Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)
Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)
9th Edition
ISBN: 9781305266292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
bartleby

Videos

Question
Book Icon
Chapter 28, Problem 43P

(a)

To determine

The potential difference across the capacitor.

(a)

Expert Solution
Check Mark

Answer to Problem 43P

The potential difference across the capacitor is 6.00V.

Explanation of Solution

Let the resistance across 1.00Ω be R1, across 4.00Ω be R2, across 8.00Ω be R3 and across 2.00Ω be R4.

When the circuit is connected with a voltage source, the capacitor starts to charge. So, it acts as an open circuit as shown in figure below.

Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses), Chapter 28, Problem 43P , additional homework tip  1

Figure (1)

Write the expression for the equivalent resistance when the resistors are connected in series.

    Req=R1+R2                                                                                                              (I)

Here, the equivalent resistance is Req and the resistors connected in series are R1 and R2.

Write the expression to current through the series connection..

    I1=VReq                                                                                                                    (II)

Here, I1 is the current flowing through the resistors in series and V is the potential difference of the battery.

Write the expression to determine the potential difference V1 in the circuit shown in figure (1)

    V1=V(I1R1)                                                                                                        (III)

Write the expression to determine the potential difference V2 in the circuit shown in figure (1)

 using the voltage division rule.

    V2=(R4R3+R4)V                                                                                                    (IV)

Write the expression to determine the potential difference across the capacitor.

    VC=V1V2                                                                                                              (V)

Here, VC is the expression to determine the potential difference across the capacitor.

Conclusion:

Substitute 1.00Ω for R1 and 4.00Ω for R2 in equation (I) to solve for Req.

    Req=1.00Ω+4.00Ω=5.00Ω

Substitute 5.00Ω for Req and 10.0V for V in equation (II) to solve for I1.

    I1=10.0V5.00Ω=2.00A

Substitute 1.00Ω for R1, 10V for V and 2.00A for I1 in equation (III) to solve for V1.

    V1=10V((2.00A)1.00Ω)=8V

Substitute 2.00Ω for R4, 10V for V and 8.00Ω for R3 in equation (IV) to solve for V2.

    V2=(2.00Ω8.00Ω+2.00Ω)(10V)=2V

Substitute 2V for V2 and 8V for V1 in equation (V) to solve for VC.

    VC=8V2V=6.00V

Therefore, the potential difference across the capacitor is 6.00V.

(b)

To determine

The time interval at which the capacitor discharges to one tenth of its initial voltage.

(b)

Expert Solution
Check Mark

Answer to Problem 43P

The time interval at which the capacitor discharges to one tenth of its initial voltage is 8.29μs.

Explanation of Solution

When the battery is disconnected, the capacitor starts discharging acting as a voltage source.

The circuit diagram is shown below after the battery is disconnected.

Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses), Chapter 28, Problem 43P , additional homework tip  2

Figure (2)

The resistors R1 and R3 are in series.

Write the expression for the equivalent resistance when the resistors are connected in series.

    Req=R1+R3                                                                                                       (VI)

Here, the equivalent resistance is Req.

Also, the resistors R2 and R4 are in series.

Write the expression for the equivalent resistance when the resistors are connected in series.

    Req=R2+R4                                                                                                    (VII)

Here, the equivalent resistance is Req.

Now the resistors Req and Req are in parallel.

Write the expression for the equivalent resistance when the resistors are connected in parallel.

    RP=Req+Req                                                                                                 (VIII)

Here, the equivalent resistance for the parallel connection is RP.

Write the expression to calculate the time interval at which the capacitor discharges to one tenth of its initial voltage.

    V(t)=V0e(tRPC)                                                                                                   (IX)

Here, V(t) is voltage across the capacitor after disconnection, V0 is the initial voltage and t is the time interval.

Substitute V010 for V(t) in equation (IX).

    V010=V0e(tRPC)1=10e(tRPC)ln1=ln10+((tRPC))(tRPC)=2.302                                                                                (X)

Conclusion:

Substitute 1.00Ω for R1 and 8.00Ω for R3 in equation (VI) to solve for Req.

    Req=1.00Ω+8.00Ω=9.00Ω

Substitute 2.00Ω for R1 and 4.00Ω for R3 in equation (VII) to solve for Req.

    Req=2.00Ω+4.00Ω=6.00Ω

Substitute 9.00Ω for Req and 6.00Ω for Req in equation (VIII) to solve for RP.

    1RP=19.00Ω+16.00ΩRP=(9.00Ω)(6.00Ω)9.00Ω+6.00Ω=3.60Ω

Substitute 3.60Ω for RP, and 1.00μF for C in equation (X) to solve for t.

    (t(3.60Ω)(1.00μF×106F1μF))=2.302t=8.29×106s(106μs1s)t=8.29μs

Therefore, the time interval at which the capacitor discharge to one tenth of its initial voltage is 8.29μs.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Three slits, each separated from its neighbor by d = 0.06 mm, are illuminated by a coherent light source of wavelength 550 nm. The slits are extremely narrow. A screen is located L = 2.5 m from the slits. The intensity on the centerline is 0.05 W. Consider a location on the screen x = 1.72 cm from the centerline. a) Draw the phasors, according to the phasor model for the addition of harmonic waves, appropriate for this location. b) From the phasor diagram, calculate the intensity of light at this location.
A Jamin interferometer is a device for measuring or for comparing the indices of refraction of gases. A beam of monochromatic light is split into two parts, each of which is directed along the axis of a separate cylindrical tube before being recombined into a single beam that is viewed through a telescope. Suppose we are given the following, • Length of each tube is L = 0.4 m. • λ= 598 nm. Both tubes are initially evacuated, and constructive interference is observed in the center of the field of view. As air is slowly let into one of the tubes, the central field of view changes dark and back to bright a total of 198 times. (a) What is the index of refraction for air? (b) If the fringes can be counted to ±0.25 fringe, where one fringe is equivalent to one complete cycle of intensity variation at the center of the field of view, to what accuracy can the index of refraction of air be determined by this experiment?
1. An arrangement of three charges is shown below where q₁ = 1.6 × 10-19 C, q2 = -1.6×10-19 C, and q3 3.2 x 10-19 C. 2 cm Y 93 92 91 X 3 cm (a) Calculate the magnitude and direction of the net force on q₁. (b) Sketch the direction of the forces on qi

Chapter 28 Solutions

Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)

Ch. 28 - Prob. 6OQCh. 28 - Prob. 7OQCh. 28 - Prob. 8OQCh. 28 - Prob. 9OQCh. 28 - Prob. 10OQCh. 28 - Prob. 11OQCh. 28 - Prob. 12OQCh. 28 - Prob. 13OQCh. 28 - Prob. 14OQCh. 28 - Prob. 15OQCh. 28 - Prob. 1CQCh. 28 - Prob. 2CQCh. 28 - Why is it possible for a bird to sit on a...Ch. 28 - Prob. 4CQCh. 28 - Prob. 5CQCh. 28 - Prob. 6CQCh. 28 - Prob. 7CQCh. 28 - Prob. 8CQCh. 28 - Is the direction of current in a battery always...Ch. 28 - Prob. 10CQCh. 28 - Prob. 1PCh. 28 - Two 1.50-V batterieswith their positive terminals...Ch. 28 - An automobile battery has an emf of 12.6 V and an...Ch. 28 - Prob. 4PCh. 28 - Prob. 5PCh. 28 - Prob. 6PCh. 28 - Prob. 7PCh. 28 - Prob. 8PCh. 28 - Prob. 9PCh. 28 - Prob. 10PCh. 28 - Prob. 11PCh. 28 - Prob. 12PCh. 28 - Prob. 13PCh. 28 - Prob. 14PCh. 28 - Prob. 15PCh. 28 - Prob. 16PCh. 28 - Prob. 17PCh. 28 - Prob. 18PCh. 28 - Prob. 19PCh. 28 - Why is the following situation impossible? A...Ch. 28 - Prob. 21PCh. 28 - Prob. 22PCh. 28 - Prob. 23PCh. 28 - Prob. 24PCh. 28 - Prob. 25PCh. 28 - The following equations describe an electric...Ch. 28 - Prob. 27PCh. 28 - Prob. 28PCh. 28 - Prob. 29PCh. 28 - Prob. 30PCh. 28 - Prob. 31PCh. 28 - Prob. 32PCh. 28 - Prob. 33PCh. 28 - Prob. 34PCh. 28 - Prob. 35PCh. 28 - Prob. 36PCh. 28 - An uncharged capacitor and a resistor are...Ch. 28 - Prob. 38PCh. 28 - Prob. 39PCh. 28 - A 10.0-F capacitor is charged by a 10.0-V battery...Ch. 28 - Prob. 41PCh. 28 - Prob. 42PCh. 28 - Prob. 43PCh. 28 - Prob. 44PCh. 28 - A charged capacitor is connected to a resistor and...Ch. 28 - Prob. 46PCh. 28 - Prob. 47PCh. 28 - Prob. 48PCh. 28 - Prob. 49APCh. 28 - Prob. 50APCh. 28 - Prob. 51APCh. 28 - Prob. 52APCh. 28 - Prob. 53APCh. 28 - Prob. 54APCh. 28 - Prob. 55APCh. 28 - Prob. 56APCh. 28 - Prob. 57APCh. 28 - Why is the following situation impossible? A...Ch. 28 - Prob. 59APCh. 28 - Prob. 60APCh. 28 - When two unknown resistors are connected in series...Ch. 28 - Prob. 62APCh. 28 - Prob. 63APCh. 28 - A power supply has an open-circuit voltage of 40.0...Ch. 28 - Prob. 65APCh. 28 - Prob. 66APCh. 28 - Prob. 67APCh. 28 - Prob. 68APCh. 28 - Prob. 69APCh. 28 - Prob. 70APCh. 28 - Prob. 71APCh. 28 - Prob. 72APCh. 28 - A regular tetrahedron is a pyramid with a...Ch. 28 - An ideal voltmeter connected across a certain...Ch. 28 - Prob. 75APCh. 28 - Prob. 76APCh. 28 - Prob. 77APCh. 28 - Prob. 78APCh. 28 - Prob. 79APCh. 28 - Prob. 80APCh. 28 - Prob. 81APCh. 28 - Prob. 82CPCh. 28 - Prob. 83CP
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers with Modern ...
Physics
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers, Technology ...
Physics
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
University Physics Volume 2
Physics
ISBN:9781938168161
Author:OpenStax
Publisher:OpenStax
DC Series circuits explained - The basics working principle; Author: The Engineering Mindset;https://www.youtube.com/watch?v=VV6tZ3Aqfuc;License: Standard YouTube License, CC-BY