Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
bartleby

Videos

Question
Book Icon
Chapter 25, Problem 76P

(a)

To determine

The equivalent resistance between a and b .

(a)

Expert Solution
Check Mark

Answer to Problem 76P

The equivalent resistance of the circuit is 4.10Ω .

Explanation of Solution

Formula used:

The expression for equivalent resistance in upper wire is given by,

  Req,1=R6+(R2+R4)(R4)(R2+R4)+R4

The expression for equivalent resistance in lower wire for is given by,

  Req,2=R822R8+R4

The expression for net equivalent resistance is given by,

  Req=Req,1Req,2Req,1+Req,2

Calculation:

The equivalent resistance in upper wire is calculated as,

  Req,1=R6+( R 2 + R 4 )( R 4 )( R 2 + R 4 )+R4=6Ω+( 2Ω+4Ω)( 4Ω)( 2Ω+4Ω)+( 4Ω)=8.4Ω

The equivalent resistance in lower wire is calculated as,

  Req,2=R822R8+R4= ( 8Ω )22( 8Ω)+4Ω=4Ω+4Ω=8Ω

The net equivalent resistance of the circuit is calculated as,

  Req=R eq,1R eq,2R eq,1+R eq,2=( 8.4Ω)( 8Ω)( 8.4Ω)+( 8Ω)=4.10Ω

Conclusion:

Therefore, the equivalent resistance of the circuit is 4.10Ω .

(b)

To determine

The current in each resistor.

(b)

Expert Solution
Check Mark

Answer to Problem 76P

The currents through upper wire 6Ω resistor, lower wire 4Ω resistor, 4Ω resistor parallel with 2Ω and 4Ω resistors, upper wire 6Ω

  =(4Ω+2Ω) resistor and, lower wire in parallel 8Ω resistor are 1.43A , 1.50A , 0.86A , 0.57A and 0.75A respectively.

Explanation of Solution

Formula used:

The expression for current in upper wire 6Ω resistor is given by,

  I6=VReq,1

The expression for current in upper wire 4Ω resistor is given by,

  I4=V(I6)(R6)R4

The expression for current in upper wire 6Ω resistor is given by,

  I6,upper=V(I6)(R6)R6

The expression for current in lower wire is given by,

  Ilower=VReq,2

The expression for current in lower wire in parallel 8Ω resistor is given by,

  I8=V(I lower)(R4)R8

Calculation:

The current in upper wire 6Ω resistor is calculated as,

  I6=VR eq,1=12.0V8.4Ω=1.43A

The current in upper wire 4Ω resistor is calculated as,

  I4=V( I 6 )( R 6 )R4=12.0V( 1.43A)( 6Ω)4Ω=0.86A

The current in upper wire 6Ω resistor is calculated as,

  I6,upper=V( I 6 )( R 6 )R6=12.0V( 1.43A)( 6Ω)6Ω=0.57A

The current in lower wire is calculated as,

  Ilower=VR eq,2=12.0V8Ω=1.5A

The current in lower wire in parallel 8Ω resistor is calculated as,

  I8=V( I lower )( R 4 )R8=12.0V( 1.5A)( 4Ω)8Ω=0.75A

Conclusion:

Therefore, the currents through upper wire 6Ω resistor, lower wire 4Ω resistor, 4Ω resistor parallel with 2Ω and 4Ω resistors, upper wire 6Ω =(4Ω+2Ω) resistor and, lower wire in parallel 8Ω resistor are 1.43A , 1.50A , 0.86A , 0.57A and 0.75A respectively.

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
(a) What is the resistance (in (2) of thirty-three 285 Q resistors connected in series? Ω (b) What is the resistance (in 2) of thirty-three 285 Q resistors connected in parallel? Ω
QUESTION 6 a) What is the time constant for the circuit shown in the figure below if the value of ε = 12.0 V, R = 21.4 ohm, and C = 39.1 mF.? b) Suppose that the switch is closed and the capacitor starts to charge. How much of the charge will be accumulated on each plate of the capacitor after 3 s of charging? Submit the value of the charge (in mC, with tho decimal places) as your answer.
A capacitor of capacitance C = 2x 10^-6 is discharged through a resistor of resistance = 10^5 ohm. When will the energy stored in a capacitor reduce to one-third of its initial value

Chapter 25 Solutions

Physics for Scientists and Engineers

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
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
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
University Physics (14th Edition)
Physics
ISBN:9780133969290
Author:Hugh D. Young, Roger A. Freedman
Publisher:PEARSON
Text book image
Introduction To Quantum Mechanics
Physics
ISBN:9781107189638
Author:Griffiths, David J., Schroeter, Darrell F.
Publisher:Cambridge University Press
Text book image
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:9780321820464
Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:Addison-Wesley
Text book image
College Physics: A Strategic Approach (4th Editio...
Physics
ISBN:9780134609034
Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:PEARSON
How To Solve Any Resistors In Series and Parallel Combination Circuit Problems in Physics; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=eFlJy0cPbsY;License: Standard YouTube License, CC-BY