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I am struggling to understand the meaning of the power rating of a resistor (e.g. 2 Ω, 40 W).

How may this power rating impact the discharge performance of a battery whose terminals are connected to the resistor?

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The power rating is how much power the resistor can dissipate before going up in smoke.

A 1 A, 12 VDC incandescent lamp is ~12 Ω resistor (at operating temp) able to dissipate 12 W while glowing white hot. Put much more voltage through (e.g., plug it into 230 VAC mains, where it would dissipate ~4,400 watts for an exceedingly brief moment), and it will become an open circuit, after an actinic flash of light.

12 V lamp, adapted from https://5fc98fa113f6897cea53-06dfa63be377ed632ae798753ae0fb3f.ssl.cf2.rackcdn.com/product_images/files/000/015/891/legacy_product_detail_large/739_eff7b1f8d62678000e123e50fce6d17753b6258b_original.jpg?1429820934

(Image adapted from: 1000Bulbs.com - Bulbrite 715710 - 10 Watt - T3.25)

However, an electric clothes dryer might have a 12 Ω element, the size of a loaf of bread, that can dissipate 4,400 watts as long as the airflow is not blocked.

Heating element, adapted from https://i5.walmartimages.com/asr/aff77e61-81de-4173-bd89-8d07ff95bbbd_1.88c834b111f424a721161a508d4bdd96.jpeg?odnHeight=612&odnWidth=612&odnBg=FFFFFF

(Image adapted from: Walmart.com - Dryer Heating Element)

Both are 12 Ω, but one is rated at 12 W, the other at 4,400 W.

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A resistor power rating (in watts) indicates how much power it can dissipate before failing or destroying itself.

Current is voltage divided by resistance (\$I = \frac{E}{R}\$). If you have a 6 volt battery connected across a 2 Ω resistor, the current will be 3 A.

Power is equal to the product of current and voltage (\$P = IE\$), which would be 36 watts in this example. If you exceed the rating of your resistor, 40 watts for the one you cited, the resistor will fail.

Power dissipation depends on some other factors as well, such as cooling. If you have active cooling you may be able to meet or exceed the component rating, but you should check the datasheet for the conditions under which the power rating was derived.

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Resistor gets hot. For instance, you have current up to 10 A, so your 2 Ω resistor will emit 10×10×2 = 200W. Too much by any means, right? So if you plan to discharge the battery at 10 A, 2 Ω is not good. You must be well below maximum power.

Besides, I guess you have your battery in an enclosure; you probably wouldn't want to emit too much heat there. It is not related to resistor absolute maximum power, but is an important consideration also.

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The resistor current rating will have no effect on the battery. A resistor ideally maintains its resistance no matter how many watts its currently dissipating.

As others have stated the wattage rating reflects how much the resistor can handle without being destroyed.

As long as the resistor is functioning properly "12ohms is 12ohms", as far as the battery is concerned.

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The power rating of a resistor is the maximum power that it can dissipate (as heat), without getting damaged.

Consider a resistor 'R' that draws a current 'I' and dissipates power 'W' when a voltage 'V' is applied to its terminals.

Applying Ohm's Law

I = V / R

The power dissipated by the resistor

W = V * I = I² * R = V² / R

Thus the permissible voltage across the resistor or the permissible current through it, may be calculated.

The permissible voltage across a 2 Ω, 40 W resistor is √(2 * 40) = 9 V and the permissible current √(40 / 2) = 4.5 A.

Battery discharge time is influenced only by the resistance of the load and not by it's power rating.

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