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Chapter 25: Problem 29

A strand of wire has resistance 5.60\(\mu \Omega\) . Find the net resistance of 120 such strands if they are (a) placed side by side to form a cable of the same length as a single strand, and (b) connected end to end to form a wire 120 times as long as a single strand.

Short Answer

Expert verified
(a) 0.0467 碌惟, (b) 672 碌惟.

Step by step solution

01

Understand the Given Problem

We are given that each strand of wire has a resistance of 5.60\(\mu \Omega\). We need to find the total resistance in two different configurations: (a) when the wires are placed side by side, and (b) when the wires are connected end to end.
02

Analyze Configuration (a) - Side by Side

When wires are placed side by side, they are in a parallel configuration. The formula for the resistance of wires in parallel is \( \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n} \). Given that each individual resistance \( R_i = 5.60\mu \Omega \) and there are 120 wires, all resistances are identical.
03

Calculate Resistance for Parallel Configuration

For 120 identical resistors in parallel, the total resistance \( R_{\text{total}} \) is given by:\[ \frac{1}{R_{\text{total}}} = 120 \times \frac{1}{5.60\mu \Omega} \]Simplifying:\[ R_{\text{total}} = \frac{5.60\mu \Omega}{120} \approx 0.0467 \mu \Omega \]
04

Analyze Configuration (b) - End to End

When wires are connected end to end, they are in a series configuration. The total resistance of resistors in series is simply the sum of the individual resistances. Thus, the total resistance \( R_{\text{total}} = R_1 + R_2 + \cdots + R_n \).
05

Calculate Resistance for Series Configuration

In series, the total resistance is:\[ R_{\text{total}} = 120 \times 5.60\mu \Omega = 672 \mu \Omega \]

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Key Concepts

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

Parallel Circuits
Parallel circuits are configurations where multiple pathways for current to flow are created by connecting electrical elements side by side. When you connect resistors, or in this case, strands of wire, in parallel, the total resistance decreases. This is because parallel paths split the total current flowing through the circuit. More paths mean less resistance to overcome for the current.

The formula for calculating total resistance in parallel is:
  • \( \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n} \)
Each term \( \frac{1}{R_i} \) refers to the reciprocal of the individual resistance. For 120 identical wires in parallel, it simplifies to:
  • \( R_{\text{total}} = \frac{5.60 \mu \Omega}{120} \)
Placing the wires in parallel greatly reduces the total resistance compared to a single strand.
Series Circuits
In series circuits, all components are connected end to end, forming a single path for current flow. The total resistance of a series circuit is simply the sum of the resistances of each component. This results in a greater overall resistance compared to the individual resistors.

The formula for the total resistance in a series circuit is:
  • \( R_{\text{total}} = R_1 + R_2 + \cdots + R_n \)
For our exercise, connecting 120 wires end to end results in the total resistance being the sum of all individual resistances:
  • \( R_{\text{total}} = 120 \times 5.60 \mu \Omega = 672 \mu \Omega \)
This configuration significantly increases the total resistance, making the circuit less capable of allowing current to flow through easily.
Resistivity
Resistivity is a fundamental property of materials that affects how they resist electrical flow. It is often represented by the symbol \( \rho \) and is measured in ohm-meters (\( \Omega\cdot m \)).

Factors influencing resistivity include:
  • Material Type: Different materials offer different levels of resistance to current flow.
  • Temperature: Typically, as temperature increases, resistivity increases for most conductive materials.
The relationship between resistance \( R \), resistivity \( \rho \), length \( L \), and cross-sectional area \( A \) is expressed as:
  • \( R = \rho \frac{L}{A} \)
Understanding resistivity is crucial when designing electrical systems, as it helps predict how certain materials will behave when placed in a circuit. Knowing this, engineers can select the most suitable materials for constructing wires and components, ensuring efficiency and safety.

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Most popular questions from this chapter

A tightly coiled spring having 75 coils, each 3.50 \(\mathrm{cm}\) in diameter, is made of insulated metal wire 3.25 \(\mathrm{mm}\) in diameter. An ohmmeter connected across its opposite ends reads 1.74\(\Omega\) . What is the resistivity of the metal?

A \(5.00-\mathrm{A}\) current runs through a 12 -gauge copper wire (diameter 2.05 \(\mathrm{mm} )\) and through a light bulb. Copper has \(8.5 \times 10^{28}\) free electrons per cubic meter. (a) How many electrons pass through the light bulb each second? (b) What is the current density in the wire? (c) At what speed does a typical electron pass by any given point in the wire? (d) If you were to use wire of twice the diameter, which of the above answers would change? Would they increase or decrease?

Copper has \(8.5 \times 10^{28}\) free electrons per cubic meter. A 71.0 - \(\mathrm{cm}\) length of 12 -gauge copper wire that is 2.05 \(\mathrm{mm}\) in diameter carries 4.85 \(\mathrm{A}\) of current. (a) How much time does it take for an electron to travel the length of the wire? (b) Repeat part (a) for 6-gauge copper wire (diameter 4.12 \(\mathrm{mm}\) ) of the same length that carries the same current. (c) Generally speaking, how does changing the diameter of a wire that carries a given amount of current affect the drift velocity of the electrons in the wire?

The current in a wire varies with time according to the relationship \(I=55 \mathrm{A}-\left(0.65 \mathrm{A} / \mathrm{s}^{2}\right) t^{2}\) . (a) How many coulombs of charge pass a cross section of the wire in the time interval between \(t=0\) and \(t=8.0 \mathrm{s} ?(\mathrm{b})\) What constant current would transport the same charge in the same time interval?

A lightning bolt strikes one end of a steel lightning rod, producing a \(15,000-\mathrm{A}\) current burst that lasts for 65\(\mu\) s. The rod is 20 \(\mathrm{m}\) long and 1.8 \(\mathrm{cm}\) in diameter, and its other end is connected to the ground by 35 \(\mathrm{m}\) of \(8.0-\mathrm{mm}\) -diameter copper wire (a) Find the potential difference between the top of the steel rod and the lower end of the copper wire during the current burst. (b) Find the total energy deposited in the rod and wire by the current burst.
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