To understand why an incandescent light bulb filament is most likely to "burn out" when it is first turned on, we need to consider the behavior of the filament as it heats up.
When the light bulb is turned on, the filament starts to heat up due to the flow of electric current through it. The resistance of the filament when it is turned off is 24 ohms, which means that it has a relatively low resistance compared to when it is operating at 240 ohms when connected to a 120-volt power source.
According to Ohm's law (V = IR), where V is the voltage, I is the current, and R is the resistance, the current flowing through the filament can be calculated using the formula I = V/R.
When the light bulb is first turned on, the current flowing through the filament is relatively high because the resistance is low. This high current leads to a rapid increase in the temperature of the filament.
As the filament heats up, its resistance increases. This is due to the phenomenon known as the positive temperature coefficient of resistance, which means that the resistance of most metals, including tungsten, increases with an increase in temperature.
As the resistance of the filament increases, the current flowing through it decreases. Consequently, the rate at which the filament heats up decreases.
However, when the light bulb is initially turned on, the rapid increase in current and subsequent temperature rise can cause a thermal shock to the filament. This rapid expansion and contraction due to the heating and cooling can lead to mechanical stress on the filament, increasing the chance of it breaking or burning out.
To calculate the power dissipated when first turned on, we can use the formula P = IV, where P is the power, I is the current, and V is the voltage.
Given that the resistance of the filament when turned off is 24 ohms, and the voltage is 120 volts, we can calculate the current using Ohm's law: I = V/R.
I = 120 V / 24 Ω
I = 5 A
Substituting the current into the power formula, we have:
P = (5 A) * (120 V)
P = 600 watts
Therefore, when the incandescent light bulb is first turned on, it dissipates a high amount of power, contributing to the increased temperature and thermal stress on the filament, making it more likely to "burn out" compared to when it is already operating at a steady state.