Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
bartleby

Concept explainers

Question
Book Icon
Chapter 20, Problem 68P

(a)

To determine

The maximum current flows through the circuit.

(a)

Expert Solution
Check Mark

Answer to Problem 68P

The maximum current flows through the circuit is 45mA.

Explanation of Solution

Write an expression for the maximum current flows through the circuit.

  I0=εR                                                                                                                     (I)

Here, I0 is the maximum current flow through the circuit, ε is the emf and R is the resistance.

Conclusion:

Substitute 9.0V for ε and 200.0Ω for R in equation (I) to find I0.

  I0=9.0V200.0Ω=(45×103A)(1mA103A)=45mA

Thus, the maximum current flows through the circuit is 45mA.

(b)

To determine

The time taken for reaching the maximum current.

(b)

Expert Solution
Check Mark

Answer to Problem 68P

The time taken for reaching the maximum current is 1.0ms.

Explanation of Solution

Write an expression for the instantaneous current.

  I'=I0(1exp(tτ))                                                                                              (II)

Here, I' is the instantaneous current, t is the time and τ is the time constant.

Rearrange equation (II) to find t.

  t=τln(1I'I0)                                                                                                   (III)

Write an expression for the time constant.

  τ=LR                                                                                                                     (IV)

Here, L is the inductance and R is the resistance.

Rearrange the equation (III) using equation (IV).

  t=LRln(1I'I0)                                                                                                  (V)

Conclusion:

Substitute 0.30H for L, 200.0Ω for R and I0/2 for I'  in equation (V) to find t.

  t=0.30H200.0Ωln(1I0/2I0)=0.30H200.0Ωln(112)=(1.0×103s)(1ms103s)=1.0ms

Thus, the time taken for reaching the maximum current is 1.0ms.

(c)

To determine

The energy stored in the inductor and the rate at which energy is dissipated in the resistor.

(c)

Expert Solution
Check Mark

Answer to Problem 68P

The energy stored in the inductor is 76μJ and the rate at which energy is dissipated in the resistor is 0.10W.

Explanation of Solution

Write an expression for the energy stored in the inductor.

  U=12LI2                                                                                                          (VI)

Here, U is the energy.

The current is half of the maximum current. Thus, rewrite equation (VI).

  U=12L(I0/2)2                                                                                                   (VII)

Write an expression for the rate at which energy is converted in the resistor.

  P=I2R                                                                                                         (VIII)

The current is half of the maximum current. Thus, rewrite equation (VIII).

  P=(I0/2)2R                                                                                                    (IX)

Conclusion:

Substitute 0.30H for L, 200.0Ω for R and 45mA for I0  in equation (IX) to find U.

  U=12(0.30H)((45mA)(1A103mA)2)2=12(0.30H)(0.045m2)2=(76×106J)(1μJ106J)=76μJ

Substitute 0.30H for L, 200.0Ω for R and 45mA for I0  in equation (IX) to find P.

  P=((45mA)(1A103mA)/2)2(200.0Ω)=(45×103A/2)2(200.0Ω)=0.10W

Thus, the energy stored in the inductor is 76μJ and the rate at which energy is dissipated in the resistor is 0.10W.

(d)

To determine

The maximum current flows through the circuit and the time taken for reaching the maximum current.

(d)

Expert Solution
Check Mark

Answer to Problem 68P

The energy stored is 110kJ.

Explanation of Solution

Write an expression for equivalent resistance.

    Req=R+rL+r                                                                                                       (X)

Here, Req is the equivalent resistance, rL  is the internal resistance of inductor and r is the internal resistance of battery.

Write an expression for the maximum current flows through the circuit.

    Imax=εReq                                                       (XI)

Write an expression for the time taken for reaching the maximum current.

    t=LReqln(1I'I0)       (XII)

Conclusion:

Substitute 75Ω for rL, 20Ω for r and 200.0Ω for R in equation (X) to find Req.

    Req=75Ω+20Ω+200.0Ω=295Ω

Substitute 9.0V for ε and 295Ω for Req in equation (XI) to find Imax.

    I0=9.0V295Ω=(31×103A)(1mA103A)=31mA

Substitute 0.30H for L, 295Ω for Req and I0/2 for I'  in equation (XII) to find t.

    t=0.30H295Ωln(1I0/2I0)=0.30H295Ωln(112)=(0.70×103s)(1ms103s)=0.70ms

Thus, the maximum current flows through the circuit is 31mA and the time taken for reaching the maximum current is 0.70ms.

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!

Chapter 20 Solutions

Physics

Ch. 20.7 - Conceptual Practice Problem 20.8 Choosing a Core...Ch. 20.9 - CHECKPOINT 20.9 Five solenoids are wound with...Ch. 20.9 - Practice Problem 20.9 Power in an Inductor The...Ch. 20.10 - Prob. 20.10CPCh. 20.10 - Prob. 20.10PPCh. 20 - Prob. 1CQCh. 20 - Prob. 2CQCh. 20 - Prob. 3CQCh. 20 - Prob. 4CQCh. 20 - Prob. 5CQCh. 20 - Prob. 6CQCh. 20 - Prob. 7CQCh. 20 - Prob. 8CQCh. 20 - Prob. 9CQCh. 20 - Prob. 10CQCh. 20 - Prob. 11CQCh. 20 - Prob. 12CQCh. 20 - Prob. 13CQCh. 20 - Prob. 14CQCh. 20 - Prob. 15CQCh. 20 - Prob. 16CQCh. 20 - Prob. 17CQCh. 20 - Prob. 18CQCh. 20 - Prob. 19CQCh. 20 - Prob. 1MCQCh. 20 - Prob. 2MCQCh. 20 - Prob. 3MCQCh. 20 - Prob. 4MCQCh. 20 - Prob. 5MCQCh. 20 - Prob. 6MCQCh. 20 - Prob. 7MCQCh. 20 - Prob. 8MCQCh. 20 - Prob. 9MCQCh. 20 - Prob. 10MCQCh. 20 - A vertical metal rod of length 20 cm moves south...Ch. 20 - Suppose that the current were to flow in the...Ch. 20 - A vertical metal rod of length 36 cm moves north...Ch. 20 - Prob. 3PCh. 20 - Prob. 4PCh. 20 - Prob. 5PCh. 20 - Prob. 6PCh. 20 - In Fig. 20.2, a metal rod of length L is sliding...Ch. 20 - Prob. 9PCh. 20 - 4. In Fig. 20.2, what would the magnitude (in...Ch. 20 - Prob. 11PCh. 20 - 6. The armature of an ac generator is a circular...Ch. 20 - Prob. 13PCh. 20 - 8. A solid copper disk of radius R rotates at...Ch. 20 - 9. A horizontal desk surface measures 1.3 m × 1.0...Ch. 20 - The magnetic field between the poles of a magnet...Ch. 20 - Prob. 36PCh. 20 - 10. A square loop of wire, 0.75 m on each side,...Ch. 20 - 11. A long straight wire carrying a steady current...Ch. 20 - 12. A long straight wire carrying a current I is...Ch. 20 - Prob. 18PCh. 20 - 14. While I1 is increasing, what is the direction...Ch. 20 - 15. While I1 is constant, does current flow in...Ch. 20 - A circular conducting loop with radius 3.40 cm is...Ch. 20 - A circular conducting loop with radius 1.8 cm is...Ch. 20 - An external magnetic field parallel to the central...Ch. 20 - An external magnetic field is parallel to the...Ch. 20 - 19. In the figure, switch s is initially open. It...Ch. 20 - 20. Crocodiles are thought to be able to detect...Ch. 20 - 21. A bar magnet approaches a coil as shown, (a)...Ch. 20 - 22. Another example of motional emf is a rod...Ch. 20 - 23. Two loops of wire are next to each other in...Ch. 20 - 24. A dc motor has coils with a resistance of 16 Ω...Ch. 20 - Prob. 33PCh. 20 - Prob. 34PCh. 20 - Prob. 35PCh. 20 - 29. A doorbell uses a transformer to deliver an...Ch. 20 - Prob. 38PCh. 20 - 31. When the emf for the primary of a transformer...Ch. 20 - 32. A transformer with a primary coil of 1000...Ch. 20 - Prob. 41PCh. 20 - An ideal transformer takes an ac voltage of...Ch. 20 - 35. A 2 m long copper pipe is held vertically....Ch. 20 - In Problem 43, the pipe is suspended from a spring...Ch. 20 - Prob. 45PCh. 20 - Prob. 46PCh. 20 - 39. A solenoid of length 2.8 cm and diameter 0.75...Ch. 20 - Prob. 48PCh. 20 - Prob. 49PCh. 20 - Prob. 50PCh. 20 - Prob. 51PCh. 20 - 44. The current in a 0.080 H solenoid increases...Ch. 20 - Prob. 53PCh. 20 - Prob. 54PCh. 20 - Prob. 55PCh. 20 - Prob. 56PCh. 20 - Refer to Problem 56. After the switch has been...Ch. 20 - Prob. 59PCh. 20 - Prob. 61PCh. 20 - Prob. 58PCh. 20 - Prob. 60PCh. 20 - Prob. 63PCh. 20 - Prob. 62PCh. 20 - Prob. 64PCh. 20 - Prob. 65PCh. 20 - Prob. 66PCh. 20 - Prob. 68PCh. 20 - Prob. 67PCh. 20 - Prob. 70PCh. 20 - Prob. 69PCh. 20 - Prob. 72PCh. 20 - Prob. 71PCh. 20 - Prob. 74PCh. 20 - Prob. 73PCh. 20 - Prob. 75PCh. 20 - Prob. 76PCh. 20 - Prob. 77PCh. 20 - Prob. 78PCh. 20 - Prob. 79PCh. 20 - 72. A uniform magnetic field of magnitude 0.29 T...Ch. 20 - Prob. 81PCh. 20 - Prob. 82PCh. 20 - Prob. 83PCh. 20 - Prob. 85PCh. 20 - Prob. 84PCh. 20 - Prob. 86PCh. 20 - Prob. 87PCh. 20 - Prob. 88PCh. 20 - Prob. 90PCh. 20 - Prob. 91PCh. 20 - Prob. 92PCh. 20 - Prob. 89PCh. 20 - Prob. 93PCh. 20 - Prob. 94PCh. 20 - Prob. 95PCh. 20 - Prob. 96PCh. 20 - Prob. 97PCh. 20 - Prob. 98PCh. 20 - Prob. 99PCh. 20 - Prob. 100P
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.
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