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Chapter 25: Problem 11

A \(1.50-m\) cylindrical rod of diameter 0.500 \(\mathrm{cm}\) is connected to a power supply that maintains a constant potential difference of 15.0 \(\mathrm{V}\) across its ends, while an ammeter measures the current through it. You observe that at room temperature \(\left(20.0^{\circ} \mathrm{C}\right)\) the ammeter reads 18.5 \(\mathrm{A}\) , while at \(92.0^{\circ} \mathrm{C}\) it reads 17.2 \(\mathrm{A}\) . You can ignore any thermal expansion of the rod. Find (a) the resistivity and (b) the temperature coefficient of resistivity at \(20^{\circ} \mathrm{C}\) for the material of the rod.

Short Answer

Expert verified
Resistivity at 20掳C is \(1.06 \times 10^{-5}\ \Omega\cdot\mathrm{m}\); temperature coefficient is \(1.06 \times 10^{-3}\ \mathrm{C}^{-1}\).

Step by step solution

01

Calculate Resistivity at 20掳C

To find the resistivity \( \rho \) of the rod's material at \( 20^{\circ} \mathrm{C} \), use Ohm's Law \( V = IR \), where \( R \) is the resistance. \( R \) can be calculated as \( R = \frac{V}{I} \). Given \( V = 15.0 \ \mathrm{V} \) and \( I_{20} = 18.5 \ \mathrm{A} \), \( R_{20} = \frac{15.0}{18.5} = 0.811 \ \Omega \). The resistivity is given by \( \rho = R \frac{A}{L} \), where \( A = \pi \left( \frac{d}{2} \right)^2 \) is the cross-sectional area with diameter \( d = 0.500 \ \mathrm{cm} = 0.005 \ \mathrm{m} \), and \( L = 1.50 \ \mathrm{m} \). Therefore, \( A = \pi \times (0.0025^2) = 1.9635 \times 10^{-5} \ \mathrm{m}^2 \). Thus, \( \rho = 0.811 \times \frac{1.9635 \times 10^{-5}}{1.50} = 1.06 \times 10^{-5} \ \Omega \cdot\mathrm{m} \).
02

Calculate Resistivity at 92掳C

Similarly, find the resistivity at \( 92^{\circ} \mathrm{C} \) using the measured current \( I_{92} = 17.2 \ \mathrm{A} \). Calculate the resistance: \( R_{92} = \frac{V}{I_{92}} = \frac{15.0}{17.2} = 0.872 \ \Omega \). Using the resistivity formula, \( \rho_{92} = 0.872 \times \frac{1.9635 \times 10^{-5}}{1.50} = 1.14 \times 10^{-5} \ \Omega \cdot \mathrm{m} \).
03

Determine the Temperature Coefficient of Resistivity

The temperature coefficient of resistivity \( \alpha \) is calculated using the formula \( \rho_{92} = \rho_{20} (1 + \alpha \Delta T) \), where \( \Delta T = 92 - 20 = 72 \ \mathrm{C} \). Rearrange the formula: \( \alpha = \frac{\rho_{92} - \rho_{20}}{\rho_{20} \times \Delta T} = \frac{1.14 \times 10^{-5} - 1.06 \times 10^{-5}}{1.06 \times 10^{-5} \times 72} = 1.06 \times 10^{-3} \ \mathrm{C}^{-1} \).

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Key Concepts

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

Ohm's Law
Ohm's Law is a fundamental concept used in the field of electrical engineering to relate voltage (V), current (I), and resistance (R). The law states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points. This is mathematically represented by the equation: \[ V = IR \] where:- \( V \) is the voltage across the conductor,- \( I \) is the current flowing through the conductor,- \( R \) is the resistance of the conductor.Applying this concept to our cylindrical rod problem, we use Ohm's Law to find the resistance of the rod at different temperatures. For example, at room temperature of 20掳C, with a voltage of 15.0 V and a current of 18.5 A, the resistance \( R \) can be calculated as \( R = \frac{V}{I} = \frac{15.0}{18.5} = 0.811 \, \Omega \). This calculated resistance is key to further finding the resistivity of the material, which measures how strongly the material opposes the flow of electric current.
Temperature Coefficient of Resistivity
The temperature coefficient of resistivity is an indicator of how a material's resistivity changes with temperature. Most conductive materials will see an increase in resistivity as the temperature rises. The coefficient \( \alpha \) is defined by the relation:\[ \rho_T = \rho_{20}(1 + \alpha \Delta T) \] Here:- \( \rho_T \) is the resistivity at temperature \( T \),- \( \rho_{20} \) is the standard resistivity at 20掳C,- \( \alpha \) is the temperature coefficient of resistivity,- \( \Delta T \) is the change in temperature from the reference 20掳C.In our scenario, we calculated the resistivity at both 20掳C and 92掳C. Using these values and the formula above, we find \( \alpha \). Substituting the known resistivities and the temperature change (72掳C), we find that \( \alpha = 1.06 \times 10^{-3} \, \mathrm{C}^{-1} \). This indicates that for each degree Celsius increase in temperature, the resistivity increases by a small factor in this particular rod material.
Cylindrical Conductors
A cylindrical conductor is a common type of shape for wires and rods used in electrical circuits. The geometry of a cylinder allows us to calculate important electrical properties such as resistivity because it provides a uniform cross-sectional area for current flow.The cross-sectional area \( A \) of a cylindrical conductor is determined using the formula:\[ A = \pi \left( \frac{d}{2} \right)^2 \]where \( d \) is the diameter of the cylinder.In the exercise, the cylindrical rod has a diameter of 0.500 cm (or 0.005 m), leading to a cross-sectional area of \( A = \pi \times (0.0025^2) = 1.9635 \times 10^{-5} \, \mathrm{m}^2 \). This area is crucial in calculating the resistivity of the material.Resistivity \( \rho \) is found using:\[ \rho = R \frac{A}{L} \]where \( L \) is the length of the cylinder. Here, the length is 1.50 m. Understanding the relation between these dimensions helps to determine how changes in current and voltage affect the material's conductivity.

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Most popular questions from this chapter

A lightning bolt strikes one end of a steel lightning rod, producing a \(15,000-\mathrm{A}\) current burst that lasts for 65\(\mu\) s. The rod is 20 \(\mathrm{m}\) long and 1.8 \(\mathrm{cm}\) in diameter, and its other end is connected to the ground by 35 \(\mathrm{m}\) of \(8.0-\mathrm{mm}\) -diameter copper wire (a) Find the potential difference between the top of the steel rod and the lower end of the copper wire during the current burst. (b) Find the total energy deposited in the rod and wire by the current burst.

The potential difference across the terminals of a battery is 8.4 \(\mathrm{V}\) when there is a current of 1.50 \(\mathrm{A}\) in the battery from the negative to the positive terminal. When the current is 3.50 \(\mathrm{A}\) in the reverse direction, the potential difference becomes 9.4 \(\mathrm{V}\) . (a) What is the internal resistance of the battery? (b) What is the emf of the battery?

A toaster using a Nichrome heating element operates on 120 \(\mathrm{V}\) . When it is switched on at \(20^{\circ} \mathrm{C}\) , the heating element carries an initial current of 1.35 \(\mathrm{A}\) . A few seconds later the current reaches the steady value of 1.23 \(\mathrm{A}\) . (a) What is the final temperature of the element? The average value of the temperature coefficient of resistivity for Nichrome over the temperature range is \(4.5 \times\) \(10^{-4}\left(\mathrm{C}^{\circ}\right)^{-1}\) . (b) What is the power dissipated in the heating element initially and when the current reaches a steady value?

The current-voltage relationship of a semiconductor diode is given by $$ I=I_{\mathrm{S}}\left[\exp \left(\frac{e V}{k T}\right)-1\right] $$ where \(I\) and \(V\) are the current through and the voltage across the diode, respectively. Is a constant characteristic of the device, \(e\) is the magnitude of the electron charge, \(k\) is the Boltzmann constant, and \(T\) is the Kelvin temperature. Such a diode is connected in series with a resistor with \(R=1.00 \Omega\) and a battery with \(\mathcal{E}=2.00 \mathrm{V} .\) The polarity of the battery is such that the current through the diode is in the forward direction (Fig. \(25.45 ) .\) The battery has negligible internal resistance. (a) Obtain an equation for \(V .\) Note that you cannot solve for \(V\) algebraically. (b) The value of \(V\) must be obtained by using a numerical method. One approach is to try a value of \(V\) , see how the left-and right-hand sides of the equation compare for this \(V\) , and use this to refine your guess for \(V\) . Using \(I_{S}=1.50 \mathrm{mA}\) and \(T=293 \mathrm{K},\) obtain a solution (accurate to three significant figures) for the voltage drop \(V\) across the diode and the current \(I\) through it.

The Tolman-Stewart experiment in 1916 demonstrated that the free charges in a metal have negative charge and provided a quantitative measurement of their charge-to-mass ratio, \(|q| / m\) . The experiment consisted of abruptly stopping a rapidly rotating spool of wire and measuring the potential difference that this produced between the ends of the wire. In a simplified model of this experment, consider a metal rod of length \(L\) that is given a uniform acceleration \(\overrightarrow{\boldsymbol{d}}\) to the right. Initially the free charges in the metal lag behind the rod's motion, thus setting up an electric field \(\overrightarrow{\boldsymbol{E}}\) in the rod. In the steady state this field exerts a force on the free charges that makes them accelerate along with the rod. (a) Apply \(\Sigma \vec{F}=m \vec{d}\) to the free charges to obtain an expression for \(|q| / m\) in terms of the magnitudes of the induced electric field \(\overrightarrow{\boldsymbol{k}}\) and the acceleration \(\overrightarrow{\boldsymbol{d}} .\) (b) If all the free charges in the metal rod have the same acceleration, the electric field \(\overrightarrow{\boldsymbol{E}}\) is the same at all points in the rod. Use this fact to rewrite the expression for \(|q| / m\) in terms of the potential \(V_{b c}\) between the ends of the rod (Fig. 25.44\()\) . (c) If the free charges have negative charge, which end of the rod, \(b\) or \(c,\) is at higher potential? (d) If the rod is 0.50 \(\mathrm{m}\) long and the free charges are electrons (charge \(q=-1.60 \times 10^{-19} \mathrm{C},\) mass \(9.11 \times 10^{-31} \mathrm{kg} ),\) what magnitude of acceleration is required to produce a potential difference of 1.0 \(\mathrm{mV}\) between the ends of the rod?(e) Discuss why the actual experiment used a rotating spool of thin wire rather than a moving bar as in our simplified analysis.
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