/*! 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 15 What multiple of the time consta... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 27: Problem 15

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 \(89.0 \%\) of its final charge?

Short Answer

Expert verified
The time is approximately 2.207 times the time constant \( \tau \).

Step by step solution

01

Understand the Charging Equation

A capacitor charging in an RC circuit is governed by the equation: \[ q(t) = Q_f (1 - e^{-t/\tau}) \] where \( q(t) \) is the charge at time \( t \), \( Q_f \) is the final charge, and \( \tau = RC \) is the time constant of the circuit.
02

Set Up the Equation for 89% Charge

We want to find the time \( t \) when the charge \( q(t) \) is 89% of the final charge \( Q_f \). Thus, we have the equation:\[ 0.89Q_f = Q_f(1 - e^{-t/\tau}) \].
03

Simplify and Solve the Equation

Cancel \( Q_f \) from both sides of the equation to get:\[ 0.89 = 1 - e^{-t/\tau} \].Rearrange to find:\[ e^{-t/\tau} = 0.11 \].
04

Solve for \( t \) in Terms of \( \tau \)

Take the natural logarithm of both sides to solve for \( t \):\[ -\frac{t}{\tau} = \ln(0.11) \].Thus, \( t = -\tau \ln(0.11) \).
05

Calculate the Multiple of \( \tau \)

Calculate \( -\ln(0.11) \) which is approximately 2.207. Therefore, the time \( t \) is:\[ t \approx 2.207\tau \].

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.

Capacitor Charging Equation
The capacitor charging equation is a fundamental concept in understanding how capacitors charge in an RC circuit. It's given by the formula: \[ q(t) = Q_f (1 - e^{-t/\tau}) \] In this equation:
  • \( q(t) \) is the charge on the capacitor at time \( t \).
  • \( Q_f \) represents the final charge that the capacitor will eventually reach.
  • \( \tau \) is the time constant, which is calculated as the product of resistance \( R \) and capacitance \( C \) in the circuit \( \tau = RC \).
  • \( e \) represents the base of the natural logarithm, approximately equal to 2.718.
The equation highlights how the charge on a capacitor changes over time as it approaches \( Q_f \). The term \( (1 - e^{-t/\tau}) \) determines how the charging progresses, starting from zero and moving towards one, causing \( q(t) \) to increase as time progresses.
Final Charge Percentage
Calculating the final charge percentage involves determining how close a capacitor is to its maximum charge after a given time period. In many cases, we use this concept to find specific time points, such as when the capacitor is 89% charged. To find when a capacitor has reached a certain percentage of its final charge \( Q_f \), set up an equation like this:\[ 0.89Q_f = Q_f (1 - e^{-t/\tau}) \]By simplifying, you determine the moment when the charge \( q(t) \) on the capacitor equals 89% of \( Q_f \).
  • Canceling \( Q_f \) from both sides, leaves \( 0.89 = 1 - e^{-t/\tau} \).
  • Solving this will help you find the required \( t \), which represents the time at that charge percentage relative to \( \tau \).
Natural Logarithm Equation
The natural logarithm equation comes into play when solving for specific times like when the capacitor is charged to a designated percentage of its final charge. The equation becomes useful in this context: Given that:\[ e^{-t/\tau} = 0.11 \]Taking the natural logarithm on both sides provides:\[ -\frac{t}{\tau} = \ln(0.11) \]This equation helps find \( t \) in relation to \( \tau \). Natural logarithms, denoted \( \ln \), are logarithms with base \( e \).
  • This operation is straightforward because the property of logarithms simplifies the exponentiation, solving \( t \) precisely.
  • The result influences how many time constants, \( \tau \), it takes for the capacitor to reach 89% of its final charge.
Exponential Decay in Circuits
Exponential decay is a crucial concept in circuits and is central to understanding how capacitors charge over time. Even though a capacitor never completely reaches its final charge instantly, it gets exponentially nearer over periods defined by its time constant, \( \tau \). The essence of exponential decay in this context derives from the equation:\[ e^{-t/\tau} \]
  • This term depicts how the difference from the final charge diminishes exponentially.
  • As \( t \) increases, \( e^{-t/\tau} \) becomes very small, meaning the charge \( q(t) \) is very close to \( Q_f \).
  • This phenomenon allows engineers to predict and design how quickly a capacitor will effectively charge in a circuit, optimizing for efficiency and functionality.
Whether the exponential term describes discharge or charging, it captures the essence of exponential behaviors making it indispensable in circuit analysis.

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

Circuit section \(A B\) absorbs energy at a rate of \(50 \mathrm{~W}\) when current \(i=2.0 \mathrm{~A}\) through it is in the indicated direction. Resistance \(R=2.0 \Omega\). (a) What is the potential difference between \(A\) and \(B\) ? Emf device \(X\) lacks internal resistance. (b) What is its emf? (c) Is point \(B\) connected to the positive terminal of \(X\) or to the negative terminal?

(a) In electron-volts, how much work does an ideal battery with a \(20.0 \mathrm{~V}\) emf do on an electron that passes through the battery from the positive to the negative terminal? (b) If \(5.17 \times 10^{18}\) electrons pass through each second, what is the power of the battery in watts?

The ideal batteries have emfs \(\&_{1}=200 \mathrm{~V}\) and \(8_{2}=50 \mathrm{~V}\) and the resistances are \(R_{1}=3.0 \Omega\) and \(R_{2}=2.0 \Omega\). If the potential at \(P\) is \(100 \mathrm{~V}\), what is it at \(Q\) ?

When the lights of a car are switched on, an ammeter in series with them reads \(10.0 \mathrm{~A}\) and a voltmeter connected across them reads \(12.0 \mathrm{~V}\) (Fig. 27-34). When the electric starting motor is turned on, the ammeter reading drops to \(8.50 \mathrm{~A}\) and the lights dim somewhat. If the internal resistance of the battery is \(0.0500 \Omega\) and that of the ammeter is negligible, what are (a) the emf of the battery and (b) the current through the starting motor when the lights are on?

In an \(R C\) series circuit, emf \(\&=12.0 \mathrm{~V}\), resistance \(R=1.40 \mathrm{M} \Omega\), and capacitance \(C=2.70 \mu \mathrm{F}\). (a) Calculate the time constant. (b) Find the maximum charge that will appear on the capacitor during charging. (c) How long does it take for the charge to build up to \(16.0 \mu \mathrm{C} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Work Energy and Power

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Scientific Method Physics

Read Explanation

Measurements

Read Explanation

Waves Physics

Read Explanation

Rotational Dynamics

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.