Physics for Scientists and Engineers with Modern Physics, Technology Update
Physics for Scientists and Engineers with Modern Physics, Technology Update
9th Edition
ISBN: 9781305401969
Author: SERWAY, Raymond A.; Jewett, John W.
Publisher: Cengage Learning
bartleby

Concept explainers

Question
Book Icon
Chapter 33, Problem 40P

(a)

To determine

The rms current in the circuit.

(a)

Expert Solution
Check Mark

Answer to Problem 40P

The rms current in the circuit is 5.42A_.

Explanation of Solution

Write expression for rms current.

    Irms=VrmsZ                                                                                                                 (I)

Here, Irms is the rms current, Vrms is the rms voltage and Z is the impedance.

Write expression for Z.

    Z=R2+(XLXC)2                                                                                           (II)

Here, R is the resistance, XL is the inductive reactance and XC is the capacitive reactance.

Write expression for XC.

    XC=1ωC                                                                                                               (III)

Here, ω is the angular frequency and C is the capacitance.

Write expression for XL.

    XL=ωL                                                                                                                (IV)

Here, L is inductance.

Write expression for ω.

    ω=2πf                                                                                                                 (V)

Here, f is the frequency.

Conclusion:

Substitute 60.0-Hz  for f in equation (V) to calculate ω.

    ω=2π(60.0-Hz)=377rad/s

Substitute 377rad/s for ω and 25.0-mH for L is equation (IV) to calculate XL.

    XL=(377rad/s)(25.0-mH)(10-3H1mH)=9.43Ω

Substitute 9.43Ω for XL, 0 for XC and 20.0-Ω for R in equation(II) to calculate Z.

    Z=202+(9.430)2=22.1Ω

Substitute 22.1Ω for Z and 120-V for Vrms in equation (I), to calculate Irms.

    Irms=120-V22.1Ω=5.42A

Therefore, the rms current in the circuit is 5.42A_.

(b)

To determine

The power factor in the circuit.

(b)

Expert Solution
Check Mark

Answer to Problem 40P

The power factor of the circuit is 0.9_.

Explanation of Solution

Write expression for power factor in the circuit.

    Powerfactor=cosϕ                                                                                              (VI)

Here, ϕ is the phase angle.

Write expression for phase angle.

    ϕ=tan1XLXCR                                                                                               (VII)

Conclusion:

Substitute 9.43H for XL, 0 for XC and 20.0-Ω for R in equation (VII), to calculate ϕ.

    ϕ=tan19.43H020.0-Ω=25.24°

Substitute 25.24° for ϕ in equation (VI), to calculate power factor.

    Powerfactor=cos(25.24°)=0.9

Therefore, the power factor of the circuit is 0.9_.

(c)

To determine

The capacitor that must be added in series to make the power factor equal to 1.

(c)

Expert Solution
Check Mark

Answer to Problem 40P

The capacitor that must be added in series to make the power factor equal to 1 is 2.81×10-4F_.

Explanation of Solution

To make the power factor equals to 1 XC and XL should be same.

Substitute 9.43Ω for XC and 377rad/s for ω in equation (III) to calculate C.

    9.43Ω=1(377rad/s)CC=2.81×10-4F

Therefore, the capacitor that must be added in series to make the power factor equal to 1 is 2.81×10-4F_.

(d)

To determine

The reduction in the supply voltage.

(d)

Expert Solution
Check Mark

Answer to Problem 40P

The reduction in the supply voltage is 108.2V_.

Explanation of Solution

Write expression for average power in terms for rms volage.

    Pavg=ΔVrms2R                                                                                                      (VIII)

Here, Pavg is the average power delivered and ΔVrms is the rms voltage.

Write expression for power in terms of power factor.

    Pavg=IrmsVrmscosϕ                                                                                                (IX)

Conclusion:

Equate equation (VIII) and equation (IX).

    ΔVrms2R=IrmsVrmscosϕ

Substitute 5.42A for Irms, 120-V for Vrms, 0.9 for cosϕ and  20.0-Ω for R in above equation, to calculate ΔVrms.

    ΔVrms220.0-Ω=(5.42A)(120-V)(0.9)ΔVrms2=11707.2ΔVrms=108.2V

Therefore, The reduction in the supply voltage if the power remains same is 108.2V_.

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
What is integrated science. What is fractional distillation What is simple distillation
19:39 · C Chegg 1 69% ✓ The compound beam is fixed at Ę and supported by rollers at A and B. There are pins at C and D. Take F=1700 lb. (Figure 1) Figure 800 lb ||-5- F 600 lb بتا D E C BO 10 ft 5 ft 4 ft-—— 6 ft — 5 ft- Solved Part A The compound beam is fixed at E and... Hình ảnh có thể có bản quyền. Tìm hiểu thêm Problem A-12 % Chia sẻ kip 800 lb Truy cập ) D Lưu of C 600 lb |-sa+ 10ft 5ft 4ft6ft D E 5 ft- Trying Cheaa Những kết quả này có hữu ích không? There are pins at C and D To F-1200 Egue!) Chegg Solved The compound b... Có Không ☑ ||| Chegg 10 וח
No chatgpt pls will upvote

Chapter 33 Solutions

Physics for Scientists and Engineers with Modern Physics, Technology Update

Ch. 33 - Prob. 4OQCh. 33 - Prob. 5OQCh. 33 - Prob. 6OQCh. 33 - Prob. 7OQCh. 33 - A resistor, a capacitor, and an inductor are...Ch. 33 - Under what conditions is the impedance of a series...Ch. 33 - Prob. 10OQCh. 33 - Prob. 11OQCh. 33 - Prob. 12OQCh. 33 - Prob. 13OQCh. 33 - Prob. 1CQCh. 33 - Prob. 2CQCh. 33 - Prob. 3CQCh. 33 - Prob. 4CQCh. 33 - Prob. 5CQCh. 33 - Prob. 6CQCh. 33 - Prob. 7CQCh. 33 - Prob. 8CQCh. 33 - Prob. 9CQCh. 33 - Prob. 10CQCh. 33 - Prob. 1PCh. 33 - (a) What is the resistance of a lightbulb that...Ch. 33 - Prob. 3PCh. 33 - Prob. 4PCh. 33 - Prob. 5PCh. 33 - Prob. 6PCh. 33 - Prob. 7PCh. 33 - Prob. 8PCh. 33 - Prob. 9PCh. 33 - Prob. 10PCh. 33 - Prob. 11PCh. 33 - Prob. 12PCh. 33 - An AC source has an output rms voltage of 78.0 V...Ch. 33 - Prob. 14PCh. 33 - Prob. 15PCh. 33 - Prob. 16PCh. 33 - Prob. 17PCh. 33 - An AC source with an output rms voltage of 86.0 V...Ch. 33 - Prob. 19PCh. 33 - Prob. 20PCh. 33 - Prob. 21PCh. 33 - Prob. 22PCh. 33 - What is the maximum current in a 2.20-F capacitor...Ch. 33 - Prob. 24PCh. 33 - In addition to phasor diagrams showing voltages...Ch. 33 - Prob. 26PCh. 33 - Prob. 27PCh. 33 - Prob. 28PCh. 33 - Prob. 29PCh. 33 - Prob. 30PCh. 33 - Prob. 31PCh. 33 - A 60.0-ft resistor is connected in series with a...Ch. 33 - Prob. 33PCh. 33 - Prob. 34PCh. 33 - A series RLC circuit has a resistance of 45.0 and...Ch. 33 - Prob. 36PCh. 33 - Prob. 37PCh. 33 - An AC voltage of the form v = 90.0 sin 350t, where...Ch. 33 - Prob. 39PCh. 33 - Prob. 40PCh. 33 - Prob. 41PCh. 33 - A series RLC circuit has components with the...Ch. 33 - Prob. 43PCh. 33 - Prob. 44PCh. 33 - A 10.0- resistor, 10.0-mH inductor, and 100-F...Ch. 33 - Prob. 46PCh. 33 - Prob. 47PCh. 33 - Prob. 48PCh. 33 - The primary coil of a transformer has N1 = 350...Ch. 33 - A transmission line that has a resistance per unit...Ch. 33 - Prob. 51PCh. 33 - Prob. 52PCh. 33 - Prob. 53PCh. 33 - Consider the RC highpass filter circuit shown in...Ch. 33 - Prob. 55PCh. 33 - Prob. 56PCh. 33 - Prob. 57APCh. 33 - Prob. 58APCh. 33 - Prob. 59APCh. 33 - Prob. 60APCh. 33 - Prob. 61APCh. 33 - Prob. 62APCh. 33 - Prob. 63APCh. 33 - Prob. 64APCh. 33 - Prob. 65APCh. 33 - Prob. 66APCh. 33 - Prob. 67APCh. 33 - Prob. 68APCh. 33 - Prob. 69APCh. 33 - (a) Sketch a graph of the phase angle for an RLC...Ch. 33 - Prob. 71APCh. 33 - Prob. 72APCh. 33 - A series RLC circuit contains the following...Ch. 33 - Prob. 74APCh. 33 - Prob. 75APCh. 33 - A series RLC circuit in which R = l.00 , L = 1.00...Ch. 33 - Prob. 77CPCh. 33 - Prob. 78CPCh. 33 - Prob. 79CPCh. 33 - Figure P33.80a shows a parallel RLC circuit. The...Ch. 33 - Prob. 81CP
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
College Physics
Physics
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
University Physics Volume 2
Physics
ISBN:9781938168161
Author:OpenStax
Publisher:OpenStax
Text book image
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning