Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 92

(a) In a charging \(R C\) circuit, how many time constants have elapsed when the capacitor has \(99.0 \%\) of its final charge? (b) How many time constants have elapsed when the capacitor has \(99.90 \%\) of its final charge? (c) How many time constants have elapsed when the current has \(1.0 \%\) of its initial value?

Short Answer

Expert verified
Answer: (a) Approximately 4.605 time constants have elapsed when the capacitor has 99.0% of its final charge. (b) Approximately 6.908 time constants have elapsed when the capacitor has 99.90% of its final charge. (c) Approximately 4.605 time constants have elapsed when the current has 1.0% of its initial value.

Step by step solution

01

(Step 1: Charging an RC circuit formula)

Recall that the charge on a capacitor as it charges in an RC circuit is given by the formula: \(Q(t) = Q_{final}(1 - e^{-t/(\tau)})\) Where \(Q(t)\) is the charge at time \(t\), \(Q_{final}\) is the final charge, and \(\tau\) is the time constant, given by the product of the resistance R and capacitance C.
02

(Step 2: Solve for number of time constants for 99% charging)

In this case, we want to find how many time constants have elapsed when the capacitor has 99% of its final charge. First, let's represent the ratio of charges as a percentage: \(\frac{Q(t)}{Q_{final}} = 0.99\) Substitute the charge formula and solve for the ratio of elapsed time to the time constant, \(t/\tau\): \(0.99 = 1 - e^{-t/(\tau)}\) To find \(t/\tau\), we rearrange and take the natural logarithm: \(t/\tau = \ln(\frac{1}{1 - 0.99}) \approx 4.605\)
03

(Step 3: Solve for number of time constants for 99.90% charging)

We'll follow the same procedure as before, but now the ratio of charges is 99.90%: \(\frac{Q(t)}{Q_{final}} = 0.999\) Substitute the charge formula and solve for the ratio of elapsed time to the time constant: \(t/\tau = \ln(\frac{1}{1 - 0.999}) \approx 6.908\)
04

(Step 4: Solve for number of time constants for 1% of initial current)

In this case, we want to find how many time constants have elapsed when the current has 1% of its initial value. First, recall the formula for current in an RC circuit: \(I(t) = I_{initial}e^{-t/\tau}\) Now, let's represent the ratio of currents as a percentage: \(\frac{I(t)}{I_{initial}} = 0.01\) Substitute the current formula and solve for the ratio of elapsed time to the time constant: \(t/\tau = \ln(\frac{I_{initial}}{0.01I_{initial}}) \approx 4.605\)
05

(Answers)

(a) Approximately 4.605 time constants have elapsed when the capacitor has 99.0% of its final charge. (b) Approximately 6.908 time constants have elapsed when the capacitor has 99.90% of its final charge. (c) Approximately 4.605 time constants have elapsed when the current has 1.0% of its initial value.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

During a "brownout," which occurs when the power companies cannot keep up with high demand, the voltage of the household circuits drops below its normal $120 \mathrm{V} .\( (a) If the voltage drops to \)108 \mathrm{V},$ what would be the power consumed by a "100-W" light-bulb (that is, a light-bulb that consumes \(100.0 \mathrm{W}\) when connected to \(120 \mathrm{V}\) )? Ignore (for now) changes in the resistance of the light-bulb filament. (b) More realistically, the light-bulb filament will not be as hot as usual during the brownout. Does this make the power drop more or less than that you calculated in part (a)? Explain.
We can model some of the electrical properties of an unmyelinated axon as an electric cable covered with defective insulation so that current leaks out of the axon to the surrounding fluid. We assume the axon consists of a cylindrical membrane filled with conducting fluid. A current of ions can travel along the axon in this fluid and can also leak out through the membrane. The inner radius of the cylinder is \(5.0 \mu \mathrm{m} ;\) the membrane thickness is \(8.0 \mathrm{nm} .\) (a) If the resistivity of the axon fluid is \(2.0 \Omega \cdot \mathrm{m},\) calculate the resistance of a 1.0 -cm length of axon to current flow along its length. (b) If the resistivity of the porous membrane is \(2.5 \times 10^{7} \Omega\).m, calculate the resistance of the wall of a 1.0 -cm length of axon to current flow across the membrane. (c) Find the length of axon for which the two resistances are equal. This length is a rough measure of the distance a signal can travel without amplification.
A \(12-\Omega\) resistor has a potential difference of \(16 \mathrm{V}\) across it. What current flows through the resistor?
If a chandelier has a label stating \(120 \mathrm{V}, 5.0 \mathrm{A},\) can its power rating be determined? If so, what is it?
A battery has a terminal voltage of \(12.0 \mathrm{V}\) when no current flows. Its internal resistance is \(2.0 \Omega .\) If a \(1.0-\Omega\) resistor is connected across the battery terminals, what is the terminal voltage and what is the current through the \(1.0-\Omega\) resistor?
See all solutions

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Radiation

Read Explanation

Atoms and Radioactivity

Read Explanation

Astrophysics

Read Explanation

Kinematics Physics

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.

Sign-up for free