Irodov 3.194: Thinner Filament Needs More Voltage – Amit & Deepali Sinha Foundation

Problem Irodov 3.194

A filament evaporates, so its diameter shrinks. To keep the filament at the temperature it had before, the voltage across it has to be raised by $\eta = 1.0\%$ . Assuming that the heat the filament gives off to its surroundings is proportional to its surface area, find by how many per cent the diameter of the filament has decreased.

A length l of the filament, of diameter d. — A length $l$ of the filament, of diameter $d$ .

Given

voltage raised by $\eta = 1.0\%$
same temperature
heat loss $\propto$ surface area
length unchanged (assumed)

Find

the relative decrease

\Delta d/d

of the diameter

Solution

Step 1Heat produced equals heat lost

Take a filament of length $l$ and diameter $d$ made of a material of resistivity $\rho$ . At a steady temperature the Joule power equals the heat given off. The resistance and the surface area are

R = \frac{\rho l}{\pi d^2/4} = \frac{4\rho l}{\pi d^2}, \qquad S = \pi d l

Calling $\alpha$ the heat given off per unit area, the balance $V^2/R = \alpha S$ reads

\frac{\pi d^2 V^2}{4\rho l} = \alpha\,\pi d l

Step 2What stays fixed

Cancelling $\pi d$ and collecting what does not change on the right:

V^2 d = 4\alpha\rho\, l^2 = \text{const}

(1)

A thinner filament makes less heat per volt squared ( $R \propto 1/d^2$ ) but also loses less heat ( $S \propto d$ ); the first effect wins, so it needs more voltage.

Step 3The decrease of the diameter

Let the diameter fall from $d$ to $d'$ while the voltage rises from $V$ to $(1 + \eta)V$ . By (1):

\frac{d'}{d} = \frac{V^2}{(1 + \eta)^2V^2} = \frac{1}{(1 + \eta)^2}

The relative decrease is $1 - d'/d$ . Numerically, $(1.010)^2 = 1.0201$ :

\frac{\Delta d}{d} = 1 - \frac{1}{1.0201} = 0.0197

\frac{\Delta d}{d} = 1 - \frac{1}{(1 + \eta)^2} \approx 1.97\,\%

(2)

For a small rise, $(1 + \eta)^{-2} \approx 1 - 2\eta$ , so the decrease is $2\eta$ , about $2\%$ , twice the relative rise of the voltage.

Check

Taking logarithmic differentials of (1) gives $2\,\delta V/V + \delta d/d = 0$ : a $1\%$ rise of $V$ must be matched by a $2\%$ fall of $d$ , the same as the exact result to first order. With $\eta = 0$ the diameter is unchanged, as it must be.

Remark

At a fixed voltage a uniformly thinned filament would run cooler: its Joule power falls as $d^2$ , its cooling area only as $d$ , which is why the voltage has to be raised. A local thin spot behaves the other way. It carries the same current as the rest of the filament, so its heat per unit length, $I^2\cdot 4\rho/\pi d^2$ , grows as it thins while its cooling area shrinks; it runs hotter, evaporates faster, and that is how a lamp filament finally burns through.