An incandescent bulb has a thin filament of tungsten that is heated to high temperature by passing a

Q: Why doesn't the whole filament just get thinner uniformly?

In reality, microscopic imperfections and temperature gradients cause some regions to evaporate slightly faster. Once a region becomes slightly thinner, it gets hotter, which accelerates the evaporation even more in a runaway process.

Q: If the resistance increases, shouldn't the power increase since $P = I^2R$?

You have to be careful about which variable is constant! The bulb is connected to a constant voltage supply (like your home's mains). Therefore, we must use $P = V^2/R$. As total resistance $R$ increases, the total power $P$ decreases.

Q: Why do we use $I^2R$ for the hot spot but $V^2/R$ for the whole bulb?

Brilliant question! The entire filament is connected across a constant voltage $V$, so $V^2/R$ is perfect for total power. However, the filament acts as a series circuit. The current $I$ is the same through the thick and thin parts. To compare the heat generated in different sections, $I^2R$ is the right tool because $I$ is constant across all sections.

Q: Does the bulb actually get dimmer over time?

Yes, overall it does! Because the total power consumed decreases, the total light output drops. However, the specific "hot spots" will glow intensely right before they break.

Q: What does Wien's Displacement Law have to do with this?

Wien's Law states that the peak wavelength of emitted light is inversely proportional to temperature ($\lambda_{max} \propto 1/T$). The hot spot reaches an extremely high temperature, shifting its light to shorter wavelengths, which correspond to higher frequencies (like shifting from red to blue-white).

Question

An incandescent bulb has a thin filament of tungsten that is heated to high temperature by passing an electric current. The hot filament emits black-body radiation. The filament is observed to break up at random locations after a sufficiently long time of operation due to non-uniform evaporation of tungsten from the filament. If the bulb is powered at constant voltage, which of the following statement(s) is (are) true?

Options:

(A) The temperature distribution over the filament is uniform.

(B) The resistance over small sections of the filament decreases with time.

(C) The filament emits more light at higher band of frequencies before it breaks up.

(D) The filament consumes less electrical power towards the end of the life of the bulb.

Sigma Percentile · Accepted Answer

The death of an incandescent light bulb is not just a simple snap of a wire; it is a fascinating interplay of thermodynamics, electricity, and quantum mechanics. Let's embark on a journey to understand exactly what happens in the final hours of a tungsten filament.

### The Anatomy of a Filament

Imagine a glowing light bulb. Over hundreds of hours of operation, the intense heat causes tungsten atoms to slowly evaporate from the surface of the filament.

However, this evaporation does not happen perfectly evenly across the entire wire. Due to microscopic imperfections and slight temperature variations, some spots lose more atoms than others. These regions become thinner, creating narrow "necks" in the wire.

This non-uniform evaporation is the root cause of the bulb's eventual demise. As the cross-sectional area $A$ decreases at these necks, the physical structure of the filament changes permanently.

### The Birth of a Hot Spot

Now, let's think about electrical resistance. We know from basic electromagnetism that resistance is inversely proportional to the cross-sectional area:

$$R = ho \frac{l}{A}$$

At the thinner sections where the area $A$ has decreased, the local resistance $R$ shoots up. This means the resistance of these small sections increases, not decreases. Therefore, statement (b) is false.

But the story doesn't end there. The entire filament is a series circuit, meaning the exact same current $I$ flows through both the thick and the thin parts. According to Joule's law of heating, the power dissipated locally is:

$$P_{	ext{local}} = I^2 R_{	ext{local}}$$

Because the thin part has a much higher resistance, it generates significantly more heat locally. It becomes an intense **hot spot**. This proves that the temperature distribution over the filament is definitely not uniform, making statement (a) false.

### The Power Paradox

What happens to the bulb as a whole? The entire filament is simply a series combination of all these small sections. Since the resistance of the thinning sections goes up, the total resistance of the bulb, $R_{	ext{total}}$, increases over time.

Because the bulb is connected to a constant voltage supply (like the mains in your house), we must evaluate the total power using the voltage formula:

$$P_{	ext{total}} = \frac{V^2}{R_{	ext{total}}}$$

A higher total resistance means the bulb draws less total power as it ages. The bulb actually becomes slightly dimmer overall! So, statement (d) is absolutely correct.

### The Final Flash

Finally, let's look at the light emitted from our super-hot spot right before it breaks.

According to **Wien's Displacement Law**, as the temperature $T$ of a black body increases, the peak wavelength $\lambda_{	ext{max}}$ of its emitted radiation shifts to shorter wavelengths:

$$\lambda_{	ext{max}} T = b$$

Shorter wavelengths correspond to higher frequencies ($
u = c/\lambda$). Therefore, as the hot spot reaches extreme temperatures just before melting, it emits a burst of light shifted towards higher frequency bands (moving from yellow-orange towards blue-white).

This makes statement (c) correct as well. The physics of a dying bulb is a beautiful, dramatic finale of localized energy release!

Select Answer:

Visualized Solution

$Non-uniform Evaporation$

$Local Resistance$

$Overall Power Consumption$

$Local Temperature$

$Black-body Radiation$

$Conclusion$

The Sigma Insight: Ohm's Law, Resistance and Electrical Power

The Anatomy of a Filament

The Birth of a Hot Spot

The Power Paradox

The Final Flash

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