Voltage effects on power dissipated

[fresno89]

Trying to find the relationship between voltage and power across a resistive something.

Using a lightbulb if the bulb was rated at 100W when 140 Volts is used, what would happen to the power if the voltage was doubled? (assuming the light bulb doesn't explode)

 

[russ_watters]

Welcome to PF.

If the light bulb doesn't explode, it obeys V=IR and P=VI.

 

[Carl Pugh]

Light bulbs aren't aware of Ohm's law.
The current would probably go up between 5% and 100%. Current would depend on rating of light bulb.
Unless of course the light bulb used a carbon filament, then the current would more than double.

 

[Windadct]

Other then electronics 101...Watts = V^2/R ... but lightbulbs are specifcally made to give off light - not be resistors.

What C-P is alluding to is that the element of the light bulb is a resistor - but the resistance varies greatly with temperature. Tungsten has a Positive Temp Coefficent - as the temp goes up the resistance goes up. A carbon element has a negative TC (as do Diodes ) so as the temp goes up the resistance goes down -

Precision resistors usually have a specific rating for this TCR.

Soooo - the better question would be - how doe the resistance of a light bulb vary over voltage - that totally depends on the lightbulb!

 

[alphy]

The relationship between voltage and power dissipated in a resistive element is described by Ohm's Law, which states that power (P) is equal to the square of the voltage (V) divided by the resistance (R). This means that as voltage increases, power dissipation also increases.

In the given scenario, if the voltage is doubled from 140 volts to 280 volts, the power dissipated by the lightbulb would quadruple. This is because doubling the voltage would result in four times the original power dissipation. Therefore, the lightbulb would now be rated at 400 watts (100W x 4).

However, it is important to note that this increase in power dissipation may cause the lightbulb to overheat and potentially fail or even explode. It is always important to follow the recommended voltage and wattage ratings for any electrical device to ensure safe operation.

In conclusion, the relationship between voltage and power dissipation is directly proportional, and doubling the voltage would result in a four-fold increase in power dissipation.

 

[alphysicist]

The relationship between voltage and power across a resistive element is described by Ohm's Law, which states that power (P) is equal to the product of voltage (V) and current (I), or P = VI. Therefore, if the voltage is doubled, the power will also double, assuming the resistance of the element remains constant.

In the case of the lightbulb, if it is rated at 100W when 140V is used, the current flowing through it can be calculated using Ohm's Law as I = P/V = 100/140 = 0.71A. If the voltage is doubled to 280V, the current will also double to 1.42A, resulting in a power of 200W (P = VI = 280 x 1.42 = 200W).

It is important to note that this relationship only holds true for resistive elements, as the power dissipated in other types of elements (such as capacitors or inductors) can be affected by factors such as frequency and phase difference between voltage and current.

Additionally, it is important to consider the limitations of the lightbulb and not exceed its rated voltage, as this can lead to overheating and potentially cause the bulb to explode. In cases where higher power is needed, it is recommended to use a higher wattage bulb rather than increasing the voltage.