½ W, 1 W) entirely:

| Rating | Margin over 3.06 W | Verdict |
|---|---|---|
| ¼ W | 0.13× | Destroys instantly |
| 1 W | 0.33× | Burns out quickly |
| 5 W | 1.6× | Technically survives, but no headroom — avoid |
| **7 W or 10 W** | 2.3× / 3.3× | **Correct choice** |

Practically, that means a **wirewound 7 W or 10 W resistor** (the axial "cigar" type, often in yellow or blue phenolic bodies). A 10 W part is the safer pick if your supply runs warm or the enclosure is hot, since derating matters more in restricted airflow.

One design note: putting a bare resistor straight across a supply for 3+ W is usually a sign it's being used as a dummy load or crude current sink. If this is meant to limit current to something else, a proper current-limiting approach (switching regulator in current-limit mode, or a series resistor sized for the real load) will be far cooler and more reliable. And if you do go with the wirewound version, leave some air around it — it'll run hot enough to be uncomfortable to touch.

=== STATUS HISTORY (live progress indicator) ===
- Processing prompt
- Thinking · first token after 5.3 s
- Thought: Simple ohm's law problem: I = 12/47 = 0.255 A, P = V²/R = 144/47 = 3.06 W. A 1/4 W … margin, 6.1 W → 7 W or 10 W). Let's run the calculation tool.
- Calling ohms_law_power
- Thinking
- Thought: Answer: 255 mA, 3.06 W. No good for a 1/4 W resistor. With a 2x derating (margin), use … type). A 5 W rating would be borderline (only 1.6x margin).
- Writing response
- Done · 1 tool call · prefill 5.1 s for 12,946 tokens · 693 generated
