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An incandescent light bulb has some amount of resistance (let's say 30 ohms) and generates a large amount of heat. It's my understanding that a resistor with the same amount of resistance (30 ohms) generates considerably less heat. If you short out a battery with a piece of copper wire, there is very little resistance, but a large amount of heat is generated. What causes these differences?

Is the physical process that resists current flow different than the physical process that generates heat?

It seems like it must be. It's my understanding that the more resistance there is in an electrical circuit, the slower a battery will drain, suggesting that the resistance is somehow decreasing the speed that energy exits the battery, not just expending all the energy as heat.

I was once told that resistors work by creating friction for the electrons passing through them and expelling the lost energy as heat, but it seems like that can't be the whole story.

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    \$\begingroup\$ P = I^2 x R or P = U^2 / R for all resistive elements. \$\endgroup\$
    – winny
    Dec 6, 2020 at 11:00
  • \$\begingroup\$ @winny I am aware of this formula, but I am still missing something. So P would be the amount of heat generated (assuming no kinetic energy is generated). And R would be the resistance, which in this example is the 30ohms. And I'm assuming we're using the same battery for all of my examples (let's say a 9 volt) But what would cause I to be higher or lower? The formula implies that more power/heat will be generated if I increases, but what could cause that? \$\endgroup\$
    – markv12
    Dec 6, 2020 at 11:15
  • \$\begingroup\$ As illustrated by Ohms law, current would be higher when voltage is higher or resistance is lower. Resistance is effectively by definition a property that obeys these laws. What the physics level mechanism of resistance in a given substance is, is a very different question. \$\endgroup\$ Dec 6, 2020 at 11:33
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    \$\begingroup\$ A 30 ohm resistor and 30 ohm lamp would draw equal amount of power from the same battery. The difference is that resistor is 100% efficient converting power to heat, and the lamp is only 98-99% efficient heating up as the rest of the energy is emitted as light. The lamp filament almost immediately starts to run at 2000K-3000K temperature so way hotter than the resistor which has more mass so temperature rises slower and levels to lower temperature. They still generate approximately the same amount of heat (if the lamp efficiency is ignored). \$\endgroup\$
    – Justme
    Dec 6, 2020 at 11:56
  • \$\begingroup\$ ”And I'm assuming we're using the same battery for all of my examples (let's say a 9 volt)” You are confusing yourself due to very high inner resistance of the battery. Please take a good hard look at Ohm’s law. Knowing it isn’t enough in electrical engineering, you must understand it and be able to apply it without thinking. Also, simulations is your friend. \$\endgroup\$
    – winny
    Dec 6, 2020 at 16:11

4 Answers 4

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An incandescent light bulb has some amount of resistance (let's say 30 ohms) and generates a large amount of heat. It's my understanding that a resistor with the same amount of resistance (30 ohms) generates considerably less heat.

No, that is incorrect. The same power is dissipated providing that they are both fed from the same voltage source. And both will draw the same current. Ohm's law prevails.

If you short out a battery with a piece of copper wire, there is very little resistance, but a large amount of heat is generated. What causes these differences?

The two scenarios (lamp/30 Ω and piece of copper wire) are incomparable. The differences are the loading of a very low ohmage piece of copper wire and the internal series resistance of the battery or voltage supply. There is no comparison to be made without considering all the circuit equivalent components and details.

It seems like it must be. It's my understanding that the more resistance there is in an electrical circuit, the slower a battery will drain, suggesting that the resistance is somehow decreasing the speed that energy exits the battery, not just expending all the energy as heat.

More resistance means less current which equals slower battery drain. However, all the energy that is drained is converted to heat (and some light as per black body radiation).

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  • \$\begingroup\$ I am starting to get it, but I am still missing something. You say that "More resistance means less current which equals slower battery drain" Which implies less energy is exiting the battery. So the more resistance you have, the less heat you will get. (keeping everything else constant) But it's my understanding that the resistance of a resistor is caused by friction against the movement of electrons. So it seems like more resistance would result in more heat. So what causes resistance and what causes heat generation must be different things. Can you explain this? \$\endgroup\$
    – markv12
    Dec 6, 2020 at 11:37
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    \$\begingroup\$ That's a physics question and not an EE question so I suggest you ask that on physics SE. However, for what it's worth, friction isn't a concept that is involved with electron movement; material conductivity is due to the number of free electrons and that depends on the conductors material which in turn is due to the atomic level stuff like valence bands in atoms and how much free space there is in the valence band. You can see now that it's not an EE question @markv12 \$\endgroup\$
    – Andy aka
    Dec 6, 2020 at 11:51
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    \$\begingroup\$ Thank you for your help! \$\endgroup\$
    – markv12
    Dec 6, 2020 at 11:52
  • \$\begingroup\$ There are probably tons of sites out there that can also cover this topic so good luck @markv12 \$\endgroup\$
    – Andy aka
    Dec 6, 2020 at 11:54
  • \$\begingroup\$ I have one more question that might get to the core of my confusion. Imagine two circuits each with a 9v battery and a single resistor. The resistor in circuit #2 has 2x the resistance of the resistor in circuit #1. Which circuit will exhaust the battery faster? \$\endgroup\$
    – markv12
    Dec 6, 2020 at 13:04
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What is the relationship between electrical resistance and heat generation?

The heat generated by a resistor would be I²R watts, where 'I' would be the current in amperes and 'R' the resistance in ohms.

The size of the resistor would decide the heat that could be safely dissipated by it. In other words, the larger the size of the resistor, the higher its rated wattage.

An incandescent light bulb has some amount of resistance (let's say 30 ohms) and generates a large amount of heat. It's my understanding that a resistor with the same amount of resistance (30 ohms) generates considerably less heat.

Your statement is incorrect. The resistor could generate the same heat as the incandescent light bulb, but under different conditions.

A 30 Ω 60 W resistor would safely dissipate 60 W at a current of 1.414 A.

(I² = 60/30 = 2 or I = √2 = 1.414)

On the other hand, a 120V 60W incandescent light bulb, with a cold resistance of 30 Ω and a working resistance of 240 Ω, would dissipate 60 W at a current of only 0.5 A.

(I² = 60/240 = 0.25 or I = √0.25 = 0.5)

Hence, a meaningful comparison would not be possible.

The effect of the heat generated, on a resistor's value, would depend on the material used. More specifically, it would depend on the material's temperature coefficient of resistance.

Carbon, used in a carbon film resistor, has a low negative temperature coefficient of resistance (-0.0005 / ° C).

A negative temperature coefficient of resistance implies a decrease in its resistance with an increase in temperature.

Nichrome, used in wire wound resistors, has a low positive temperature coefficient of resistance (+0.0004 / ° C).

A positive temperature coefficient of resistance implies an increase in its resistance, with an increase in temperature.

Tungsten, used for the filament of an incandescent lamp, has a high positive temperature coefficient of resistance (+0.0045 / ° C, ten times that of nichrome). That explains the reason for its much lower cold resistance, when compared to its working resistance.

Incandescent lamps work at very high temperatures required for illumination, their filaments being protected from damage, due to oxidation, by their vacuum or inert gas working environment.

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The 30 ohms in your case represent the load, so in the example you mentioned (incandescent light bulb) the voltage is higher.
Watt's law says power (watts) =voltage (V) * current (I)
enter image description here

as you can see this generates a 403.33 watts. (Images from https://ohmslawcalculator.com/ohms-law-calculator)

In the second example a battery of 12 volts and 30 ohms enter image description here This generates only 4.8 watts the difference is huge.

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  • \$\begingroup\$ Why did you use different voltages for the two cases? There will be no difference if both 30 ohm loads use the same voltage. \$\endgroup\$
    – Justme
    Dec 6, 2020 at 13:23
  • \$\begingroup\$ @Justme exactly \$\endgroup\$
    – user140351
    Dec 6, 2020 at 13:52
  • \$\begingroup\$ Well why did you not use 110V for both then? Or 12V for both? In the original post there was no mention of different voltages being applied to 30 ohm loads. \$\endgroup\$
    – Justme
    Dec 6, 2020 at 14:10
  • \$\begingroup\$ Please provide a link to the source of the graphics you included in your answer. \$\endgroup\$ Dec 6, 2020 at 14:22
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The physical process responsible for electrical resistivity is exactly the same process that generates heat in the resistor. It's electron-phonon interaction and to fully understand it we need to go full quantum theory.

But your question can be reasonably answered in the context of classical theory. What you seem confused about is the relationship between the amount of heat transferred to a body and the temperature rise that comes from it. The temperature depends on a lot of factors, but most importantly on the mass of the body that is being heated.

So, let's consider two (cylindrical with square section) resistors made with the same material (which means same resistivity but also same thermal capacity and same thermal conductivity), and in the same cooling conditions (we can imagine to put them in a vacuum to simplify things). You know that the resistance is a function of the resistivity of the material (which is the same here) and of the ratio length/area

\$R = \rho \frac{l}{S}\$

we can have the same resistance, for example with a piece of thin wire (l=10mm, d=0.1mm, S=0.01 mm^2) and with a massive thick rod (l=1 m, d=1 cm , S= 1 cm^2). Same resistance means same energy per unit time dissipated. But do you expect the thick rod to glow red as the thin wire? I doubt it. The l/S ratio is constant, but their product is not. And the product is the volume, and the volume leads to mass.

Moreover: lamp filaments are hair-thin and placed in a vacuum (mainly for chemical, but also thermal reasons) while power resistors are bulky (both in their resistive material and the mass adding white cement that is in direct contact to 'steal' heat by thermal conduction) and their heat exchange via convection is facilitated by fins. You can't expect them to reach the same temperature, given the same electrical conditions.

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  • \$\begingroup\$ It's a bit of a mistake to claim that it is mass which matters; that would only be true for a very brief pulse-type experiment. In the kind of steady state being asked about, it's rather surface area which matters, since the temperature is that at which the rate of energy lost to the environment equals the rate of electrical energy dissipated in resistance. \$\endgroup\$ Dec 6, 2020 at 21:59
  • \$\begingroup\$ @ChrisStratton Fair enough, but you still have to increase the internal energy of the body to reach the steady temperature and, depending on the mass, the transient could be so slow that steady state would not be reached in 'experimental times' (make the rod longer and thicker for example, and if you connect it to a battery it will drain the battery before reaching steady state). To be pedantic, a real steady state would require considering the interaction with the environment as well (the room gets hotter and energy exchange will... change). \$\endgroup\$ Dec 6, 2020 at 22:32

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