Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 26: Problem 44

In the movie Back to the Future, time travel is made possible by a flux capacitor, which generates 1.21 GW of power. Assuming that a 1.00 - F capacitor is charged to its maximum capacity with a \(12.0-\mathrm{V}\) car battery and is discharged through a resistor, what resistance is necessary to produce a peak power output of 1.21 GW in the resistor? How long would it take for a \(12.0-\mathrm{V}\) car battery to charge the capacitor to \(90.0 \%\) of its maximum capacity through this resistor?

Short Answer

Expert verified
Answer: The required resistance is approximately \(1.19 \times 10^{-7}\,\Omega\) to produce a peak power output of 1.21 GW. It would take approximately \(2.57 \times 10^{-7}\,\mathrm{s}\) (257 nanoseconds) to charge the capacitor to 90% of its maximum capacity.

Step by step solution

01

Find the Resistance

First, we will use the power formula to find the resistance (R) required to produce a peak power output of 1.21 GW (1.21 x 10^9 W). The power formula can be given as: \(P = \dfrac{V^2}{R}\) Where P is the power, V is the voltage, and R is the resistance. Now, we can find the resistance (R) by rearranging the formula: \(R = \dfrac{V^2}{P}\) Let's plug in the voltage and power values: \(R = \dfrac{(12.0\,\mathrm{V})^2}{1.21 \times 10^9\,\mathrm{W}}\)
02

Calculate the Resistance

Now, let's calculate the resistance: \( R = \dfrac{144\,\mathrm{V^2}}{1.21 \times 10^9\,\mathrm{W}} = 1.19 \times 10^{-7}\,\Omega\) Therefore, the required resistance is approximately \(1.19 \times 10^{-7}\,\Omega\) to produce a peak power output of 1.21 GW in the resistor.
03

Find the Time to Charge the Capacitor

Next, we need to find the time it takes for the capacitor to charge to 90% of its maximum capacity through this resistor. To do this, we can use the charging equation for a capacitor: \(t = -RC\ln(1-\dfrac{Q}{Q_{max}})\) Where \(t\) is the time, \(R\) is the resistance, \(C\) is the capacitance, \(Q\) is the charge at time \(t\), and \(Q_{max}\) is the maximum charge. In this case, we want the charge to be 90% of the maximum charge, so: \(\dfrac{Q}{Q_{max}} = 0.9\) We can now plug in the values already known: \(t = -(1.19\times 10^{-7}\,\Omega) (1.00\,\mathrm{F}) \ln(1-0.9)\)
04

Calculate the Time to Charge the Capacitor

Now, let's calculate the time it takes for the capacitor to charge to 90% of its maximum capacity: \(t \approx -(1.19\times 10^{-7}\,\Omega) (1.00\,\mathrm{F}) \ln(0.1)\) \(t \approx 2.57 \times 10^{-7}\,\mathrm{s}\) Therefore, it would take approximately \(2.57 \times 10^{-7}\,\mathrm{s}\) (257 nanoseconds) for a \(12.0\,\mathrm{V}\) car battery to charge the capacitor to 90% of its maximum capacity through this resistor.

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

Two capacitors in series are charged through a resistor. Identical capacitors are instead connected in parallel and charged through the same resistor. How do the times required to fully charge the two sets of capacitors compare?

Explain why the time constant for an \(\mathrm{RC}\) circuit increases with \(R\) and with \(C\). (The answer "That's what the formula says" is not sufficient.)

How can you light a \(1.0-\mathrm{W}, 1.5-\mathrm{V}\) bulb with your \(12.0-V\) car battery?

Two resistors, \(R_{1}\) and \(R_{2},\) are connected in series across a potential difference, \(\Delta V_{0}\). Express the potential drop across each resistor individually, in terms of these quantities. What is the significance of this arrangement?

Two parallel plate capacitors, \(C_{1}\) and \(C_{2},\) are con nected in series with a \(60.0-\mathrm{V}\) battery and a \(300 .-\mathrm{k} \Omega\) resistor, as shown in the figure. Both capacitors have plates with an area of \(2.00 \mathrm{~cm}^{2}\) and a separation of \(0.100 \mathrm{~mm}\). Capacitor \(C_{1}\) has air between its plates, and capacitor \(C_{2}\) has the gap filled with a certain porcelain (dielec-
See all solutions

Recommended explanations on Physics Textbooks

Waves Physics

Read Explanation

Fields in Physics

Read Explanation

Electromagnetism

Read Explanation

Physics of Motion

Read Explanation

Astrophysics

Read Explanation

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.

Sign-up for free