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An incandescent light bulb uses a coiled filament of tungsten that is \(580 \mathrm{~mm}\) long with a diameter of \(46.0 \mu \mathrm{m} .\) At \(20.0^{\circ} \mathrm{C}\) tungsten has a resistivity of \(5.25 \times 10^{-8} \Omega \cdot \mathrm{m} .\) Its temperature coefficient of resistivity is \(0.0045\left(\mathrm{C}^{\circ}\right)^{-1},\) and this remains accurate even at high temperatures. The temperature of the filament increases linearly with current, from \(20^{\circ} \mathrm{C}\) when no current flows to \(2520^{\circ} \mathrm{C}\) at 1.00 A of current. (a) What is the resistance of the light bulb at \(20^{\circ} \mathrm{C} ?\) (b) What is the current through the light bulb when the potential difference across its terminals is \(120 \mathrm{~V} ?\) (Hint: First determine the temperature as a function of the current; then use this to determine the resistance as a function of the current. Substitute this result into the equation \(V=I R\) and solve for the current \(I .\) ) (c) What is the resistance when the potential is \(120 \mathrm{~V} ?\) (d) How much energy does the light bulb dissipate in 1 min when \(120 \mathrm{~V}\) is supplied across its terminals? (e) How much energy does the light bulb dissipate in 1 min when half that voltage is supplied?

Step by step solution

01

Calculate the Resistance at 20°C

Start by using the fomula for resistivity \( R = \frac {\rho L}{A}\), where \( \rho \) is resistivity, \( L \) is length, \( A \) is cross-sectional area. The cross-sectional area of the tungsten coil is given by \( A = \pi r^2 \), where \( r \) is the radius of the cross section. Calculate this value.

02

Determine the temperature as a function of current

In order to determine the temperature as a function of the current, create a linear equation using the provided temperatures at endpoints current 0 and 1 A. \( T(I) = T_0 + \alpha I \), where \( T(I) \) is temperature at any current \( I \), \( T_0 \) is the initial temperature, and \( \alpha \) is the rate of change of temperature with respect to current.

03

Determine resistance as a function of current

Now, calculate the resistance as a function of current using the equation \( R(I) = R_0 [1 + \beta (T(I) - T_0)] \), where \( \beta \) is the temperature coefficient of resistivity, \( T_0 \) is the initial temperature, and \( T(I) \) is the temperature at current \( I \).

04

Calculate the current when potential difference is 120V

Substitute the resistance function obtained in step 3 into Ohm's law to solve for the current \( I \).

05

Calculate the resistance at the potential difference of 120 V

Using the current found in step 4, find the resistance at this current by substituting this current into the resistance function created in step 3.

06

Calculate the energy dissipated at the potential difference of 120 V and half of it

The energy dissipated can be calculated by using the formula \( E = P \times t \) where \( P = IV \) is the power and \( t \) is time. For the potential of 120 V and half of it, calculate the energy dissipated in 1 minute (or 60 seconds).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understand Electric Resistance

Electric resistance is a fundamental property of materials which measures how much the material opposes the flow of electric current. It's governed by Ohm's law and is calculated using the formula
\[ R = \frac{\rho L}{A} \]
where \( R \) is the resistance, \( \rho \) is the resistivity of the material, \( L \) is the length through which the current flows, and \( A \) is the cross-sectional area perpendicular to the current.

In the context of our exercise, tungsten's resistance at room temperature is calculated by taking into account its resistivity, the length of the filament, and its cross-sectional area. For a real-world application such as an incandescent light bulb, this measurement is crucial as it determines how much voltage is needed for a certain current to flow through the filament and in turn, how much light the bulb can emit.

Temperature Coefficient of Resistivity

As temperature affects electric resistance, the temperature coefficient of resistivity (often denoted as \( \beta \)) quantifies this effect. This coefficient tells us how much the resistivity of a material changes per degree change in temperature. The formula to describe the change in resistance with temperature is:
\[ R(T) = R_0 [1 + \beta (T - T_0)] \]
where \( R(T) \) is the resistance at temperature \( T \), \( R_0 \) is the original resistance, \( \beta \) is the temperature coefficient, and \( T_0 \) is the original temperature.

Tungsten, used in our exercise, has a relatively high temperature coefficient, meaning its resistance changes significantly as the filament heats up. This is a critical aspect when designing electrical components that must operate across a range of temperatures, such as light bulbs, which must be calculated to ensure optimal performance throughout the entire range of operational temperatures.

Applying Ohm's Law

Ohm's Law is a basic principle in electrical engineering, stating that the current through a conductor is directly proportional to the voltage across it and inversely proportional to its resistance. It can be written as:
\[ V = I \times R \]
where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance. In practice, this law allows us to predict one electrical parameter if the other two are known.

In the example with the incandescent light bulb, Ohm's Law is used to establish a relationship between voltage, current, and resistance. The law helps in finding out the current running through the bulb when a specific voltage is applied across it. For energy efficiency and safety, knowing how to apply Ohm's Law is essential for any electrical application, including the proper functioning of household lightning. It also serves as a foundation for more complex electronic theories and circuits.