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Keeping the temperature constant, if you increase the voltage across a filament lamp and a metallic conductor, the filament lamp gets hot but the metallic conductor doesn't. Both are made of metal so I don't see why this should happen. Is it because a filament lamp is made of coiled-up length of metal wire? (But even then I don't see exactly how this makes a difference.)

Thanks

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    $\begingroup$ When and how did you perform that experiment? In my experience, wires get warmer when they carry current. If you run enough current through a copper wire, you can make it explode. (Don't ask me how I know.) The amount of power dissipated is equal to $I^2R$ or $\frac{V^2}{R}$ where $I$ is the current in the wire, $V$ is the voltage drop, and $R$ is the resistance. In most applications (e.g., the power cord for a lamp), we try to keep $R$ (and therefore $V$) very low. In the lamp filament however, $R$ (and therefore $V$) are much higher. $\endgroup$ – Solomon Slow May 11 '15 at 16:28
  • $\begingroup$ @jameslarge: Sorry, I think I described it wrong. I am currently having difficult with Ohm's law,please see the comment under docscience's answer. $\endgroup$ – user45220 May 11 '15 at 17:03
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    $\begingroup$ "...but the metallic conductor doesn't..." That part is not true. Any conductor (except for a superconductor) will dissipate heat when you force current through it. The lamp filament and the lamp cord both carry the same current, but the resistance of the lamp filament is much greater than the resistance of an equal length of the cord. Therefore, the amount of heat generated in the filament is much greater than the amount of heat generated in the cord. But SOME heat is generated in the cord. $\endgroup$ – Solomon Slow May 11 '15 at 18:35
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    $\begingroup$ Try this experiment: You'll need a long extension cord, and a portable heater or a hair dryer. Coil the long cord, plug it into an electrical outlet, plug the heater into the coiled cord, and turn the heat on. Position the heater so it does not blow hot air onto the cord. Also, DO NOT LEAVE IT RUNNING UNATTENDED. After ten minutes or so, feel the cord or, if you want to get really scientific, put a thermometer in the coils, and measure its temperature before and after running the heater. $\endgroup$ – Solomon Slow May 11 '15 at 18:39
  • $\begingroup$ @jameslarge: Thank you very much for these explanations! Unfortunately I don't think I can do the experiment though, but it looks interesting. $\endgroup$ – user45220 May 11 '15 at 18:51
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Heat losses (aka $I^2R$ losses) occur because charge has a hard time getting through the conducting element.

In a light bulb the filament is purposely made with a higher resistance, $R$, compared to the resistance of the metallic leads on which it is welded to.

Heat is energy and power is the rate at which energy is transfered

$$P=I^2R$$

and so with a larger $R$ in the filament you can expect to see larger power and rate of energy loss.

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  • $\begingroup$ If the resistance of the filament is too high for the design of the bulb, heat will be transferred not only in the form of light and heat in the bulb, but back through the metallic leads and into the circuit. If the filament is more robust than the circuit breaker, a fire could start. $\endgroup$ – Ernie May 11 '15 at 16:14
  • $\begingroup$ That's why filament leads are usually engineered to be long, allowing some of the heat conducted by the leads to bleed off through the base or to radiate to the ambient. I should also have mentioned that ohms law doesn't strictly apply to a filament . It's more complex. The R actually increases as temperature increases. $\endgroup$ – docscience May 11 '15 at 16:19
  • $\begingroup$ @docscience, If you are going to say that Ohm's Law doesn't apply, then I think you owe a deeper explanation to beginners who may be lurking here. I would start by throwing out $R$ because resistance really is defined by Ohm's Law: If you model a device with Ohm's Law, then $R$ is constant. If Ohm's Law is not an adequate model of the device over some range of operating conditions, then there is no "resistance", there is only $\frac{dV}{dI}$. $\endgroup$ – Solomon Slow May 11 '15 at 16:48
  • $\begingroup$ @jameslarge: I think my trouble is with Ohm's law itself. Would this be a correct definition? : The current through a component is proportional to the potential difference across it provided that its temperature doesn't change. Thanks $\endgroup$ – user45220 May 11 '15 at 17:02
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    $\begingroup$ @user45220, Ohm's Law, like many scientific laws, is a simplified model. There is no "temperature" in Ohm's Law even though temperature does have a real effect on conductors. In most experiments, the temperature effect is negligible, and Ohm's Law is an adequate model, but in the case of an incandescent light bulb, the absolute "hot" temperature can be more than $10\times$ the absolute cold temperature. That can't be ignored, and that's why docscience said that "Ohm's Law does not strictly apply." $\endgroup$ – Solomon Slow May 11 '15 at 18:48

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