/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 A strand of wire has resistance ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 µΩ, (b) 672 µΩ.

Step by step solution

01

Understanding the Problem

We are given the resistance of a single strand of wire as 5.60 \(\mu \Omega\). We are asked to find the net resistance of 120 such strands in two different configurations: (a) placed side by side, (b) connected end to end. Each configuration affects how the total resistance is calculated.
02

Parallel Configuration (Side by Side)

When the strands are placed side by side, they form a parallel configuration. In a parallel circuit, the total or equivalent resistance \(R_{\text{eq}}\) is given by \(\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots + \frac{1}{R_n}\). For 120 identical strands, the formula becomes \(\frac{1}{R_{\text{eq}}} = 120 \times \frac{1}{5.60}\). Solving this yields \(R_{\text{eq}} = \frac{5.60}{120} \mu \Omega\).
03

Calculating Parallel Resistance

Substitute the given value into the formula: \(R_{\text{eq}} = \frac{5.60}{120} \mu \Omega\). This simplifies to \(R_{\text{eq}} = 0.0467 \mu \Omega\). So, the net resistance for the strands placed side by side is 0.0467 \(\mu \Omega\).
04

Series Configuration (End to End)

In a series configuration, resistances simply add up. This is expressed as \(R_{\text{eq}} = R_1 + R_2 + \ldots + R_n\). For 120 identical strands, \(R_{\text{eq}} = 120 \times 5.60 \mu \Omega\).
05

Calculating Series Resistance

Calculate the total resistance for the series configuration: \(R_{\text{eq}} = 120 \times 5.60 \mu \Omega = 672 \mu \Omega\). So, the net resistance when connecting the strands end to end is 672 \(\mu \Omega\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Parallel circuits
Parallel circuits are arrangements where multiple paths exist for electricity to flow. Each path, or branch, carries current separately. One of the key benefits of parallel circuits is that they maintain the same voltage across each component, while the current may differ. When it comes to resistors in parallel, the goal is often to reduce the overall resistance. The overall or equivalent resistance, denoted as \(R_{\text{eq}}\), can be found using the formula: \[ \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots + \frac{1}{R_n} \]This formula means that adding more conductive paths reduces the total resistance, which allows more current to flow. In the case of 120 identical strands of wire placed parallel, the equivalent resistance becomes very small because the inverse of a large sum is small. In the given problem, each strand has a resistance of 5.60 \(\mu \Omega\). Placing 120 such strands side by side results in a total or equivalent resistance of 0.0467 \(\mu \Omega\). This reflects how electricity follows the path of least resistance, distributing the load among the various pathways.
Series circuits
Series circuits organize components one after another along a single path. In this setup, the entire current flows sequentially through each component without branching off. As a result, each component in a series circuit experiences the same current.In a series configuration, the total resistance is simply the sum of all individual resistances because the current has only one path to travel. The formula for the total equivalent resistance \(R_{\text{eq}}\) is: \[ R_{\text{eq}} = R_1 + R_2 + \ldots + R_n \]Adding resistors in series increases the total resistance, which reduces the overall current flowing through the circuit. For example, if 120 wire strands, each with a resistance of 5.60 \(\mu \Omega\), are connected end to end, the total resistance becomes quite large, exactly 672 \(\mu \Omega\). This is because the resistances add up in a linear fashion, making electricity take a longer path, which inherently has more resistance.
Electrical resistance
Electrical resistance is a measure of how much a material opposes the flow of electric current. Resistance is influenced by several factors:
  • Material: Different materials have different resistivities, affecting how easily current flows through them.
  • Length: Resistance is directly proportional to the length of the conductor; longer conductors provide more resistance.
  • Cross-sectional area: A larger cross-section allows more current to pass through, reducing resistance.
  • Temperature: Resistance typically increases with temperature due to increased vibration of atoms obstructing the flow of electrons.
This concept is crucial for designing circuits, as the resistance determines the amount of current that can pass through a given component. Ohm's law, \(V = IR\), describes the relationship between voltage \(V\), current \(I\), and resistance \(R\) in a simple form. In practical applications, understanding resistance helps in assessing how components in a circuit impact the overall current flow and voltage distribution; whether they are set up in series or parallel, each configuration impacts the total resistance differently.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 1.50-m cylinder of radius 1.10 cm is made of a complicated mixture of materials. Its resistivity depends on the distance \(x\) from the left end and obeys the formula \(\rho (x) = a + bx^2\), where a and b are constants. At the left end, the resistivity is 2.25 \(\times\) 10\(^{-8} \Omega\) \(\cdot\) m, while at the right end it is 8.50 \(\times\) 10\(^{-8}\Omega\) \(\cdot \) m. (a) What is the resistance of this rod? (b) What is the electric field at its midpoint if it carries a 1.75-A current? (c) If we cut the rod into two 75.0-cm halves, what is the resistance of each half?

A copper transmission cable 100 km long and 10.0 cm in diameter carries a current of 125 A. (a) What is the potential drop across the cable? (b) How much electrical energy is dissipated as thermal energy every hour?

Unlike the idealized ammeter described in Section 25.4, any real ammeter has a nonzero resistance. (a) An ammeter with resistance \(R_A\) is connected in series with a resistor \(R\) and a battery of emf \(\varepsilon\) and internal resistance r. The current measured by the ammeter is \(I_A\). Find the current through the circuit if the ammeter is removed so that the battery and the resistor form a complete circuit. Express your answer in terms of \(I_A\), \(r\), \(R_A\), and \(R\). The more "ideal" the ammeter, the smaller the difference between this current and the current IA. (b) If \(R\) = 3.80 \(\Omega\), \(\varepsilon\) = 7.50 V, and \(r\) = 0.45 \(\Omega\), find the maximum value of the ammeter resistance \(R_A\) so that \(I_A\) is within 1.0% of the current in the circuit when the ammeter is absent. (c) Explain why your answer in part (b) represents a \(maximum\) value.

A heart defibrillator is used to enable the heart to start beating if it has stopped. This is done by passing a large current of 12 A through the body at 25 V for a very short time, usually about 3.0 ms. (a) What power does the defibrillator deliver to the body, and (b) how much energy is transferred?

The battery for a certain cell phone is rated at 3.70 V. According to the manufacturer it can produce \(3.15 \times 10^4\) J of electrical energy, enough for 5.25 h of operation, before needing to be recharged. Find the average current that this cell phone draws when turned on.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.