The typical resistance of the tungsten filament of a 60 watt incandescent light bulb when it is turned off is 24 ohms and its resistance when operating from a 120 volt power source is 240 ohms. Generally, an incandescent light bulb filament is most likely to “burn out” when it is first turned on than at any other time. Explain why? HINT: Calculate the power dissipated when first turned on.
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 tungsten filament in the bulb.
When the light bulb is turned off, the resistance of the tungsten filament is given as 24 ohms. This resistance represents the filament at a relatively low temperature.
When the light bulb is turned on and operates from a 120-volt power source, the resistance of the filament increases to 240 ohms. This increased resistance indicates that the filament is hotter and has a higher temperature.
The resistance of an object, in this case, the tungsten filament, is influenced by its temperature. As the temperature of a conductor increases, its resistance also increases. This is known as positive temperature coefficient.
When the light bulb is first turned on, the cold tungsten filament has a lower resistance, which allows a higher current to flow through it. This results in a rapid increase in temperature as the filament heats up.
As the filament heats up further, its resistance also increases. This, in turn, limits the current flowing through the filament, preventing it from overheating and burning out.
However, during the initial moments of turning on the light bulb, the rapid increase in temperature can cause a surge in the current. This surge in current can overload the filament before it reaches its normal operating temperature and equilibrium resistance value.
Therefore, the filament in an incandescent light bulb is most likely to "burn out" when it is first turned on due to the surge in current and the resultant rapid increase in temperature, which exceeds the filament's capability to dissipate heat effectively.
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.
With less resistance at turnon, before the filament heats up to operating temperature, there is a current and power surge that stresses the delicate tungsten filament. Temperature gradients during rapid heating can cause thermal stress fracture.
You can answer this the way the hint suggested: what is V^2/R when turned on as compared to later?
But the reason for failing is not this, it has to do with a metal problem mainly. Tungsten when hot then cooled crystallizes. After many times, the filiament when cold is brittle in spots, very unductile. This brittleness in spots offers high resistace in just those spots, so a lot of ohmic losses occur in those spots, and failure occurs. Yes, the surge currents cause it, however these spots are the ones which fail.