Physics for Scientists and Engineers, Vol. 1
6th Edition
ISBN: 9781429201322
Author: Paul A. Tipler, Gene Mosca
Publisher: Macmillan Higher Education
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 25, Problem 114P
(a)
To determine
The support to assert that a leaky capacitor can be modeled as a capacitor that has an infinite resistance in parallel with a resistor.
(b)
To determine
The proof that time constant for discharging the capacitor is given by
(c)
To determine
The time it takes for the charge of mica-filled capacitor to decrease to
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Find the effective capacitance of a 5 uf capacitor and a 12 uf capacitor connected in (a) series and (b) parallel.
A parallel-plate vacuum capacitor is connected to a battery and charged until the stored electric energy is U. The battery is removed, and then a dielectric material with dielectric constant K is inserted into the capacitor, filling the space between the plates. Finally, the capacitor is fully discharged through a resistor (which is connected across the capacitor terminals).
Find Ur, the the energy dissipated in the resistor.
Consider the same situation as in the previous part, except that the charging battery remains connected while the dielectric is inserted.(Figure 2) The battery is then disconnected and the capacitor is discharged. For this situation, what is Ur, the energy dissipated in the resistor?
The potential difference across a charged capacitor is 16 V. The capacitor discharges through a fixed resistor. After a time equal to the time constant, the potential difference has reduced to V. The magnitude of V is:
Chapter 25 Solutions
Physics for Scientists and Engineers, Vol. 1
Ch. 25 - Prob. 1PCh. 25 - Prob. 2PCh. 25 - Prob. 3PCh. 25 - Prob. 4PCh. 25 - Prob. 5PCh. 25 - Prob. 6PCh. 25 - Prob. 7PCh. 25 - Prob. 8PCh. 25 - Prob. 9PCh. 25 - Prob. 10P
Ch. 25 - Prob. 11PCh. 25 - Prob. 12PCh. 25 - Prob. 13PCh. 25 - Prob. 14PCh. 25 - Prob. 15PCh. 25 - Prob. 16PCh. 25 - Prob. 17PCh. 25 - Prob. 18PCh. 25 - Prob. 19PCh. 25 - Prob. 20PCh. 25 - Prob. 21PCh. 25 - Prob. 22PCh. 25 - Prob. 23PCh. 25 - Prob. 24PCh. 25 - Prob. 25PCh. 25 - Prob. 26PCh. 25 - Prob. 27PCh. 25 - Prob. 28PCh. 25 - Prob. 29PCh. 25 - Prob. 30PCh. 25 - Prob. 31PCh. 25 - Prob. 32PCh. 25 - Prob. 33PCh. 25 - Prob. 34PCh. 25 - Prob. 35PCh. 25 - Prob. 36PCh. 25 - Prob. 37PCh. 25 - Prob. 38PCh. 25 - Prob. 39PCh. 25 - Prob. 40PCh. 25 - Prob. 41PCh. 25 - Prob. 42PCh. 25 - Prob. 43PCh. 25 - Prob. 44PCh. 25 - Prob. 45PCh. 25 - Prob. 46PCh. 25 - Prob. 47PCh. 25 - Prob. 48PCh. 25 - Prob. 49PCh. 25 - Prob. 50PCh. 25 - Prob. 51PCh. 25 - Prob. 52PCh. 25 - Prob. 53PCh. 25 - Prob. 54PCh. 25 - Prob. 55PCh. 25 - Prob. 56PCh. 25 - Prob. 57PCh. 25 - Prob. 58PCh. 25 - Prob. 59PCh. 25 - Prob. 60PCh. 25 - Prob. 61PCh. 25 - Prob. 62PCh. 25 - Prob. 63PCh. 25 - Prob. 64PCh. 25 - Prob. 65PCh. 25 - Prob. 66PCh. 25 - Prob. 67PCh. 25 - Prob. 68PCh. 25 - Prob. 69PCh. 25 - Prob. 70PCh. 25 - Prob. 71PCh. 25 - Prob. 72PCh. 25 - Prob. 73PCh. 25 - Prob. 74PCh. 25 - Prob. 75PCh. 25 - Prob. 76PCh. 25 - Prob. 77PCh. 25 - Prob. 78PCh. 25 - Prob. 79PCh. 25 - Prob. 80PCh. 25 - Prob. 81PCh. 25 - Prob. 82PCh. 25 - Prob. 83PCh. 25 - Prob. 84PCh. 25 - Prob. 85PCh. 25 - Prob. 86PCh. 25 - Prob. 87PCh. 25 - Prob. 88PCh. 25 - Prob. 89PCh. 25 - Prob. 90PCh. 25 - Prob. 91PCh. 25 - Prob. 92PCh. 25 - Prob. 93PCh. 25 - Prob. 94PCh. 25 - Prob. 95PCh. 25 - Prob. 96PCh. 25 - Prob. 97PCh. 25 - Prob. 98PCh. 25 - Prob. 99PCh. 25 - Prob. 100PCh. 25 - Prob. 101PCh. 25 - Prob. 102PCh. 25 - Prob. 103PCh. 25 - Prob. 104PCh. 25 - Prob. 105PCh. 25 - Prob. 106PCh. 25 - Prob. 107PCh. 25 - Prob. 108PCh. 25 - Prob. 109PCh. 25 - Prob. 110PCh. 25 - Prob. 111PCh. 25 - Prob. 112PCh. 25 - Prob. 113PCh. 25 - Prob. 114PCh. 25 - Prob. 115PCh. 25 - Prob. 116PCh. 25 - Prob. 117PCh. 25 - Prob. 118PCh. 25 - Prob. 119PCh. 25 - Prob. 120P
Knowledge Booster
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
- A block of weakly-conducting material with conductivity sigma has two pieces of metalembedded inside, which form a capacitor with capacitance C.(a) Find the resistance R between the two pieces of metal.(b) At t = 0, the two pieces of metal have a potential difference V0, and at future times t not equal to 0the potential is V (t) = V0e-t/TFind Tarrow_forwardA frictionless device changes the gap width, and hence the capacitance, of an air-filled parallel-plate capacitor. A switch allows a 375V voltage source to be connected to and disconnected from the capacitor. With the initial gap width, the capacitance is 821.3pF. The switch is closed allowing the capacitor to become fully charged. The switch is then opened, and the gap width is changed until the capacitance is 210.2pF. a. What is the magnitude of the work, in millijoules, that is required to change the gap width from the initial value to the final value? b. What is the ratio of the final gap width to the initial gap width? c. With the final gap width, what is the potential difference, in volts, across the capacitor?arrow_forwardA capacitor with a capacitance of 3.5 uF is initially uncharged. It is connected in series with a switch of negligible resistance, a resistor with a resistance of 10.5 kOhm, and a battery that has a potential difference of 105V. (a) Immediately after the switch is closed, what is the voltage drop VC, in volts, across the capacitor? (b) Immediately after the switch is closed, what is the voltage drop VR, in volts, across the resistor? (c) Immediately after the switch is closed, what is the current, in amperes, through the resistor? (d) Find an expression for the time after the switch is closed when the current in the resistor equals half its maximum value. (e) What is the charge Q, in microcoulombs, on the capacitor when the current in the resistor equals one half its maximum value.arrow_forward
- A capacitor of capacitance C = 4.5 uF is initially uncharged. It is connected in series with a switch of negligible resistance, a resistor of resistance R = 8.5 kOhm, and a battery which provides a potential difference of VB = 105V. (a) Calculate the time constant t for the circuit in seconds. (b) After a very long time after the switch has been closed, what is the voltage drop VC across the capacitor in terms of VB? (c) Calculate the charge Q on the capacitor a very long time after the switch has been closed in C. (d) Calculate the current I a very long time after the switch has been closed in A. (e) Calculate the time t after which the current through the resistor in one-third of its maximum value in s. (f) Calculate the cahrge Q on the capacitor when the current in the resistor equals one third its maximum value in C.arrow_forwardThe capacitive network shown in the figure is assembled with initially uncharged capacitors. A potential difference, Vab = +100V, is applied across the network. The switch S in the network is initially open but is then closed. Assume that all the capacitances shown are accurate to two significant figures. What is the equivalent capacitance between ab a) With the switch S open? b) With the switch S closed? A) (a) 14 micro-Farad; (b) 17 micro-Farad B) (a) 11 micro-Farad; (b) 17 micro-Farad C) (a) 8 micro-Farad; (b) 71 micro-Farad D) (a) 99 micro-Farad; (b) 67 micro-Farad E) (a) 4 micro-Farad; (b) 7 micro-Faradarrow_forwarda load of 300kW,with power factor 0.65 lagging,has the power factor improved to 0.90lagging by parallel capacitors.How many kvar must these capacitors furnish,and what is the resulting percent reduction in apparent power?arrow_forward
- A capacitor with initial charge q0 is discharged through a resistor. What multiple of the time constant τ gives the time the capacitor takes to lose (a) the first 1/5-th of its charge and (b)4/5-th of its charge?arrow_forwardPlease asaparrow_forwardA capacitor of capacitance C = 7.5 μF is initially uncharged. It is connected in series with a switch of negligible resistance, a resistor of resistance R = 11.5 kΩ, and a battery which provides a potential difference of VB = 110 V. (a) Calculate the time constant τ for the circuit in seconds. (b) After a very long time after the switch has been closed, what is the voltage drop VC across the capacitor in terms of VB? (c) Calculate the charge Q on the capacitor a very long time after the switch has been closed in C.arrow_forward
- A 13.0 kΩ resistor and a capacitor are connected in series and then a 20.0 V potential difference is suddenly applied across them. The potential difference across the capacitor rises to 5.0 V in 4.8 μs. Find the capacitance of the capacitor. (Your result must be in multiples of 10-10 Farads and include 2 digit after the decimal point. That means if, for example, you get a result of a 9.22x10-10 just type 9.22 in the answer box. Maximum of 5% of error is accepted in your answer. )arrow_forwardcan you please ans (a), (b) & (c)?arrow_forwardConsider an initially uncharged capacitor in an RC circuit. The resistance is 20,000 Ohms and the capacitance is 500 x10• F. If the source potential being used to charge the capacitor is 15 V, how long after charging begins will the power delivered to the resistor be 25 % of its maximum value?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
College PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning
University Physics (14th Edition)PhysicsISBN:9780133969290Author:Hugh D. Young, Roger A. FreedmanPublisher:PEARSON
Introduction To Quantum MechanicsPhysicsISBN:9781107189638Author:Griffiths, David J., Schroeter, Darrell F.Publisher:Cambridge University Press
Physics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
Lecture- Tutorials for Introductory AstronomyPhysicsISBN:9780321820464Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina BrissendenPublisher:Addison-Wesley
College Physics: A Strategic Approach (4th Editio...PhysicsISBN:9780134609034Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart FieldPublisher:PEARSON
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
University Physics (14th Edition)
Physics
ISBN:9780133969290
Author:Hugh D. Young, Roger A. Freedman
Publisher:PEARSON
Introduction To Quantum Mechanics
Physics
ISBN:9781107189638
Author:Griffiths, David J., Schroeter, Darrell F.
Publisher:Cambridge University Press
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:9780321820464
Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:Addison-Wesley
College Physics: A Strategic Approach (4th Editio...
Physics
ISBN:9780134609034
Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:PEARSON
Ohm's law Explained; Author: ALL ABOUT ELECTRONICS;https://www.youtube.com/watch?v=PV8CMZZKrB4;License: Standard YouTube License, CC-BY