# What does it mean when we say that power of a bulb is 10 W? Since $V/I=$ resistance is a constant, how can power $=VI$ be a constant?

My question is simple. In Ideal situation, at constant temperature, we know that normal appliances like a filament bulb has straight Voltage vs Current graph, meaning its resistance is constant or voltage is directly proportional to current.

Now, we also have bulbs of desired power available. eg. 10W, 20W, 100W bulbs etc. Since I understand that power = V x Current, the power for a bulb can not be a constant if its resistant is assumed a constant. A normal mathematical thinking can confirm that.

So, what does it means when we say the a certain bulb is a 10W bulb? Does it simply means that it would consume 2 times energy at a given voltage if it replaces a 5W bulb?

Actually, we don't know that "filament bulb has straight Volatage vs Current graph": "The actual resistance of the filament is temperature dependent. The cold resistance of tungsten-filament lamps is about 1/15 the hot-filament resistance when the lamp is operating. For example, a 100-watt, 120-volt lamp has a resistance of 144 ohms when lit, but the cold resistance is much lower (about 9.5 ohms)." (http://en.wikipedia.org/wiki/Incandescent_light_bulb )

The power rating given on lightbulbs always refers to the power at a specified operational voltage (which is always given together with the power or implied by the type of socket). The power at different voltages is not easily predictable as the resistance of the filament will vary strongly in dependence of temperature (which depends on the dissipated power).

Furthermore, fluorescent lamps and LED lamps which have electronic components to control the lamp will usually not work at all at other voltages than the specified operational one, so a power rating that does not refer to the nominal operational voltage will not make any sense here anyway.

Since I understand that power = V x Current, the power for a bulb can not be a constant if its resistant is assumed a constant. A normal mathematical thinking can confirm that.

If the voltage and current don't change, then the power is constant. The electricity supplied from the wall is at 115V (more or less). If the resistance of the bulb is 1322 ohms, then the current is 87 milliamps. This give the power for the bulb as 10 Watts. The resistance of the bulb (at a certain temperature) is nearly constant, therefore the power calculated is nearly constant.

Why, because the voltage is constant and the resistance is constant, which means the current is constant, ....

The bulb is intended to operate using a particular source voltage and frequency. (For example, 120 V of 60 Hz single-phase AC.) Under these conditions, the bulb is expected to use an average of 10 W of electrical power. (The power usage might be different while the bulb is warming up after being turned on, or while the filament is exploding while burning out.)

This power rating is very important to the designer of the bulb, the specifier of the bulb, the installer of the bulb, and the owner of the bulb.

• The power rating determines how much heat must be safely conducted and/or convected and/or radiated out from the bulb to a safe place. If the any of the materials between the filament and the final heat sink get too hot, there could be serious trouble. (For example, the filament might burn out, the bulb might shatter, the wiring's insulation might melt, a home's blown-in cellulose insulation might catch fire, et cetera.) To avoid such trouble, these designers, specifiers, and installers need to make sure that properly rated materials and devices are used around the bulb.
• Furthermore, the power used by the bulb affects the building's energy usage. (It helps satisfy the need for heat in the winter, but it increases the need for air conditioning in the summer. It is a particular problem inside of refrigerators.)
• Electricity is expensive. The direct operating cost of the bulb is proportional to the input power (while the bulb is on) times the amount of time the bulb is on.

The power rating is more important to these people than the details of the resistance of the bulb. As other answerers have pointed out, these details are far more complicated than "Ohm's Approximation" might suggest. ("Ohm's Law" is a bit of a misnomer.)