Why Does a Lightbulb Burn Brighter Prior to Burning Out?

[pharaoh]

According to ohm's law, why would a light bulb burn brighter the instant prior to it burning out?

answer:
ohm's law is R=V/I
the ratio of potential difference to current is a constant for a given conductor. any resistance that does not change with temperature, voltage or the direction of charges obey ohm's law.

at the instant the flow of the potential drop be higher but then due to the energy transferring to thermal the light bulb started to lose its brightens. the brighten of the bulb depends on the current that pass through, so I will assume that at the beginning the current flow is larger until its reach other resistor causing the voltage to split into difference value depending on the resistance.

Is it right?

 

[anathema]

Yes, you are correct. As the light bulb starts to burn out, the resistance of the filament increases due to the heating of the wire. This causes the current to decrease, according to ohm's law. However, since the voltage remains constant, the decrease in current leads to a decrease in power, which results in the light bulb burning less brightly. So, the instant before the bulb burns out, the current is at its highest and the bulb appears to be the brightest.

 

[m2003]

Yes, you are correct. Ohm's law states that the current flowing through a conductor is directly proportional to the voltage applied and inversely proportional to the resistance of the conductor. As a lightbulb nears the end of its lifespan, the resistance of the filament increases due to wear and tear. This causes the current to decrease and the voltage to increase, resulting in a brighter light for a brief moment before the bulb burns out.