EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
16th Edition
ISBN: 8220100546716
Author: Katz
Publisher: CENGAGE L
bartleby

Videos

Question
Book Icon
Chapter 33, Problem 58PQ
To determine

The expression for average power supplied.

Expert Solution & Answer
Check Mark

Answer to Problem 58PQ

The expression for average power supplied is Pavg=εrms×Irms×cosφ.

Explanation of Solution

Write the expression for average power.

    Pavg=0TdU0Tdt                                            (I)

Here, Pavg is the average power, U is the energy dissipated in time t and T is one complete time period.

Write the expression for current to the circuit connected to an alternating source of emf.

    I=Imaxsinωt                                (II)

Here, I is the current in the circuit, Imax is the peak value of the current and ω is the angular frequency.

Write the expression for emf in the circuit.

    ε=εmaxsin(ωt+φ)                       (III)

Here, εmax is the peak value of emf, ε is the induced emf, and φ is the phase difference.

Write the expression for energy dissipated in time t.

    dU=εIdt                      (IV)

Substitute εmaxsin(ωt+φ) for ε and Imaxsinωt for I in the equation (IV) to solve for expression for dU.

    dU=εmaxsin(ωt+φ)Imaxsinωtdt=εmaxImax(sin(ωt+φ)sinωtdt)=εmaxImax(cos(ωtωtφ)+cos(ωt+ωt+φ))dt=εmaxImax(cos(φ)+cos(2ωt+φ))dt

Substitute εmaxImax(cos(φ)+cos(2ωt+φ))dt for dU in equation (I) to calculate the expression for Pavg.

    Pavg=0TεmaxImax(cos(φ)+cos(2ωt+φ))dt0Tdt

Integrate the  expression between the limits 0 and T.

    Pavg=0TεmaxImax2(cos(φ)+cos(2ωt+φ))dt0Tdt=εmaxImax2[cos(φ)0Tdt+0Tcos(2ωt+φ)dt0Tdt]=εmaxImax2[cos(φ)[t]0T+12ω[sin(2ωt+φ)]0T[t]0T]=εmaxImax2[cos(φ)T+12ω[sin(2ωT+φ)sinφ]T]                     (V)

Now,

    ω=2πT

Substitute, 2πT for ω in the equation (V).

    Pavg=εmaxImax2[cos(φ)T+122πT[sin(22πTT+φ)sinφ]T]=εmaxImax2[cos(φ)T+T4π[sin(4π+φ)sinφ]T]=εmaxImax2TT(cos(φ)+14π[sin(φ)sinφ])=εmaxImax2cos(φ)                            (VI)

The above expression can be rewritten as.

    Pavg=εmax2×Imax2×cosφ                                                                                    (VII)

Also, rms value of current is given by.

    Irms=Imax2

Here, Irms is the rms value of current.

And,

The rms value of emf is given by.

    εrms=εmax2

Here, εrms is the rms value of induced emf.

Substitute εrms for εmax2 and Irms for Irms2 in equation (VII).

    Pavg=εrmsIrmscosφ

Therefore, the expression for average power supplied is Pavg=εrmsIrmscosφ.

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
Can someone help me
Need help on the following questions on biomechanics. (Please refer to images below)A gymnast weighing 68 kg attempts a handstand using only one arm. He plants his handat an angle resulting in the reaction force shown.A) Find the resultant force (acting on the Center of Mass)B) Find the resultant moment (acting on the Center of Mass)C) Draw the resultant force and moment about the center of mass on the figure below. Will the gymnast rotate, translate, or both? And in which direction?
Please help me on the following question (Please refer to image below)An Olympic lifter (m = 103kg) is holding a lift with a mass of 350 kg. The barexerts a purely vertical force that is equally distributed between both hands. Each arm has amass of 9 kg, are 0.8m long and form a 40° angle with the horizontal. The CoM for each armis 0.5 m from hand. Assuming the lifter is facing us in the diagram below, his right deltoidinserts 14cm from the shoulder at an angle of 13° counter-clockwise from the humerus.A) You are interested in calculating the force in the right deltoid. Draw a free body diagramof the right arm including the external forces, joint reaction forces, a coordinate system andstate your assumptions.B) Find the force exerted by the right deltoidC) Find the shoulder joint contact force. Report your answer using the magnitude and directionof the shoulder force vector.

Chapter 33 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

Ch. 33 - Prob. 4PQCh. 33 - Prob. 5PQCh. 33 - Prob. 6PQCh. 33 - Prob. 7PQCh. 33 - Prob. 8PQCh. 33 - Prob. 9PQCh. 33 - Prob. 10PQCh. 33 - Prob. 11PQCh. 33 - At one instant, a current of 6.0 A flows through...Ch. 33 - Prob. 13PQCh. 33 - Prob. 14PQCh. 33 - Prob. 15PQCh. 33 - In Figure 33.9A (page 1052), the switch is closed...Ch. 33 - Prob. 17PQCh. 33 - Prob. 18PQCh. 33 - Prob. 19PQCh. 33 - Prob. 20PQCh. 33 - Prob. 21PQCh. 33 - Prob. 22PQCh. 33 - In the LC circuit in Figure 33.11, the inductance...Ch. 33 - A 2.0-F capacitor is charged to a potential...Ch. 33 - Prob. 26PQCh. 33 - Prob. 27PQCh. 33 - Prob. 28PQCh. 33 - For an LC circuit, show that the total energy...Ch. 33 - Prob. 30PQCh. 33 - Prob. 31PQCh. 33 - Prob. 32PQCh. 33 - Prob. 33PQCh. 33 - Suppose you connect a small lightbulb across a DC...Ch. 33 - Prob. 35PQCh. 33 - Prob. 36PQCh. 33 - Prob. 37PQCh. 33 - Prob. 38PQCh. 33 - Prob. 39PQCh. 33 - Prob. 40PQCh. 33 - Prob. 41PQCh. 33 - Prob. 42PQCh. 33 - Prob. 43PQCh. 33 - In an ideal AC circuit with capacitance, there is...Ch. 33 - Prob. 45PQCh. 33 - Prob. 46PQCh. 33 - Prob. 47PQCh. 33 - Prob. 48PQCh. 33 - Prob. 49PQCh. 33 - An AC generator with an rms emf of 15.0 V is...Ch. 33 - Prob. 51PQCh. 33 - Prob. 52PQCh. 33 - Prob. 53PQCh. 33 - Prob. 54PQCh. 33 - Prob. 55PQCh. 33 - Prob. 56PQCh. 33 - Prob. 57PQCh. 33 - Prob. 58PQCh. 33 - Prob. 59PQCh. 33 - An AC source of angular frequency is connected to...Ch. 33 - An RLC series circuit is constructed with R =...Ch. 33 - Prob. 62PQCh. 33 - A series RLC circuit driven by a source with an...Ch. 33 - Prob. 64PQCh. 33 - Prob. 65PQCh. 33 - Prob. 66PQCh. 33 - Prob. 67PQCh. 33 - Prob. 68PQCh. 33 - Prob. 69PQCh. 33 - Prob. 70PQCh. 33 - Problems 71 and 72 paired. Figure P33.71 shows a...Ch. 33 - Prob. 72PQCh. 33 - Prob. 73PQCh. 33 - Prob. 74PQCh. 33 - Prob. 75PQCh. 33 - In a series RLC circuit with a maximum current of...Ch. 33 - Prob. 77PQCh. 33 - Two coaxial cables of length with radii a and b...Ch. 33 - Prob. 79PQCh. 33 - Prob. 80PQCh. 33 - Prob. 81PQCh. 33 - Prob. 82PQCh. 33 - Prob. 83PQCh. 33 - Prob. 84PQ
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:9781285737027
Author:Raymond A. Serway, Chris Vuille
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: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Text book image
College Physics
Physics
ISBN:9781305952300
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
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Introduction To Alternating Current; Author: Tutorials Point (India) Ltd;https://www.youtube.com/watch?v=0m142qAZZpE;License: Standard YouTube License, CC-BY