# Calculating the rate of heat dissipation in a tungsten filament

The formula for the rate of heat dissipation in a resistor is iR^2 for an Ohmic resistor. If this Ohmic resistor is replaced with a tungsten filament, which has a resistance that varies linearly with temperature, how does one calculate the rate at which heat is dissipated by the electrons in this case?

My thoughts:

Once the tungsten filament is connected to the circuit, a current will be established at all points within the circuit instantaneously, then the power dissipated by the resistor can be calculated by the formula iR^2. However, during the very small transient period while the current reestablishes, heat will still be dissipated to the resistor, so its resistance will have increased even more by the time the current has decreased.

• What do you mean by “during the very small transient period while the current reestablishes”? How many times is the current connected? In addition, do you know the resistance of the filament as a function of temperature, or are you going to use the relation for pure tungsten? Commented Jul 26, 2022 at 1:22
• @Chemomechanics , the current will become less since the voltage source is constant and since the resistance has increased. However, while the current reaches steady state, electrons will continue to pass through the resistor, leading to an increase in the resistance by the time the current has changed. I am using the relation for pure tungsten. Commented Jul 26, 2022 at 1:26
• The temperature of the W depends on how heat is lost (atmosphere, radiation, etc.). Commented Jul 26, 2022 at 1:31
• It seems that the key aspect of the question is then the heat transfer model of the filament, as the instantaneous local energy dissipated can be integrated over the filament and over time. Commented Jul 26, 2022 at 1:43

We have the connection between the filament temperature $$T$$ and its resistivity $$\rho(T)$$ (namely, the temperature-dependent resistivity of pure tungsten), so if we model the filament temperature, we can introduce the local heat generation $$J^2\rho\,dV$$, where $$J$$ is the current density and $$dV$$ is an infinitesimal volume element, and integrate this term over the filament to obtain the filament heat dissipation rate.

To start, assume the ends of the filament remain at a constant $$T_\infty$$ because of the large thermal mass of the mount. (This assumption may be violated if the base heats up substantially.) Assume also that the filament heats up as a 1D object; that is, the filament is long and thin, and there's little temperature variation between the center and the surface. We'll also assume that the filament is straight for simplicity; a helix may require more complex treatment.

An energy balance yields

$$k\frac{d^2T(x,t)}{dx^2}+J^2\rho-\frac{2h}{R}[T(x,t)-T_\infty]-\frac{2h}{R}\sigma\epsilon[T(x,t)^4-T_\infty^4]=pc\dot T(x,t)$$

where $$k$$ is the tungsten thermal conductivity, $$T(t)$$ is the time-dependent temperature, $$x$$ is the distance along the filament, $$h$$ is the convective coefficient, $$R$$ is the filament radius, $$T_\infty$$ is the ambient temperature, $$\sigma$$ is the Stefan–Boltzmann coefficient, $$\epsilon$$ is the emissivity, $$p$$ is the density, and $$c$$ is the specific heat. Here, we're saying that the total of conduction energy gains, Joule heating, convection losses, and radiative losses equals the rate of temperature increase as mediated by various material properties. The $$2h/R$$ terms arise because we're dividing a surface area $$2\pi R L$$ by a volume $$\pi R^2 L$$.

The boundary conditions are $$T(0,t)=T(L,t)=T(x,0)=T_\infty$$, where $$L$$ is the filament length.

Now, it may be the case that some of these terms are negligible; you have to analyze these on your own for your specific conditions. Then, you can solve for the time-dependent temperature analytically (example here) or numerically, probably using an iterative process, calculate the location-dependent resistivity $$\rho(x,t)$$, and then calculate the location- and time-dependent heat generation.

I'm not sure if this type of analysis is what you're looking for?

The formula for the rate of heat dissipation in a resistor is iR^2 for an Ohmic resistor.

The general formula for the electrical energy consumed by any two-terminal device is $$P = IV.$$ Your formula is a special case that applies, as you said, to linear resistors. Ohm's Law is used to replace the $$V$$ term by $$IR$$, giving $$P=I^2R$$.

So if you have an appropriate ammeter and voltmeter, you can easily measure the power consumption of your tungsten filament by simultaneously measuring the current through it and the voltage across it. Unless the filament is very tiny (microscopic), meters are readily available that will burden the circuit little enough to not worry about the effect of one meter affecting the measurement of the other meter.