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A ductile metal wire has resistance \(R\). What will be the resistance of this wire in terms of \(R\) if it is stretched to three times its original length, assuming that the density and resistivity of the material do not change when the wire is stretched? (Hint: The amount of metal does not change, so stretching out the wire will affect its cross-sectional area.)

Short Answer

Expert verified
The resistance of the wire when stretched to three times its original length, assuming the density and resistivity remain constant, is \(9 \times R\).

Step by step solution

01

Recognize the formula for resistance

The resistance \(R\) of a wire can be calculated using the formula \(R=\rho * \frac{l}{A}\) where: \( \rho\) is the resistivity of the material, \( l \) is the length and \( A \) is the cross-sectional area of the wire.
02

Understand the change in wire dimension

When the wire is stretched to three times its original length, its volume remains constant. That is, the original length (\(l\)) times the original cross-sectional area (\(A\)) is equal to the new length (3\(l\)) times the new cross-sectional area (\(A'\)) . This gives us: \( l*A = 3*l*A'\). Solving this for \(A'\) we find that \(A' = \frac{A}{3}\).
03

Calculate the New Resistance

The new resistance \(R'\) is obtained by substituting for \(l\) and \(A'\) in the formula obtained in step 1, giving \(R' = \rho * \frac{3l}{\frac{A}{3}} = R * 9\).

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Key Concepts

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

Resistivity
Resistivity is an intrinsic property of a material that measures its opposition to the flow of electric current. It is denoted by the symbol \(\rho\). Imagine resistivity as the inherent friction an electric charge feels as it travels through a wire.
This property is not affected by the size or shape of the material itself. It depends solely on the type of material from which the wire is made. Metals like copper and aluminum have low resistivity, which makes them excellent conductors.
  • Resistivity is measured in ohm-meters (\(\Omega \cdot m\)).
  • It is affected by temperature, usually increasing as the temperature rises.
In our exercise, when the wire is stretched, the resistivity remains unchanged, which simplifies calculations.
Cross-sectional area
The cross-sectional area of a wire represents the size of its face when cut perpendicular to its length. Think of it as the thickness of the wire. This area plays a crucial role because it determines how much space is available for electrical current to pass through.
In the provided exercise, when the ductile wire is stretched to three times its original length, the cross-sectional area is significantly reduced. This is because the volume of the wire remains constant — stretching increases length while decreasing the cross-sectional area.Understanding this, we find that if the length is tripled, then the new cross-sectional area \(A'\) is reduced to a third of the original (\(A' = \frac{A}{3}\)). Consequently, the smaller the cross-sectional area, the higher the resistance.
Ductile metal wire
A ductile metal wire is one that can be easily stretched or drawn into thin threads without breaking. Metals like copper, aluminum, and gold are examples that exhibit high ductility, making them perfect for wiring and cables.
  • Ductility involves the ability to withstand tensile stress.
  • It is a property that differs from resistivity, focusing on the material's physical flexibility instead.
In the context of the exercise, the fact that the wire can be stretched to three times its original length means it must be quite ductile. When stretched, the physical dimensions change, but the amount of material remains constant. This affects the geometry but not the material's inherent properties, such as resistivity.
Electrical resistance calculation
Electrical resistance is a measure of how much a conductor opposes the flow of electric current. The formula for resistance \(R\) is \(R = \rho \cdot \frac{l}{A}\), where \(\rho\) is the resistivity, \(l\) is the length of the conductor, and \(A\) is the cross-sectional area.
To calculate the new resistance when a wire is stretched, we must adjust the values of \(l\) and \(A\) accordingly. In our exercise:
  • Original length is \(l\); new length is \(3l\).
  • Original area is \(A\); new area is \(\frac{A}{3}\).
Substituting these into the resistance formula gives us the new resistance \(R'\), which is \(R' = \rho \cdot \frac{3l}{\frac{A}{3}} = 9R\). Hence, stretching the wire increases its resistance ninefold.

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Most popular questions from this chapter

A typical small flashlight contains two batteries, each having an emf of \(1.5 \mathrm{~V},\) connected in series with a bulb having resistance \(17 \Omega .\) (a) If the internal resistance of the batteries is negligible, what power is delivered to the bulb? (b) If the batteries last for \(5.0 \mathrm{~h}\), what is the total energy delivered to the bulb? (c) The resistance of real batteries increases as they run down. If the initial internal resistance is negligible, what is the combined internal resistance of both batteries when the power to the bulb has decreased to half its initial value? (Assume that the resistance of the bulb is constant. Actually, it will change somewhat when the current through the filament changes, because this changes the temperature of the filament and hence the resistivity of the filament wire.

A typical cost for electrical power is 0.120 dollar per kilowatthour. (a) Some people leave their porch light on all the time. What is the yearly cost to keep a \(75 \mathrm{~W}\) bulb burning day and night? (b) Suppose your refrigerator uses \(400 \mathrm{~W}\) of power when it's running, and it runs 8 hours a day. What is the yearly cost of operating your refrigerator?

On your first day at work as an electrical technician, you are asked to determine the resistance per meter of a long piece of wire. The company you work for is poorly equipped. You find a battery, a voltmeter, and an ammeter, but no meter for directly measuring resistance (an ohmmeter). You put the leads from the voltmeter across the terminals of the battery, and the meter reads \(12.6 \mathrm{~V}\). You cut off a \(20.0 \mathrm{~m}\) length of wire and connect it to the battery, with an ammeter in series with it to measure the current in the wire. The ammeter reads 7.00 A. You then cut off a \(40.0 \mathrm{~m}\) length of wire and connect it to the battery, again with the ammeter in series to measure the current. The ammeter reads 4.20 A. Even though the equipment you have available to you is limited, your boss assure you of its high quality: The ammeter has very small resistance, and the voltmeter has very large resistance. What is the resistance of 1 meter of wire?

(a) Estimate the maximum volume of water the hot-water heater in your home can hold. (b) How much heat would be required to raise the temperature of that water from \(20^{\circ} \mathrm{C}\) to a standard household hot-water temperature of \(45^{\circ} \mathrm{C}\) ? (Water has a density of \(1.00 \mathrm{~kg} / \mathrm{L}\) and a heat capacity of \(4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K} .\) ) (c) Suppose the water should be fully heated in \(1.5 \mathrm{~h}\). To what power output does this correspond? (d) If the element has a potential difference of \(220 \mathrm{~V},\) what current is required? (e) What should be the resistance of the element?

A battery has emf \(\mathcal{E}\) and internal resistance \(r=2.00 \Omega\). A \(12.0 \Omega\) resistor is connected to the battery, and the resistor consumes electrical power at a rate of \(96.0 \mathrm{~J} / \mathrm{s}\). What is the emf of the battery?

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