Doubling voltage and power in a light bulb

[daysrunaway]

Homework Statement

A light bulb is designed for use in a 120 V circuit. If by mistake it is plugged into a 240 V circuit, what will happen to the power of the light bulb? Explain.

Homework Equations

I = V/R
P = VI

The Attempt at a Solution

Since I = V/R and the resistance of the lightbulb does not change, doubling V also doubles I. P_f = 2V2I = 4P₀. Alternatively, P = V²/R, so doubling V quadruples P. Where I'm getting stuck is on explaining this. What are the practical implications of quadrupling the power? What is the physical representation of quadrupling it - I mean, aside from the equations, WHY is it quadrupled? I'm really stuck on this.

 

[poor mystic]

Tungsten actually has a pronounced positive change in resistance with increasing heat, so that a cold lamp suffers an 'inrush' current as it comes to its working temperature and resistance.
If the voltage is doubled from the rated voltage, less than 4 times the rated power is dissipated in the filament. However, experience shows that commercial tungsten filament lamps are not designed to withstand such treatment: we know that the bulb will fail due to melting of the filament and consequent lack of a conduction path.

I hate to leave you feeling stuck on the matter of why. All I can say is that if you had to carry twice as much water, twice as high, to generate the potential to run some machine, you'd soon understand what 'quadrupling' the power means.

 

[wwshr87]

Ok, so let's say that the light bulb is 100 ohms. At 120V it would consume 1.2A and 144 Watts. We connected by mistake to 240V, then it consumes 2.4A and 576 Watts. Most likely the light bulb is not designed to take that much current and it fries.

 

[Dickfore]

Light bulbs are heated to a high temperature and their resistance is much greater than the resistance of a "cold" light bulb. Therefore, assuming R = const. is not valid.

Let us assume that the resistance is proportional to the aboslute temperature and that the temperature determines the radiated power according to Stefan's Law, i.e. the radiated power is proportional to the 4th power of the absolute temperature. These two proportionalities would mean that the resistance is proporional to the fourth root of the radiated power.

In equilibrium, the radiated power is equal to the power consumed. Using the two above equations to eliminate I, you should get:

$$I = \frac{V}{R}$$

$$P = V I = \frac{V^{2}}{R}$$

or

$$P R = V^{2}$$

But, $R \propto P^{1/4}$, so:

$$P^{5/4} \propto V^{2}$$

$$P \propto V^{8/5}$$

So, if you double the voltage, this "improved" model would predict that the power should increase by a factor of:

$$2^{8/5} = 2 \, 2^{3/5} = 3.03$$

One should perform an experiment to see if the above scaling arguments holds.

 

[rwisz]

I can explain the implications of doubling voltage and power in a light bulb in the following way:

When a light bulb is designed for use in a 120 V circuit, it means that the internal components and wiring of the bulb are optimized to handle a voltage of 120 V. This voltage is also known as the "operating voltage" or "nominal voltage" of the bulb. When the bulb is plugged into a 240 V circuit, which is double its operating voltage, the voltage across the bulb will also double.

According to Ohm's law (I = V/R), the current flowing through the bulb will also double since the resistance of the bulb remains constant. This means that the bulb will draw twice as much current as it was designed for. In terms of power (P = VI), this translates to four times the original power.

In practical terms, this means that the bulb will emit four times the amount of light and produce four times the amount of heat. This can be dangerous as the internal components of the bulb may overheat and potentially cause a fire. Additionally, the sudden increase in power can cause the bulb to burn out quickly, shortening its lifespan.

To understand why doubling the voltage quadruples the power, we can look at the equation P = V2/R. Since R is constant, doubling V will directly result in quadrupling P. This is because voltage and power have a direct relationship - as one increases, the other also increases proportionally. The physical representation of quadrupling power can be seen in the increased brightness and heat produced by the bulb, as well as the increased strain on its internal components.

In conclusion, plugging a light bulb designed for a 120 V circuit into a 240 V circuit will result in a quadrupling of power, which can have dangerous consequences for the bulb and its surroundings. It is important to always use bulbs in circuits with the correct operating voltage to ensure safe and efficient operation.