/*! 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 59 What multiple of the time consta... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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?

Short Answer

Expert verified
The time taken is approximately 4.605 times the time constant \(\tau\).

Step by step solution

01

Understand the Charging Formula

The charge on a capacitor as a function of time in an R C circuit is given by the formula: \[ q(t) = Q_f \left( 1 - e^{-t/RC} \right) \] where \(Q_f\) is the final charge, \(R\) is the resistance, and \(C\) is the capacitance. The product \(RC\) is the time constant \(\tau\). Our objective is to find the time \(t\) when the capacitor reaches \(99.0\%\) of its final charge.
02

Set Up the Equation for 99% Charge

We know that \(q(t) = 0.99 Q_f\). Substituting this into the charging formula, we get:\[ 0.99 Q_f = Q_f \left( 1 - e^{-t/\tau} \right) \]Dividing both sides by \(Q_f\), we have:\[ 0.99 = 1 - e^{-t/\tau} \]
03

Solve for Exponential Term

Rearrange the equation:\[ e^{-t/\tau} = 1 - 0.99 = 0.01 \]
04

Solve for t/Ï„ using Natural Logarithms

Take the natural logarithm of both sides to solve for \(t/\tau\):\[ -\frac{t}{\tau} = \ln(0.01) \]Hence,\[ \frac{t}{\tau} = -\ln(0.01) \]
05

Calculate t/Ï„

Calculate the value of \(-\ln(0.01)\). Since \(\ln(0.01) = -4.605\), we have:\[ \frac{t}{\tau} = 4.605 \] This means that the time taken for the capacitor to charge to \(99.0\%\) of its final charge is approximately \(4.605\) times the time constant \(\tau\).

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 91Ó°ÊÓ!

Key Concepts

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

Capacitor Charging
Capacitor charging occurs in RC circuits when an electric current flows into the capacitor, causing it to store electrical energy. As the capacitor charges, the voltage across its terminals rises until it reaches its maximum, or final, charge. The process of charging a capacitor can be understood through the charging formula:\[ q(t) = Q_f \left( 1 - e^{-t/RC} \right) \]where:
  • \(q(t)\) is the charge at time \(t\),
  • \(Q_f\) is the final charge the capacitor will reach,
  • \(R\) and \(C\) are the resistance and capacitance of the circuit respectively,
  • \(e\) is the base of the natural logarithm.
In this formula, the term \(RC\) represents the time constant \(\tau\), which is a key factor that dictates how quickly the capacitor will charge. The higher the resistance \(R\) or the capacitance \(C\), the longer it will take to charge the capacitor. This is why the time constant \(\tau = RC\) is so significant; it helps predict how fast or slow the capacitor will reach near its full charge. Understanding this helps you manage how electrical devices are powered and their efficiency in circuits.
Exponential Decay
Exponential decay describes how the charge on a capacitor changes over time as it approaches its final value. In the equation for a charging capacitor:\[ q(t) = Q_f \left( 1 - e^{-t/RC} \right) \]The exponential term \(e^{-t/RC}\) is responsible for the decay aspect. It shows how the difference between the current charge \(q(t)\) and the final charge \(Q_f\) decreases exponentially over time. Key points to understand about exponential decay:
  • The charge doesn't increase linearly; instead, it starts rapidly and slows down as it approaches the maximum charge.
  • When \(t = \tau\), the charge is about 63% of \(Q_f\), which signifies one time constant's worth of charging. Each subsequent \(\tau\) fills more of the remaining gap towards \(Q_f\).
  • This decay pattern ensures that after five time constants \(5\tau\), the capacitor is charged to more than 99% of its full capacity.
Through exponential decay, engineers and scientists can understand how different components in circuits slow down the charging process and can make informed adjustments to optimize performance.
Natural Logarithms
Natural logarithms, denoted as \(\ln\), are used to simplify equations involving exponential decay, like those found in capacitor charging scenarios. Like in our solution, natural logarithms help to isolate variables and solve for unknowns.In the exercise provided, we needed to determine the multiple of the time constant \(\tau\) that corresponds to the capacitor reaching 99% of its final charge. Here’s how natural logarithms played a role:
  • We started with the equation \(0.99 = 1 - e^{-t/\tau}\), representing 99% of the final charge.
  • Rearranging it, we had: \(e^{-t/\tau} = 0.01\).
  • Applying the natural logarithm to both sides, we solved: \(-\frac{t}{\tau} = \ln(0.01)\).
  • Knowing \(\ln(0.01) = -4.605\), we concluded: \(\frac{t}{\tau} = 4.605\), meaning it takes approximately 4.605 times the time constant to reach 99% charge.
Natural logarithms provide the mathematical means to transform and solve complex exponential equations, making them approachable and helping students and engineers find precise answers quickly.

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

Two identical batteries of emf \(\mathscr{E}=12.0\) \(V\) and internal resistance \(r=0.200 \Omega\) are to be connected to an external resistance \(R\), either in parallel (Fig. \(27-50\) ) or in series (Fig. 27-51). If \(R=2.00 r\), what is the current \(i\) in the external resistance in the (a) parallel and (b) series arrangements? (c) For which arrangement is \(i\) greater? If \(R=r / 2.00\), what is \(i\) in the external resistance in the (d) parallel and (e) series arrangements? (f) For which arrangement is \(i\) greater now?

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

An initially uncharged capacitor \(C\) is fully charged by a device of constant emf \(\mathscr{E}\) connected in series with a resistor \(R\) (a) Show that the final energy stored in the capacitor is half the energy supplied by the emf device. (b) By direct integration of \(i^{2} R\) over the charging time, show that the thermal energy dissipated by the resistor is also half the energy supplied by the emf device.

A \(1.0 \mu \mathrm{F}\) capacitor with an initial stored energy of \(0.50 \mathrm{~J}\) is discharged through a \(1.0 \mathrm{M} \Omega\) resistor. (a) What is the initial charge on the capacitor? (b) What is the current through the resistor when the discharge starts? Find an expression that gives, as a function of time \(t,(\mathrm{c})\) the potential difference \(V_{C}\) across the capacitor, (d) the potential difference \(V_{R}\) across the resistor, and (e) the rate at which thermal energy is produced in the resistor.

(a) In electron-volts, how much work does an ideal battery with a \(12.0 \mathrm{~V}\) emf do on an electron that passes through the battery from the positive to the negative terminal? (b) If \(3.40 \times 10^{18}\) electrons pass through each second, what is the power of the battery in watts?

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.