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Chapter 26: Problem 24

24\. Heating Element A heating element is made by maintaining a potential difference of \(75.0 \mathrm{~V}\) across the length of a Nichrome wire that has a \(2.60 \times 10^{-6} \mathrm{~m}^{2}\) cross section. Nichrome has a resistivity of \(5.00 \times 10^{-7} \Omega \cdot \mathrm{m} .\) (a) If the element dissipates \(5000 \mathrm{~W}\), what is its length? (b) If a potential difference of \(100 \mathrm{~V}\) is used to obtain the same dissipation rate, what should the length be?

Short Answer

Expert verified
Length is 5.85 m for 75V potential difference and 10.4 m for 100V potential difference.

Step by step solution

01

Understand the power dissipation formula

The power dissipation in a resistor can be expressed as: \[ P = \frac{V^2}{R} \] where \(P\) is the power, \(V\) is the potential difference, and \(R\) is the resistance.
02

Relate resistance to resistivity

The resistance \(R\) of a wire can be calculated using its resistivity \(\rho\), length \(L\), and cross-sectional area \(A\): \[ R = \frac{\rho L}{A} \].
03

Combine the formulas

Using the formula for power and resistance, we combine them: \[ P = \frac{V^2 A}{\rho L} \] Rearranging for length \(L\), we get: \[ L = \frac{V^2 A}{P \rho} \]
04

Calculate the length for 75V potential difference

Plug in the given values for the situation with a potential difference of 75V: \[ L = \frac{(75.0 \, \text{V})^2 (2.60 \times 10^{-6} \, \text{m}^2)}{5000 \, \text{W} (5.00 \times 10^{-7} \, \text{惟鈰卪})} \] Calculate the numerator and denominator separately, then divide to find \(L\).
05

Solution for 75V

Calculate the following: \[ V^2 = (75.0)^2 = 5625 \, \text{V}^2 \] \[ P \rho = 5000 \, \text{W} \times 5.00 \times 10^{-7} \, \text{惟鈰卪} = 2.5 \times 10^{-3} \, \text{W鈰呂┾媴m} \] \[ L = \frac{5625 \times 2.60 \times 10^{-6}}{2.5 \times 10^{-3}} = 5.85 \, \text{m} \]
06

Calculate the length for 100V potential difference

Repeat the same steps for a potential difference of 100V: \[ L = \frac{(100 \, \text{V})^2 (2.60 \times 10^{-6} \, \text{m}^2)}{5000 \, \text{W} (5.00 \times 10^{-7} \, \text{惟鈰卪})} \]
07

Solution for 100V

Calculate the following: \[ V^2 = (100)^2 = 10000 \, \text{V}^2 \] \[ P \rho = 5000 \, \text{W} \times 5.00 \times 10^{-7} \, \text{惟鈰卪} = 2.5 \times 10^{-3} \, \text{W鈰呂┾媴m} \] \[ L = \frac{10000 \times 2.60 \times 10^{-6}}{2.5 \times 10^{-3}} = 10.4 \, \text{m} \]

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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 fundamental in understanding electric circuits. It states that the current (\text{I}) flowing through a conductor between two points is directly proportional to the voltage (\text{V}) across the two points and inversely proportional to the resistance (\text{R}). This relationship can be expressed through the formula:
\( V = IR \).
This means that for a given voltage, the current is determined by the resistance of the wire. Higher resistance means less current flow, and vice versa. In the context of the heating element, Ohm's Law helps in understanding how the potential difference (voltage) influences the power dissipation in the Nichrome wire.
Resistivity
Resistivity is a property of a specific material that quantifies how strongly it resists electric current. It is represented by the symbol \( \rho \). The resistivity of a material depends on its composition and structure, and is given in ohm-meters (\(\Omega \cdot \text{m}\)).
For a wire, resistance (\text{R}) can be determined using the formula:
\( R = \frac{\rho L}{A} \)
where \(L\) is the length of the wire and \(A\) is its cross-sectional area. In our exercise, we use the resistivity of Nichrome, \(5.00 \times 10^{-7} \Omega \cdot \text{m}\). By understanding resistivity, we can see how different materials and dimensions influence the overall resistance and thus the current and power dissipation in the wire.
Electrical Power
Electrical power refers to the rate at which electrical energy is transferred by an electric circuit. It can be calculated from the voltage and the current in the circuit using the formula:
\( P = IV \).
However, for resistive elements, it can also be derived from the voltage and resistance:
\( P = \frac{V^2}{R} \).
In our problem, we calculated the length of the Nichrome wire by combining the power dissipation formula with resistivity and geometric properties to get:
\( L = \frac{V^2 A}{P \rho} \).
This formula relates all the electrical and physical parameters to find the length needed to achieve a specific power dissipation. In the context of heating elements, this ensures that we can design wires that heat to the desired temperature given specific voltage conditions.

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Most popular questions from this chapter

\begin{array}{l} \text { 49. Increases Over Time The charge passing through a conductor }\\\ \text { increases over time as } q(t)=\left(1.5 \mathrm{C} / \mathrm{s}^{3}\right) t^{3}-\left(4.5 \mathrm{C} / \mathrm{s}^{2}\right) t^{2}+(2 \mathrm{C} / \mathrm{s}) t, \end{array} where \(t\) is in seconds. (a) What equation describes the current in the circuit as a function of time? (b) What is the current in the conductor at \(t=0.0 \mathrm{~s}\) and at \(t=1.0 \mathrm{~s}\) ?

12\. A Certain Wire 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?

34\. Near Earth Near the Earth, the density of protons in the solar wind (a stream of particles from the Sun) is \(8.70 \mathrm{~cm}^{-3}\), and their speed is \(470 \mathrm{~km} / \mathrm{s}\). (a) Find the current density of these protons. (b) If the Earth's magnetic field did not deflect them, the protons would strike the planet. What total current would the Earth then receive?

27\. Linear Accelerator A linear accelerator produces a pulsed beam of electrons. The pulse current is \(0.50 \mathrm{~A}\), and each pulse has a duration of \(0.10 \mu \mathrm{s}\) (a) How many electrons are accelerated per pulse? (b) What is the average current for an accelerator operating at 500 pulses/s? (c) If the electrons are accelerated to an energy of \(50 \mathrm{MeV}\), what are the average and peak powers of the accelerator?

26\. \(100 \mathrm{~W}\) Lightbulb A \(100 \mathrm{~W}\) lightbulb is plugged into a standard \(120 \mathrm{~V}\) outlet. (a) How much does it cost per month to leave the light turned on continuously? Assume electric energy costs \(12 \mathrm{~d} / \mathrm{kW} \cdot \mathrm{h} .\) (b) What is the resistance of the bulb? (c) What is the current in the bulb? (d) Is the resistance different when the bulb is turned off?
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