# Derivation and theory for $I = kV^n$ [closed]

I performed an experiment in college to study the nature of a filament of a lamp, determine $$k$$ and $$n$$ in the relation mentioned, and study the variation of wattage of a lamp. I assume ohm's law is not followed here, as temperature does not remain constant (correct me if I'm wrong). But I want to know the theory and derivation behind the relation. And the expected value of n is 0.5 for my metal filament lamp. Why? Please help.

• Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
– Community Bot
Apr 21 at 21:43
• Possibly relevant: physics.stackexchange.com/a/649609/27732 Apr 21 at 22:17
• "As it's currently written, it's hard to tell exactly what you're asking." Seems completely clear to me what's being asked. Apr 24 at 8:13

For temperatures up to a few hundred kelvin, the resistance of a metal wire is roughly proportional to its kelvin temperature, $$T$$, so $$R=\alpha T,\ \ \ \ \ \text{that is}\ \ \ \ \ \tfrac VI=\alpha T \ \ \ \ \ \ [\alpha\ \text{is a constant}].$$ If we assume, as a bold approximation, that, in equilibrium, heat equal to the electrical power input is radiated away per unit time according to Stefan's law, $$VI=A\sigma T^4\ \ \ \ \ \ [A\sigma\ \text{is constant}].$$ Eliminating $$T$$ we find $$\ \ \ I^5=\tfrac{\sigma A}{\alpha^4}V^3\ \ \ \ \ \text {that is}\ \ \ \ \ \ I\propto V^{3/5}$$.
This simple treatment is flawed because for temperatures of a few hundred kelvin, heat loss from the filament by conduction and possibly convection will be comparable with heat loss by radiation, while at higher temperatures (filament glowing orange or white) its resistance will no longer be nearly proportional to $$T$$. Indeed, an empirical formula for the resistance of a tungsten wire, usable up to 3600 K, is $$R=\beta\ T^{1.20},\ \ \ \ \ \text{that is}\ \ \ \ \ \tfrac VI=\beta\ T^{1.20} \ \ \ \ \ \ [\beta\ \text{is a constant}].$$ Eliminating $$T$$ between this and the Stefan equation, we get $$VI=\frac{A\sigma}{\beta^{\tfrac 4{1.20}}}\left(\frac VI \right)^{\tfrac 4{1.20}}$$ So we have $$I\propto V^n \ \ \ \ \text{in which}\ \ \ \ n=\frac{\tfrac 4{1.20}-1}{\tfrac 4{1.20}+1}=0.54$$ We'd expect this to be a better value for $$n$$ than the 0.60 found earlier, at least when the filament is glowing visibly and losing much more heat by radiation than by conduction and convection.
• @DocAi I'd be interested to know what value of $n$ emerges from your experiment, if indeed you find a relationship of the form $I=kV^n$. Apr 21 at 22:56
• Thank you for reporting your value. Have you considered asking your lab teacher why he/she thinks that the value of $n$ should be 0.5 ? May 3 at 16:39
• Good plan. As I implied in my answer, the $I$ against $V$ relationship for a filament lamp is unlikely to be mathematically neat (for example $I\propto V^n$ in which $n=\tfrac 12=0.50000...$), because of the various factors that need to be taken into account. It's not as if we're dealing with some basic law of nature. May 3 at 18:55