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Chapter 19: Problem 76

A piece of wire has a resistance \(R\) . It is cut into three pieces of equal length, and the pieces are twisted together parallel to each other. What is the resistance of the resulting wire in terms of \(R ?\)

Short Answer

Expert verified
The resistance of the resulting wire is \( \frac{R}{9} \).

Step by step solution

01

Determine the Length of Each Piece

The wire is cut into three pieces of equal length. Therefore, each piece will have a length equal to one third of the original length of the wire. This will affect the resistance since resistance is directly proportional to the length of a conductor.
02

Calculate the Resistance of Each Piece

The resistance of a wire is proportional to its length. If the original wire has a resistance of \( R \), each piece will have a resistance of \( \frac{R}{3} \), as each piece is one-third of the total length.
03

Wire Them in Parallel

When resistors are twisted together in parallel, the total resistance \( R_{total} \) can be calculated using the formula for resistors in parallel: \[ \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \]Since all pieces have the same resistance \( \frac{R}{3} \):\[ \frac{1}{R_{total}} = \frac{1}{\frac{R}{3}} + \frac{1}{\frac{R}{3}} + \frac{1}{\frac{R}{3}} \]
04

Solve for Total Resistance

Substitute each term in the equation with their values:\[ \frac{1}{R_{total}} = \frac{3}{R} + \frac{3}{R} + \frac{3}{R} \]Simplifying gives:\[ \frac{1}{R_{total}} = \frac{9}{R} \]To find \( R_{total} \), take the reciprocal:\[ R_{total} = \frac{R}{9} \]

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Key Concepts

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

Parallel Resistance
In the world of electricity, understanding how resistors function in a circuit is fundamental. When dealing with parallel resistance, the resistors are arranged such that each one is connected across the same two points, forming junctions at every connection. This arrangement offers multiple paths for the current to travel, allowing it to divide accordingly.
  • Parallel wiring means the total resistance will always be less than the smallest individual resistance in the circuit.
  • The formula for calculating the total parallel resistance, when dealing with multiple resistors, is: \( \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots \)
Understanding this formula is key to solving many practical problems in electricity and electronics. Remember, in parallel circuits, increasing the number of resistors decreases the overall resistance, which increases the flow of current through the circuit.
Cutting Wires
Cutting wires plays an integral role in modifying their resistance. If you imagine a single, resistive wire and cut it into multiple pieces, the resistance of each piece changes proportionally to its new length. Resistance in a wire is directly proportional to its length, following the formula \( R = \rho \frac{L}{A} \) where \( R \) is resistance, \( \rho \) is resistivity, \( L \) is length, and \( A \) is cross-sectional area.
  • Reducing the length of a wire reduces its resistance proportionally.
  • If a wire with a total resistance \( R \) is cut into three equal lengths, each shorter piece will have a resistance of \( \frac{R}{3} \).
This knowledge sets the ground for further manipulations, such as combining these shorter wires again in different configurations, to explore the resulting changes in total resistance.
Resistors in Parallel
When resistors are connected in parallel, they're wired so that the same voltage is applied across each one. In our exercise, the three cut pieces of wire are effectively connected in such a manner. The current has multiple direct pathways, leading to a reduction in the total resistance.
  • The combined resistance of several resistors in parallel is always less than the resistance of the smallest individual resistor.
  • For identical resistors, like the wire pieces in the exercise, the total resistance simplifies further: \( R_{total} = \frac{R}{n} \) where \( n \) is the number of identical resistors.
Incorporating this concept allows for efficient circuit designs and for a thorough understanding of the exercise solution where three \( \frac{R}{3} \) resistors yield \( R_{total} = \frac{R}{9} \).
Physics Problem Solving
Physics problems, like the one involving cutting wires and calculating their resistance, require both a logical approach and a solid grasp of the fundamental concepts. Apply a systematic methodology to such problems.
  • First, break down the problem step-by-step, clarifying each component and how they relate to the core principles involved.
  • Second, utilize relevant formulas and equations effectively. Ensure to understand the why behind formulas like those for parallel resistance.
  • Third, recheck your calculations to confirm their logical consistency and accuracy with the physical concepts.
By mastering clear problem-solving techniques, handling these physics tasks becomes intuitive, allowing you to tackle more complex electrical problems confidently.

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Most popular questions from this chapter

In an ionic solution, a current consists of \(\mathrm{Ca}^{2+}\) ions (of charge \(+2 e )\) and \(\mathrm{Cl}^{-}\) ions (of charge \(-e )\) traveling in opposite directions. If \(5.11 \times 10^{18} \mathrm{Cl}^{-}\) ions go from \(A\) to \(B\) every 0.50 min, while \(3.24 \times 10^{18} \mathrm{Ca}^{2+}\) ions move from \(B\) to \(A\), what is the current (in mA) through this solution, and in which direction \((\) from \(A\) to \(B\) or from \(B\) to \(A)\) is it going?

Transmission of nerve impulses. Nerve cells transmit electric signals through their long tubular axons. These signals propagate due to a sudden rush of \(\mathrm{Na}^{+}\) ions, each with charge \(+e,\) into the axon. Measurements have revealed that typically about \(5.6 \times 10^{11} \mathrm{Na}^{+}\) ions enter each meter of the axon during a time of 10 \(\mathrm{ms}\) . What is the current during this inflow of charge in a meter of axon?

Current in the body. The resistance of the body varies from approximately 500 \(\mathrm{k} \Omega\) (when it is very dry) to about 1 \(\mathrm{k} \Omega\) (when it is wet). The maximum safe current is about 5.0 \(\mathrm{mA} .\) At 10 \(\mathrm{mA}\) or above, muscle contractions can occur that may be fatal. What is the largest potential difference that a person can safely touch if his body is wet? Is this result within the range of common household voltages?

Copper has \(8.5 \times 10^{28}\) electrons per cubic meter. (a) How many electrons are there in a 25.0 \(\mathrm{cm}\) length of 12 -gauge copper wire (diameter 2.05 \(\mathrm{mm} ) ?\) (b) If a current of 1.55 \(\mathrm{A}\) is flowing in the wire, what is the average drift speed of the electrons along the wire? (There are \(6.24 \times 10^{18}\) electrons in 1 coulomb of charge.)

A 6.00 \(\mathrm{V}\) lantern battery is connected to a 10.5\(\Omega\) lightbulb, and the resulting current in the circuit is 0.350 A. What is the internal resistance of the battery?
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