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Chapter 19: Problem 15

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
(a) Resistivity at 20掳C: \(1.061 \times 10^{-5} \ \Omega\cdot\mathrm{m}\); (b) Temperature coefficient: \(1.045 \times 10^{-3} \ \mathrm{掳C}^{-1}\).

Step by step solution

01

Calculate the Resistance at Room Temperature

Use Ohm's law, which states that the resistance \( R \) is given by \( R = \frac{V}{I} \). Here, \( V = 15.0 \) V and \( I = 18.5 \) A (at 20掳C). Thus, the resistance \( R_{20} \) at 20掳C is:\[ R_{20} = \frac{15.0}{18.5} \approx 0.811 \ \Omega \]
02

Calculate the Resistance at Higher Temperature

Similarly, at 92.0掳C, \( I = 17.2 \) A. Use Ohm's law to find the resistance, \( R_{92} \):\[ R_{92} = \frac{15.0}{17.2} \approx 0.872 \ \Omega \]
03

Calculate the Resistivity at Room Temperature

The resistivity \( \rho \) is related to resistance by the formula \( R = \rho \frac{L}{A} \), where \( L \) is the length and \( A \) is the cross-sectional area of the rod. The area \( A \) is given by \( A = \pi \left(\frac{d}{2}\right)^2 \) with \( d = 0.500 \) cm. Convert \( d \) to meters and calculate:\[ A = \pi \left(0.0025 \right)^2 \approx 1.9635 \times 10^{-5} \ \text{m}^2 \]Thus, the resistivity \( \rho_{20} \) is:\[ \rho_{20} = R_{20} \frac{A}{L} = 0.811 \frac{1.9635 \times 10^{-5}}{1.5} \approx 1.061 \times 10^{-5} \ \Omega \cdot \text{m} \]
04

Calculate the Temperature Coefficient of Resistivity

The temperature coefficient of resistivity \( \alpha \) is given by:\[ R_{92} = R_{20} (1 + \alpha (T_{92} - T_{20})) \]Substituting the known values:\[ 0.872 = 0.811 (1 + \alpha (92 - 20)) \]This simplifies to:\[ \alpha = \frac{0.872 - 0.811}{0.811 \times 72} \approx 1.045 \times 10^{-3} \ \text{掳C}^{-1} \]

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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 principle in electronics and physics, stating that the current flowing through a conductor between two points is directly proportional to the voltage across the two points. The formula is expressed as:\[ R = \frac{V}{I} \]where:
  • \( R \) is the resistance (in ohms, \( \Omega \))
  • \( V \) is the voltage (in volts, V)
  • \( I \) is the current (in amperes, A)
In the original exercise, Ohm's Law is applied to calculate the resistance at room temperature and at a higher temperature. By knowing the voltage (15.0 V) and the current at two different temperatures, we can find how the resistance in the cylindrical rod changes with temperature. This is accomplished by rearranging Ohm's Law to solve for resistance, which helps in determining other properties, like resistivity.
Temperature Coefficient of Resistivity
The temperature coefficient of resistivity is a measure of how the resistivity of a material changes with temperature. It is denoted by \( \alpha \) and helps in determining the sensitivity of the material's resistivity to temperature changes. The relationship between resistance and the temperature coefficient is given by:\[ R_T = R_0 (1 + \alpha (T - T_0)) \]where:
  • \( R_T \) is the resistance at temperature \( T \)
  • \( R_0 \) is the resistance at a reference temperature \( T_0 \)
  • \( \alpha \) is the temperature coefficient of resistivity
  • \( T \) and \( T_0 \) are temperatures in degrees Celsius
In the exercise, after calculating resistance at two different temperatures, the formula is rearranged to solve for the temperature coefficient \( \alpha \). This coefficient is crucial for predicting how the resistance of materials like metals will change as temperatures vary. Understanding \( \alpha \) allows engineers to design circuits that can operate under different environmental conditions.
Cylindrical Rod
A cylindrical rod is a geometric shape that is characterized by a circular cross-section and a constant diameter along its entire length. In the context of electricity and resistivity, the properties of the cylindrical rod affect how electric current flows through it. The cross-sectional area \( A \) of the rod is critical and can be calculated with:\[ A = \pi \left( \frac{d}{2} \right)^2 \]where \( d \) is the diameter of the rod.The resistivity \( \rho \) of a material is related to resistance and is calculated using:\[ R = \rho \frac{L}{A} \]where \( L \) is the length of the rod and \( A \) is its cross-sectional area.In the problem, the cylindrical rod's dimensions (length and diameter) are used to determine the area, which is then applied to calculate the resistivity of the material at room temperature. Knowing the resistivity helps in understanding how conductive the material is in its cylindrical form, which is essential for designing efficient electrical components.

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Most popular questions from this chapter

Lightbulbs in series, I. The power rating of a lightbulb is the power it consumes when connected across a 120 \(\mathrm{V}\) out-let. (a) If you put two 100 \(\mathrm{W}\) bulbs in series across a 120 \(\mathrm{V}\) outlet, how much power would each consume if its resistance were constant? (b) How much power does each one consume if you connect them in parallel across a 120 \(\mathrm{V}\) outlet?

A 500.0\(\Omega\) resistor is connected in series with a capacitor. What must be the capacitance of the capacitor to produce a time constant of 2.00 \(\mathrm{s} ?\)

Treatment of heart failure. A heart defibrillator is used to enable the heart to start beating if it has stopped. This is done by passing a large current of 12 A through the body at 25 \(\mathrm{V}\) for a very short time, usually about 3.0 \(\mathrm{ms}\) (a) What power does the defibrillator deliver to the body, and (b) how much energy is transferred?

Electrical safety. This procedure is not recommended! You'll see why after you work the problem. You are on an aluminum ladder that is standing on the ground, trying to fix an electrical connection with a metal screwdriver having a metal handle. Your body is wet because you are sweating from the exertion; therefore, it has a resistance of 1.0 \(\mathrm{k} \Omega\) . (a) If you accidentally touch the "hot" wire connected to the 120 \(\mathrm{V}\) line, how much current will pass through your body? Is this amount enough to be dangerous? (The maximum safe current is about 5 \(\mathrm{mA}\). (b) How much electrical power is delivered to your body?

A capacitor is charged to a potential of 12.0 \(\mathrm{V}\) and is then connected to a voltmeter having an internal resistance of 3.40 \(\mathrm{M\Omega} .\) After a time of 4.00 \(\mathrm{s}\) the voltmeter reads 3.0 \(\mathrm{V}\) What are (a) the capacitance and (b) the time constant of the circuit?
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