Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
bartleby

Concept explainers

Question
Book Icon
Chapter 32, Problem 32.64AP

(a)

To determine

The inductor behaves like an open circuit or short circuit or a resister of some particular resistance or none of those choices before the switch is opened.

(a)

Expert Solution
Check Mark

Answer to Problem 32.64AP

The inductor behaves as the short circuit because of no resistance.

Explanation of Solution

Given info: The induced voltage is 10.0V , the resistance of the resistor is 4.00Ω and the inductance of the inductor is 1.00H .

The inductor has no resistance. If the switch is closed for a long time, then inductor will reach saturation and voltage passes through the inductor. Hence, it behaves as a short circuit.

Conclusion:

Therefore, the inductor behaves as the short circuit because of no resistance.

(b)

To determine

The current carried by the inductor.

(b)

Expert Solution
Check Mark

Answer to Problem 32.64AP

The current carried by the inductor is 0.500A .

Explanation of Solution

Given info: The induced voltage is 10.0V , the resistance of the resistor is 4.00Ω and the inductance of the inductor is 1.00H .

The figure of the circuit diagrammed referred from P31.15 is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 32, Problem 32.64AP , additional homework tip  1

Figure (1)

The net resistance for parallel combination is,

1R'=1R+12RR'=2R3

Here,

R' is the net resistance for parallel combination.

R is the resistance of the circuit.

The net resistance is connected in series with R for parallel combination is,

Rnet=R+R' (1)

Here,

Rnet is the net resistance is connected in series.

Substitute 2R3 for R' in equation (1).

Rnet=R+2R3 (2)

Substitute 4.00Ω for R in equation (2) to find the Rnet .

Rnet=4.00Ω+2(4.00Ω)3=6.67Ω

Formula to calculate the current of battery is,

Ib=εRnet

Here,

Ib is the battery current.

ε is the induced voltage.

Substitute 10.0V for ε and 6.67Ω for Rnet in equation (3) to find the Ib .

Ib=10.0V6.67Ω=1.499A1.50A

The voltage across the parallel combination of resistors is,

V=εIbR (6)

Substitute 1.50A for Ib , 10.0V for ε and 4.00Ω for R in equation (6) to find the V .

V=10.0V(1.50A)(4.00Ω)=4.00V

Formula to calculate the current though the inductor is,

Ii=V2R (7)

Substitute 4.00V for V and 4.00Ω for R in equation (7) to find the Ii .

Ii=4.00V2(4.00Ω)=0.500A

Conclusion:

Therefore, the current carried by the inductor for t<0 is 0.500A .

(c)

To determine

The energy stored in the inductor.

(c)

Expert Solution
Check Mark

Answer to Problem 32.64AP

The energy stored in the inductor for t<0  is 0.125J .

Explanation of Solution

Given info: The induced voltage is 10.0V , the resistance of the resistor is 4.00Ω and the inductance of the inductor is 1.00H .

Formula to calculate the energy stored in the inductor is,

Ui=12LIi2

Substitute 1.00H for L and 0.500A for Ii to find the Ui .

Ui=12(1.00H)(0.500A)2=0.125J

Thus, the energy stored in the inductor for t<0 is 0.125J .

Conclusion:

Therefore, the energy stored in the inductor for t<0 is 0.125J .

(d)

To determine

The energy previously stored in the inductor after the switch is opened.

(d)

Expert Solution
Check Mark

Answer to Problem 32.64AP

The energy becomes 0.125J of the additional internal energy in the R and 2R . in the middle branch after switch is opened.

Explanation of Solution

Given info: The induced voltage is 10.0V , the resistance of the resistor is 4.00Ω and the inductance of the inductor is 1.00H .

When switch is opened, the energy stored in the inductor will dissipate through resistor R and 2R . Thus, the energy becomes 0.125J of the additional internal energy in the R and 2R . in the middle branch.

Conclusion:

Therefore, the energy becomes 0.125J of the additional internal energy in the R and 2R . in the middle branch after switch is opened.

(e)

To determine

To draw: The graph of the current in the inductor for t0 with the initial and final values and the time constant.

(e)

Expert Solution
Check Mark

Answer to Problem 32.64AP

Answer The graph of the current in the inductor for t0 and label the initial and final values and the time constant.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 32, Problem 32.64AP , additional homework tip  2

Explanation of Solution

Introduction:

The graph of the current verses time shows the variation of the current in the circuit with time and tells the nature of the current.

Explanation:

Given info: The induced voltage is 10.0V , the resistance of the resistor is 4.00Ω and the inductance of the inductor is 1.00H .

After time t=0 , the resistance of the circuit will be,

R1=R+2R=3R

Formula to calculate the time constant is,

t=LR1 (8)

Substitute 3R for R1 and 1.00H for L to find the t .

t=1.00H3R (9)

Substitute 4.0Ω for R in equation (9) to find the t .

t=1.00H3×4.0Ω=0.083s

The current flowing through the inductor at time t is,

I=I0eRLt

Substitute 0.500A for I0 and 12.0Ω for R1 and 0.083s for t in equation (10).

I=(0.500A)e(12.0Ω)1.0H(0.083s)

Thus, the graph of the current (the initial and final values) in the inductor for t0 and the time constant from equation (11) is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 32, Problem 32.64AP , additional homework tip  3

Figure (2)

The graph shows that current decays with exponentially with time constant. The current decreases from 0.500A towards zero because time constant goes to zero so, the final value of the current gets zero.

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
The switch in the circuit shown in the figure has been open for a long time. The resistors in the circuit each have a value of 60 Ω, the inductors each have a value of 60 mH, and  V = 10 V. (a) What is the total equivalent inductance of the circuit between points A and B? (b) What is the current through R1 the instant after the switch is closed? (c) What is the current through R1 long after the switch has been closed?
After being closed for a long time, the switch S in the circuit shown in the figure below is thrown open at  t = 0.  In the circuit,   = 24.0 V,   RA = 4.50 kΩ,   RB = 7.10 kΩ,  and  L = 513 mH.   (a) What is the emf across the inductor immediately after the switch is opened? V(b) When does the current in the resistor RB have a magnitude of 1.00 mA? s
For physics lab, two students constructed an RL circuit similar to the one shown in the figure, with = 6.00 V, L = 5.00 mH, and R = 7.60 0. + E ms S (a) What is the inductive time constant of the circuit (in ms)? R (b) Calculate the current in the circuit (in A) 250 μs after the switch is closed. A (c) What is the value of the final steady-state current (in A)? A ms (d) After what time interval (in ms) does the current reach 80.0% of its maximum value?

Chapter 32 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

Ch. 32 - Prob. 32.6OQCh. 32 - Prob. 32.7OQCh. 32 - Prob. 32.1CQCh. 32 - Prob. 32.2CQCh. 32 - A switch controls the current in a circuit that...Ch. 32 - Prob. 32.4CQCh. 32 - Prob. 32.5CQCh. 32 - Prob. 32.6CQCh. 32 - The open switch in Figure CQ32.7 is thrown closed...Ch. 32 - After the switch is dosed in the LC circuit shown...Ch. 32 - Prob. 32.9CQCh. 32 - Discuss the similarities between the energy stored...Ch. 32 - Prob. 32.1PCh. 32 - Prob. 32.2PCh. 32 - Prob. 32.3PCh. 32 - Prob. 32.4PCh. 32 - An emf of 24.0 mV Ls induced in a 500-turn coil...Ch. 32 - Prob. 32.6PCh. 32 - Prob. 32.7PCh. 32 - Prob. 32.8PCh. 32 - Prob. 32.9PCh. 32 - Prob. 32.10PCh. 32 - Prob. 32.11PCh. 32 - A toroid has a major radius R and a minor radius r...Ch. 32 - Prob. 32.13PCh. 32 - Prob. 32.14PCh. 32 - Prob. 32.15PCh. 32 - Prob. 32.16PCh. 32 - Prob. 32.17PCh. 32 - Prob. 32.18PCh. 32 - Prob. 32.19PCh. 32 - When the switch in Figure P32.18 is closed, the...Ch. 32 - Prob. 32.21PCh. 32 - Show that i = Iiet/ is a solution of the...Ch. 32 - Prob. 32.23PCh. 32 - Consider the circuit in Figure P32.18, taking =...Ch. 32 - Prob. 32.25PCh. 32 - The switch in Figure P31.15 is open for t 0 and...Ch. 32 - Prob. 32.27PCh. 32 - Prob. 32.28PCh. 32 - Prob. 32.29PCh. 32 - Two ideal inductors, L1 and L2, have zero internal...Ch. 32 - Prob. 32.31PCh. 32 - Prob. 32.32PCh. 32 - Prob. 32.33PCh. 32 - Prob. 32.34PCh. 32 - Prob. 32.35PCh. 32 - Complete the calculation in Example 31.3 by...Ch. 32 - Prob. 32.37PCh. 32 - A flat coil of wire has an inductance of 40.0 mH...Ch. 32 - Prob. 32.39PCh. 32 - Prob. 32.40PCh. 32 - Prob. 32.41PCh. 32 - Prob. 32.42PCh. 32 - Prob. 32.43PCh. 32 - Prob. 32.44PCh. 32 - Prob. 32.45PCh. 32 - Prob. 32.46PCh. 32 - In the circuit of Figure P31.29, the battery emf...Ch. 32 - A 1.05-H inductor is connected in series with a...Ch. 32 - A 1.00-F capacitor is charged by a 40.0-V power...Ch. 32 - Calculate the inductance of an LC circuit that...Ch. 32 - An LC circuit consists of a 20.0-mH inductor and a...Ch. 32 - Prob. 32.52PCh. 32 - Prob. 32.53PCh. 32 - Prob. 32.54PCh. 32 - An LC circuit like the one in Figure CQ32.8...Ch. 32 - Show that Equation 32.28 in the text Ls Kirchhoffs...Ch. 32 - In Figure 31.15, let R = 7.60 , L = 2.20 mH, and C...Ch. 32 - Consider an LC circuit in which L = 500 mH and C=...Ch. 32 - Electrical oscillations are initiated in a series...Ch. 32 - Review. Consider a capacitor with vacuum between...Ch. 32 - Prob. 32.61APCh. 32 - An inductor having inductance I. and a capacitor...Ch. 32 - A capacitor in a series LC circuit has an initial...Ch. 32 - Prob. 32.64APCh. 32 - When the current in the portion of the circuit...Ch. 32 - At the moment t = 0, a 24.0-V battery is connected...Ch. 32 - Prob. 32.67APCh. 32 - Prob. 32.68APCh. 32 - Prob. 32.69APCh. 32 - At t = 0, the open switch in Figure P31.46 is...Ch. 32 - Prob. 32.71APCh. 32 - Prob. 32.72APCh. 32 - Review. A novel method of storing energy has been...Ch. 32 - Prob. 32.74APCh. 32 - Review. The use of superconductors has been...Ch. 32 - Review. A fundamental property of a type 1...Ch. 32 - Prob. 32.77APCh. 32 - In earlier times when many households received...Ch. 32 - Assume the magnitude of the magnetic field outside...Ch. 32 - Prob. 32.80CPCh. 32 - To prevent damage from arcing in an electric...Ch. 32 - One application of an RL circuit is the generation...Ch. 32 - Prob. 32.83CP
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
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 with Modern ...
Physics
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
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
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: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning