/*! 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 84 A charging \(R C\) circuit contr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 18: Problem 84

A charging \(R C\) circuit controls the intermittent windshield wipers in a car. The emf is \(12.0 \mathrm{V}\). The wipers are triggered when the voltage across the 125 - \(\mu\) F capacitor reaches \(10.0 \mathrm{V} ;\) then the capacitor is quickly discharged (through a much smaller resistor) and the cycle repeats. What resistance should be used in the charging circuit if the wipers are to operate once every \(1.80 \mathrm{s} ?\)

Short Answer

Expert verified
Answer: The required resistance for the charging circuit is approximately \(1143.72 \, \Omega\).

Step by step solution

01

Identify the variables in the problem

The given values are: Voltage when the capacitor in the charging circuit is triggered (V_c): 10.0V Initial voltage across the capacitor (V_0): 12.0V Capacity of the capacitor (C): 125碌F Time it takes for the wiper to operate (t): 1.80s Resistance (R): Unknown
02

Re-write the formula in terms of R

We are trying to find R, so we need to rewrite the equation in terms of R. Since we have: \(V_c(t) = V_0(1 - e^{-t/(RC)})\) We can divide both sides by V_0 to isolate the exponential part: \(\frac{V_c}{V_0} = 1 - e^{-t/(RC)}\) We can now make it easier to solve for R: \(e^{-t/(RC)} = 1 - \frac{V_c}{V_0}\) \(-t/(RC) = \ln(1 - \frac{V_c}{V_0})\) \(R = -t / (C \ln(1 - \frac{V_c}{V_0}))\)
03

Plug in the values and solve for R

Now, we can plug in the values to find R: \(R = -1.80 / (125 \times 10^{-6} \ln(1 - \frac{10.0}{12.0}))\) Calculate the value of R: \(R \approx 1143.72 \, \Omega\) Therefore, a resistance of approximately \(1143.72 \, \Omega\) should be used in the charging circuit for the windshield wipers to operate once every 1.80 seconds.

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!

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 galvanometer is to be turned into a voltmeter that deflects full scale for a potential difference of \(100.0 \mathrm{V}\) What size resistor should be placed in series with the galvanometer if it has an internal resistance of $75 \Omega\( and deflects full scale for a current of \)2.0 \mathrm{mA} ?$
A 2.00 - \(\mu\) F capacitor is charged using a \(5.00-\mathrm{V}\) battery and a 3.00 - \(\mu\) F capacitor is charged using a \(10.0-\mathrm{V}\) battery. (a) What is the total energy stored in the two capacitors? (b) The batteries are disconnected and the two capacitors are connected together \((+\text { to }+\) and \- to \(-\) ). Find the charge on each capacitor and the total energy in the two capacitors after they are connected. (c) Explain what happened to the "missing" energy. [Hint: The wires that connect the two have some resistance.]

A common flashlight bulb is rated at \(0.300 \mathrm{A}\) and \(2.90 \mathrm{V}\) (the values of current and voltage under operating conditions). If the resistance of the bulb's tungsten filament at room temperature \(\left(20.0^{\circ} \mathrm{C}\right)\) is \(1.10 \Omega,\) estimate the temperature of the tungsten filament when the bulb is turned on.
An ammeter with a full scale deflection for \(I=10.0 \mathrm{A}\) has an internal resistance of \(24 \Omega .\) We need to use this ammeter to measure currents up to 12.0 A. The lab instructor advises that we get a resistor and use it to protect the ammeter. (a) What size resistor do we need and how should it be connected to the ammeter, in series or in parallel? (b) How do we interpret the ammeter readings?
A strip of doped silicon \(260 \mu \mathrm{m}\) wide contains $8.8 \times 10^{22}$ conduction electrons per cubic meter and an insignificant number of holes. When the strip carries a current of \(130 \mu \mathrm{A},\) the drift speed of the electrons is \(44 \mathrm{cm} / \mathrm{s} .\) What is the thickness of the strip?
See all solutions

Recommended explanations on Physics Textbooks

Circular Motion and Gravitation

Read Explanation

Translational Dynamics

Read Explanation

Particle Model of Matter

Read Explanation

Fundamentals of Physics

Read Explanation

Modern Physics

Read Explanation

Solid State 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.