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Mobile App Chapter 25: Problem 68
What is (a) the conductance and (b) the radius of a \(3.5-\mathrm{m}\) -long iron heating element for a \(110-\mathrm{V}, 1500-\mathrm{W}\) heater?
Short Answer
Expert verified
Answer: The conductance of the heating element is approximately 0.124 S, and the radius is about 2.30 x 10^-4 m.
Step by step solution
01
1. Calculate the current (I)
We will use the relation between power (P), voltage (V), and current (I), which is \(P=VI\). Rearrange the formula to find the current: \(I = \frac{P}{V}\). Given \(P=1500W\) and \(V=110V\), plug in the values and calculate: \(I = \frac{1500}{110} \approx 13.64A\).
02
2. Calculate the resistance (R)
We will use Ohm's law, \(V = IR\), to find the resistance (R). Rearrange the formula to find the resistance: \(R = \frac{V}{I}\). Use the values \(V = 110V\) and \(I = 13.64A\) and calculate: \(R = \frac{110}{13.64} \approx 8.07\,\Omega\).
03
3. Calculate the conductance (G)
Conductance (G) is the reciprocal of resistance (R), so: \(G = \frac{1}{R}\). Use the value \(R = 8.07\,\Omega\) and calculate: \(G = \frac{1}{8.07} \approx 0.124\,\mathrm{S}\).
04
4. Calculate the cross-sectional area (A)
The resistance (R) of a cylindrical wire is given by the formula: \(R = \frac{ρL}{A}\), where ρ is the resistivity, L is the length, and A is the cross-sectional area. We will use the resistivity of iron, which is \(ρ = 9.71\times10^{-8}\,\Omega m\). Rearrange the formula to find the cross-sectional area: \(A = \frac{ρL}{R}\). Use the values \(ρ = 9.71\times10^{-8}\,\Omega m\), \(L= 3.5m\), and \(R=8.07\,\Omega\) and calculate: \(A = \frac{9.71\times10^{-8}\, (\SI{3.5}{\meter})}{8.07} \approx 1.67\times10^{-7}\,\mathrm{m}^2\).
05
5. Calculate the radius (r)
The cross-sectional area (A) of a cylindrical wire is given by the formula: \(A = \pi r^2\), where r is the radius. Rearrange the formula to find the radius: \(r = \sqrt{\frac{A}{\pi}}\). Use the value \(A = 1.67\times10^{-7}\,\mathrm{m}^2\) and calculate: \(r = \sqrt{\frac{1.67\times10^{-7}}{\pi}} \approx 2.30\times10^{-4}\,\SI{}{m}\).
The conductance of the heating element is \(0.124\,\mathrm{S}\) and its radius is approximately \(2.30\times10^{-4}\,\SI{}{m}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Ohm's Law
Understanding Ohm's Law is essential when diving into the world of electrical circuits. Put simply, it relates the voltage across a resistor with the current flowing through it and its resistance. The law is usually expressed in the equation:
\[ V = IR \]
where \( V \) is the voltage in volts, \( I \) is the current in amperes (amps), and \( R \) is the resistance in ohms (\( \Omega \)). Ohm's Law is the starting point for many calculations related to electric circuits. For instance, if we know the voltage supplied to a heater and the power it uses, we can reverse-engineer the current using Ohm's Law as part of an equation rearrangement process. Likewise, if the current and voltage are known, the resistance can be found, as seen in our exemplary problem.
\[ V = IR \]
where \( V \) is the voltage in volts, \( I \) is the current in amperes (amps), and \( R \) is the resistance in ohms (\( \Omega \)). Ohm's Law is the starting point for many calculations related to electric circuits. For instance, if we know the voltage supplied to a heater and the power it uses, we can reverse-engineer the current using Ohm's Law as part of an equation rearrangement process. Likewise, if the current and voltage are known, the resistance can be found, as seen in our exemplary problem.
Electric Power Formula
The electric power formula is another key piece of the puzzle when working with electrical systems. It links the electrical power in watts (W) to the current flowing through a device and the voltage across it, as per the equation:
\[ P = VI \]
Here \( P \) stands for the electric power, \( V \) for the voltage, and \( I \) for the current. You can visualize electric power as the rate at which electrical energy is converted into another form of energy, such as heat in your heater. In our example, knowing the heater's power output and operating voltage allows us to calculate the flowing current. This becomes especially useful when conducting energy consumption analyses or designing electrical components to handle specific power levels.
\[ P = VI \]
Here \( P \) stands for the electric power, \( V \) for the voltage, and \( I \) for the current. You can visualize electric power as the rate at which electrical energy is converted into another form of energy, such as heat in your heater. In our example, knowing the heater's power output and operating voltage allows us to calculate the flowing current. This becomes especially useful when conducting energy consumption analyses or designing electrical components to handle specific power levels.
Resistivity and Resistance
Resistivity and resistance are critical concepts that help define how easily electricity can flow through a material. Resistance (\( R \)) measures how much a specific object opposes the flow of electric current and is measured in ohms (\( \Omega \)). Higher resistance means less current flow for a given voltage, according to Ohm's Law.
Resistivity (\( \rho \)), on the other hand, is an intrinsic property of materials that quantifies how strongly a material opposes the flow of electric current. It is expressed in ohm-meters (\( \Omega\cdot m \)). Each material has a unique resistivity value, which doesn't change regardless of the shape and size of the material. The resistance of a wire or other object is directly proportional to its resistivity and length but inversely proportional to its cross-sectional area as shown by the formula:
\[ R = \frac{\rho L}{A} \]
Where \( L \) is the length and \( A \) is the cross-sectional area of the object. When calculating dimensions of heating elements like in our problem, understanding the difference between both terms is crucial.
Resistivity (\( \rho \)), on the other hand, is an intrinsic property of materials that quantifies how strongly a material opposes the flow of electric current. It is expressed in ohm-meters (\( \Omega\cdot m \)). Each material has a unique resistivity value, which doesn't change regardless of the shape and size of the material. The resistance of a wire or other object is directly proportional to its resistivity and length but inversely proportional to its cross-sectional area as shown by the formula:
\[ R = \frac{\rho L}{A} \]
Where \( L \) is the length and \( A \) is the cross-sectional area of the object. When calculating dimensions of heating elements like in our problem, understanding the difference between both terms is crucial.
Cross-Sectional Area of Wire
The cross-sectional area (\( A \)) of a wire is a measure of the size of the face through which the current flows, perpendicular to the flow direction. The formula for finding the resistance directly ties it to the cross-sectional area, showing that thicker wires have less resistance to current flow.
Calculating the cross-sectional area is fundamental when working with properties of wires and cables. For cylindrical wires, the area is determined using the equation:
\[ A = \pi r^2 \]
where \( r \) is the radius of the wire. A larger cross-sectional area allows more electrons to flow with less resistance, facilitating higher current capacity. This concept is applied when determining the necessary wire thickness for electrical systems, ensuring they are safe and functional as demonstrated by the problem we're analyzing.
Calculating the cross-sectional area is fundamental when working with properties of wires and cables. For cylindrical wires, the area is determined using the equation:
\[ A = \pi r^2 \]
where \( r \) is the radius of the wire. A larger cross-sectional area allows more electrons to flow with less resistance, facilitating higher current capacity. This concept is applied when determining the necessary wire thickness for electrical systems, ensuring they are safe and functional as demonstrated by the problem we're analyzing.
Conductance and Resistance Relationship
Finally, let's explore the relationship between conductance (\( G \)) and resistance (\( R \)). Conductance is the measure of how easily electric current can flow through a conductor. The higher the conductance, the easier it is for the current to pass through. The relation is straightforward:
\[ G = \frac{1}{R} \]
This tells us that conductance is the reciprocal of resistance and is measured in siemens (\( S \)). A low resistance conductor has high conductance and vice versa. This relationship is important when designing and selecting materials for electrical components to ensure they have the appropriate conductivity for the application. In our exemplary problem, we calculated the resistance first and then found the conductance, demonstrating the direct relationship between these two properties.
\[ G = \frac{1}{R} \]
This tells us that conductance is the reciprocal of resistance and is measured in siemens (\( S \)). A low resistance conductor has high conductance and vice versa. This relationship is important when designing and selecting materials for electrical components to ensure they have the appropriate conductivity for the application. In our exemplary problem, we calculated the resistance first and then found the conductance, demonstrating the direct relationship between these two properties.
