Problem 21
Question
A common flashlight bulb is rated at \(0.30 \mathrm{~A}\) and \(2.9 \mathrm{~V}\) (the values of the current and voltage under operating conditions). If the resistance of the tungsten bulb filament at room temperature \(\left(20^{\circ} \mathrm{C}\right)\) is \(1.1 \Omega,\) what is the temperature of the filament when the bulb is on?
Step-by-Step Solution
Verified Answer
The filament temperature is approximately \(1751.31^{\circ}C\) when the bulb is on.
1Step 1: Understanding the Problem
We need to find the temperature of the tungsten filament when the bulb is on, given the initial resistance at room temperature, the current, and the voltage when operating.
2Step 2: Use Ohm's Law to Find Operating Resistance
Ohm's Law states that \( V = IR \), where \( V \) is voltage, \( I \) is current, and \( R \) is resistance. We can solve for resistance using the formula: \[R = \frac{V}{I} = \frac{2.9}{0.30} = 9.67\, \Omega\]Thus, the operating resistance of the filament is \(9.67 \Omega\).
3Step 3: Relationship Between Resistance and Temperature
For tungsten, the resistance changes with temperature according to the formula: \[R_t = R_0 (1 + \alpha \Delta T)\]where \( R_0 \) is the initial resistance, \( R_t \) is the resistance at temperature \( T \), \( \alpha \) is the temperature coefficient of resistance for tungsten (approximately \(0.0045\, ^{\circ}C^{-1}\)), and \( \Delta T \) is the change in temperature.
4Step 4: Solve for Temperature Change
We rearrange the formula to solve for the change in temperature \( \Delta T \): \[\Delta T = \frac{R_t - R_0}{R_0 \cdot \alpha} = \frac{9.67 - 1.1}{1.1 \times 0.0045}\]Simplify to find \( \Delta T \):\[\Delta T = \frac{8.57}{0.00495} \approx 1731.31\, ^{\circ}C\]
5Step 5: Calculate Absolute Temperature
The absolute temperature of the filament when the bulb is on is: \[T = 20^{\circ}C + 1731.31^{\circ}C = 1751.31^{\circ}C\]Thus, the temperature of the filament is approximately \(1751.31^{\circ}C\) when the bulb is operating.
Key Concepts
Tungsten FilamentTemperature Coefficient of ResistanceResistance and Temperature Relationship
Tungsten Filament
Tungsten filaments are widely used in light bulbs due to their excellent properties. Tungsten has a very high melting point, which makes it suitable for withstanding high temperatures without melting. This is essential in applications like bulbs, where the filament needs to emit light steadily over long periods.
Tungsten has excellent electrical conductivity, which helps it function efficiently as a filament.
Tungsten has excellent electrical conductivity, which helps it function efficiently as a filament.
- It offers high resistance to erosion from electric currents.
- It emits a bright and consistent glow when heated.
- Its long lifespan makes it economical for use in different environments.
Temperature Coefficient of Resistance
The temperature coefficient of resistance (\( \alpha \)) is a crucial factor in understanding how materials behave under temperature changes. It defines how the resistance of a material changes with temperature.
For tungsten, this value is approximately \( 0.0045 \, ^{\circ}C^{-1} \). This means that for each degree Celsius increase in temperature, the resistance increases by a factor of about 0.0045.
For tungsten, this value is approximately \( 0.0045 \, ^{\circ}C^{-1} \). This means that for each degree Celsius increase in temperature, the resistance increases by a factor of about 0.0045.
- The higher the temperature coefficient, the more significant the change in resistance with temperature.
- This factor allows us to predict resistance changes in applied conditions like heated filaments.
Resistance and Temperature Relationship
Resistance and temperature share a direct relationship. This relationship, especially significant in conductive materials like tungsten, can be expressed by the equation:
\[ R_t = R_0 (1 + \alpha \Delta T) \]
In this formula:
Thus, understanding this relationship helps in calculating the operational temperature of devices like light bulbs, ensuring they function within safe and efficient parameters.
\[ R_t = R_0 (1 + \alpha \Delta T) \]
In this formula:
- \( R_0 \) is the initial resistance (at room temperature).
- \( R_t \) is the resistance at a given higher temperature.
- \( \alpha \) is the temperature coefficient of resistance.
- \( \Delta T \) is the change in temperature.
Thus, understanding this relationship helps in calculating the operational temperature of devices like light bulbs, ensuring they function within safe and efficient parameters.
Other exercises in this chapter
Problem 19
What is the resistivity of a wire of \(1.0 \mathrm{~mm}\) diameter, \(2.0 \mathrm{~m}\) length, and \(50 \mathrm{~m} \Omega\) resistance?
View solution Problem 20
A certain wire has a resistance \(R .\) What is the resistance of a second wire, made of the same material, that is half as long and has half the diameter?
View solution Problem 22
The legend that Benjamin Franklin flew a kite as a storm approached is only a legend-he was neither stupid nor suicidal. Suppose a kite string of radius \(2.00
View solution Problem 23
When \(115 \mathrm{~V}\) is applied across a wire that is \(10 \mathrm{~m}\) long and has a \(0.30 \mathrm{~mm}\) radius, the magnitude of the current density i
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