/*! 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 57 A capacitor is charged until it ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 25: Problem 57

A capacitor is charged until it holds \(5.0 \mathrm{J}\) of energy, then connected across a \(10-\mathrm{k} \Omega\) resistor. In \(8.6 \mathrm{ms}\), the resistor dissipates 2.0 J. Find the capacitance.

Short Answer

Expert verified
The capacitance of the capacitor is \( C = \frac{2 \times (E1 - E2)}{(V1^2 - V2^2)} \), where \( E1 = 5.0 \) J is the initial energy in the capacitor, \( E2 = 2.0 \) J is the energy dissipated by the resistor, \( V1 \) is the initial voltage on the capacitor, and \( V2 \) is the voltage dropped over the resistor calculated in Step 3. Substitute the known values to find the value of C.

Step by step solution

01

Calculate the initial voltage across the capacitor

In the first step, rearrange the formula for the energy stored in a capacitor to solve for the voltage. So, in this case, we have \( V = \sqrt{\frac{2E_c}{C}} \). But we do not yet know the capacitance C. So far, all we can say is that the voltage when the capacitor was fully charged was equal to the square root of two times its stored energy divided by its capacitance.
02

Determine the energy dissipated by the resistor

The energy dissipated by the resistor over a given time period can be calculated as \( E = P \times t \). However, we only have energy and time, not power. We also know that power, in terms of resistance and voltage, is given by \( P = \frac{V^2}{R} \). Thus, we can determine the voltage that had dissipated 2J of energy over 8.6ms through the \(10k \Omega\) resistor.
03

Find Voltage Dropped by the Resistor

Rearrange \( E = \frac{V^2t}{R} \) to solve for voltage. The derived equation is \( V = \sqrt{\frac{ER}{t}} \). Substitute E=2J (energy dissipated), R=10000 ohm (resistor), and t=0.0086s (time) to find the voltage.
04

Find the Capacitance

In this step, subtract the voltage dropped over the resistor from the initial voltage on the capacitor. This value is equal to the square root of two times the energy remaining in the capacitor divided by the capacitance (from step 1). Solve this equation for the capacitance C and that's our final answer.

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 capacitor used to provide steady voltages in the power supply of a stereo amplifier charges rapidly to \(35 \mathrm{V}\) every \(1 / 60\) second. It must then hold that voltage to within \(1.0 \mathrm{V}\) for the next \(1 / 60 \mathrm{s}\) while it discharges through the amplifier. If the amplifier draws 1.2 A from the \(35-\mathrm{V}\) supply, (a) what's its effective resistance, and (b) what capacitance is needed?

A \(6.0-\mathrm{V}\) battery has internal resistance \(2.5 \Omega .\) If the battery is short-circuited, what's the rate of energy dissipation in its internal resistance?

Work all circuit problems to two significant figures even if component values are given with one significant figure. Sketch a circuit diagram for a circuit that includes a resistor \(R_{1}\) connected to the positive terminal of a battery, a pair of parallel resistors \(R_{2}\) and \(R_{3}\) connected to the lower-voltage end of \(R_{1}\) and then returned to the battery's negative terminal, and a capacitor across \(R_{2}\).

A student who's confused about voltage and current tries to measure the voltage across a lighted light bulb by inserting a voltmeter in series with the bulb. What happens to the bulb? Explain.

Two identical resistors in series dissipate equal power. How can this be, when electric charge loses energy in flowing through the first resistor?
See all solutions

Recommended explanations on Physics Textbooks

Electrostatics

Read Explanation

Measurements

Read Explanation

Physical Quantities And Units

Read Explanation

Quantum Physics

Read Explanation

Work Energy and Power

Read Explanation

Electricity and Magnetism

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.