This article will tell you how much power is dissipated in a 75W light bulb.

This will also include computations on how to get power dissipation levels.

Light bulbs are one of the most essential electronics that a house needs. There are various types of light bulbs sold in the market, from Candelabra Bulbs** **to** **LED Flame Bulbs mostly determined by their composition as well as the watts per voltage they produce to generate an ample amount of light to illuminate our houses, rooms, etc. While LED and incandescent light produce different brightness and luminescence, the watt-voltage rating of a light bulb is also a determinant in how much light it produces per volt of electricity accepted.

Knowing the dissipated power in a light bulb in any watt ratings is important for those who are either managing their electricity bills or simply controlling the power rating of every electronics in their household. For machine makers, knowing the dissipated power levels keeps the electric currents in check to avoid overcharges and short circuitries, especially for those who are making machines with light bulb indicators.

**How to Get Dissipated Power in a Light Bulb?**

The key to understanding **how much power is dissipated in a light bulb that is normally rated at 75W** is how the dissipation of power is computed. We can start by deriving the formula from Ohm’s Law, where Power (P) is the quotient between Voltage (V) raised to the squared power, and resistance (R), therefore:

**P = V**^{2}**/R**

As the given is a light bulb rated in watts, the measurement of power (P) is given but not the resistance of the light bulb. The resistance value is import because this will determine the dissipated power of the given light bulb and the voltage of electricity where the bulb was attached. This is the new formula to get the resistance (R) in Ohms.

**R = V**^{2}** / P**

You will immediately get the resistance value once you have both the Voltage (raised to the power of 2) and the power in Watts. Once you have the given values, you can now compute the dissipated power over a given power rating of a light bulb.

**How Much Power is Dissipated in an Average 75W Light Bulb?**

To compute, we will use the first formula given above. Let us also say that the voltage in the household is at 120V.

P = V^{2}/R

R = V^{2}/P

R = (120)^{2} V / 75W

R = 14,400V / 75W

**R = 192Ω**

Now, we have resistance value at 192Ω. Now, we can solve the dissipated power value using the first equation, which is P = V2/R.

P = V^{2} / R

P = (120)^{2}V / 192Ω

P = 14,400V / 192Ω

**P = 75W**

This means that at a regular household voltage, the power dissipated in the light bulb is the average rating, which is 75W. What if we use the standard industry voltage rating (220V?)

P = V2 / R

P = (220V)^{2}V / 192Ω

P = 48,400V / 192Ω

**P = 252W**

In this equation, your 75W light bulb would have blown up. Should the voltage go lower than standard, chances are the light bulb would emit only a small amount of light. Overall, it all depends on the square of the voltage.

**Conclusion**

In knowing **how much power is dissipated in a light bulb that is normally rated at 75W**, it is good to know first the resistance level of the bulb as well as the voltage output of your electrical outlet or power source. Should the light bulb not meet or exceed the required voltage for power to be dissipated, the light bulb would either dim or blows itself up respectfully. The formula shown in this article is the best way for you to know first if your 75W lightbulb is fit for your luminescence needs.

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