/*! 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 86 Two resistors \(R_{1}\) and \(R_... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 27: Problem 86

Two resistors \(R_{1}\) and \(R_{2}\) may be connected either in series or in parallel across an ideal battery with emf \(\mathscr{E}\). We desire the rate of energy dissipation of the parallel combination to be five times that of the series combination. If \(R_{1}=100 \Omega\), what are the (a) smaller and (b) larger of the two values of \(R_{2}\) that result in that dissipation rate?

Short Answer

Expert verified
The smaller value of \( R_2 \) is approximately 52.36 \( \Omega \), and the larger value is approximately 247.64 \( \Omega \).

Step by step solution

01

Understanding Energy Dissipation

The rate of energy dissipation in a circuit is given by the power formula, where power \( P \) can be expressed as \( P = \frac{V^2}{R} \) for a resistor with resistance \( R \) connected to a voltage \( V \). When resistors are in series, total resistance \( R_{series} = R_1 + R_2 \). When resistors are in parallel, total resistance \( R_{parallel} = \frac{R_1 R_2}{R_1 + R_2} \).
02

Formula for Power in Series and Parallel

For resistors in series, the power \( P_s \) dissipated is \( P_s = \frac{\mathscr{E}^2}{R_1 + R_2} \). For resistors in parallel, the power \( P_p \) dissipated is \( P_p = \frac{\mathscr{E}^2}{\frac{R_1 R_2}{R_1 + R_2}} \). The problem states that \( P_p = 5P_s \).
03

Setting up the Ratio Equation

Substitute the power expressions into the given condition \( P_p = 5P_s \): \[ \frac{\mathscr{E}^2}{\frac{R_1 R_2}{R_1 + R_2}} = 5 \cdot \frac{\mathscr{E}^2}{R_1 + R_2} \]. This simplifies (by canceling \( \mathscr{E}^2 \)) to: \[ \frac{R_1 + R_2}{R_1 R_2} = 5 \cdot \frac{1}{R_1 + R_2} \].
04

Simplifying the Equation

Multiply both sides by \( (R_1R_2)(R_1 + R_2) \) to eliminate the fractions: \[ (R_1 + R_2)^2 = 5R_1R_2 \]. Substitute \( R_1 = 100 \) into the equation: \[ (100 + R_2)^2 = 5 \cdot 100 \cdot R_2 \] which simplifies to: \[ 10000 + 200R_2 + R_2^2 = 500R_2 \].
05

Solving the Quadratic Equation

Rearrange terms to form a quadratic equation: \[ R_2^2 - 300R_2 + 10000 = 0 \]. Using the quadratic formula \( R_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), set \( a = 1 \), \( b = -300 \), and \( c = 10000 \). Substitute these into the formula: \[ R_2 = \frac{300 \pm \sqrt{300^2 - 4 \cdot 1 \cdot 10000}}{2} \].
06

Calculating the Roots of the Quadratic

Calculate the discriminant: \( 300^2 - 4 \times 1 \times 10000 = 90000 - 40000 = 50000 \). Then, \( \sqrt{50000} = 100\sqrt{5} \). Substitute back to find \( R_2 \): \[ R_2 = \frac{300 \pm 100\sqrt{5}}{2} \].
07

Determining the Resistor Values

Calculate the two possible values for \( R_2 \):- Smaller value: \( R_2 = \frac{300 - 100\sqrt{5}}{2} \approx 52.36 \ \Omega \).- Larger value: \( R_2 = \frac{300 + 100\sqrt{5}}{2} \approx 247.64 \ \Omega \).

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 91影视!

Key Concepts

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

Series and Parallel Circuits
Understanding how resistors behave in series and parallel circuits is key in solving problems related to electrical circuits. In a series circuit, resistors are arranged one after another along the same path. This means that the total resistance in a series circuit is simply the sum of all resistances:
  • Series Total Resistance: \( R_{series} = R_1 + R_2 + ... + R_n \)
Because the current flows through each resistor sequentially, any increase in resistance is additive. This results in a voltage drop across each resistor.
On the other hand, in a parallel circuit, resistors are connected alongside each other in different paths. This configuration allows multiple current paths, effectively reducing the overall resistance within the circuit. The formula for total resistance in a parallel circuit is:
  • Parallel Total Resistance: \( \frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n} \)
A clear understanding of these concepts allows for the correct calculation of resistance, which is essential in exploring energy dissipation and power in circuits.
Power Dissipation Formula
Energy dissipation in electrical circuits is often discussed in terms of power, which is the rate of energy loss over time. The power dissipation formula is fundamental in circuitry, given as:
  • Power \( P = \frac{V^2}{R} \)
  • Alternatively, \( P = I^2R \) when current is known.
  • Or \( P = VI \) when voltage and current are given.
This equation shows that power is inversely related to resistance when the voltage is constant across a circuit. Therefore, higher resistance will lead to lower power dissipation if all other factors remain constant
.It's important to remember that in problem-solving scenarios, like with this exercise, knowing how to manipulate and compare power in series versus parallel can provide insights into how resistors will behave when configurations change. Additionally, using the relationship between power and energy loss helps in designing circuits for specific tasks.
Quadratic Equation in Physics
Quadratic equations often arise in physics problems involving resistors because they provide a way to solve for unknowns such as resistance and other parameters when power conditions are specified. The standard form of a quadratic equation is:
  • \( ax^2 + bx + c = 0 \)
In this exercise, the step-by-step solution involves solving a quadratic equation that comes from equating power dissipations in series and parallel configurations to a specified ratio. Solving such equations requires using the quadratic formula:
  • \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
This formula helps find the roots, or solutions, of the equation, which in turn provides the resistor values. In the context of the exercise, solving the quadratic equation yields the values of \( R_2 \) needed to achieve the desired power ratio. Thus, familiarity with quadratics is not just academic but practical when optimizing circuit designs.

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

Suppose that, while you are sitting in a chair, charge separation between your clothing and the chair puts you at a potential of \(200 \mathrm{~V},\) with the capacitance between you and the chair at \(150 \mathrm{pF}\). When you stand up, the increased separation between your body and the chair decreases the capacitance to \(10 \mathrm{pF}\). (a) What then is the potential of your body? That potential is reduced over time, as the charge on you drains through your body and shoes (you are a capacitor discharging through a resistance). Assume that the resistance along that route is \(300 \mathrm{G} \Omega\). If you touch an electrical component while your potential is greater than \(100 \mathrm{~V},\) you could ruin the component. (b) How long must you wait until your potential reaches the safe level of \(100 \mathrm{~V} ?\) If you wear a conducting wrist strap that is connected to ground, your potential does not increase as much when you stand up; you also discharge more rapidly because the resistance through the grounding connection is much less than through your body and shoes. (c) Suppose that when you stand up, your potential is \(1400 \mathrm{~V}\) and the chair-to-you capacitance is \(10 \mathrm{pF}\). What resistance in that wrist-strap grounding connection will allow you to discharge to \(100 \mathrm{~V}\) in \(0.30 \mathrm{~s},\) which is less time than you would need to reach for, say, your computer?

The potential difference between the plates of a leaky (meaning that charge leaks from one plate to the other) \(2.0 \mu \mathrm{F}\) capacitor drops to one-fourth its initial value in \(2.0 \mathrm{~s}\). What is the equivalent resistance between the capacitor plates?

A \(5.0 \mathrm{~A}\) current is set up in a circuit for \(6.0 \mathrm{~min}\) by a rechargeable battery with a \(6.0 \mathrm{~V}\) emf. By how much is the chemical energy of the battery reduced?

What multiple of the time constant \(\tau\) gives the time taken by an initially uncharged capacitor in an \(R C\) series circuit to be charged to \(99.0 \%\) of its final charge?

A capacitor with initial charge \(q_{0}\) is discharged through a resistor. What multiple of the time constant \(\tau\) gives the time the capacitor takes to lose (a) the first one-third of its charge and (b) two-thirds of its charge?
See all solutions

Recommended explanations on Physics Textbooks

Quantum Physics

Read Explanation

Modern Physics

Read Explanation

Thermodynamics

Read Explanation

Force

Read Explanation

Solid State Physics

Read Explanation

Wave Optics

Read Explanation
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.