Laboratory Manual for Introductory Circuit Analysis
Laboratory Manual for Introductory Circuit Analysis
13th Edition
ISBN: 9780133923780
Author: Robert L. Boylestad, Gabriel Kousourou
Publisher: PEARSON
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Textbook Question
Chapter 14, Problem 1P

Plot the following waveform versus time showing one clear. Complete cycle. Then determine the derivative of the waveform using Eq. (14.1) and sketch one complete cycle of the derivative directly under the Original waveform. Compare the magnitude of the derivative at various points versus the slope of the original sinusoidal function.

v = 4 sin 62.8 t

Expert Solution & Answer
Check Mark
To determine

To plot:

The graph of the waveform versus time for one complete cycle. Calculate the derivative of the waveform and then draw one complete cycle of it. Also, compare the magnitude of the derivative at various points versus the slope of the original sinusoidal function.

Explanation of Solution

Given:

The given equation is v=4sin62.8t.

Calculation:

Let us consider the sinusoidal expression of voltage as:

  v(t)=4sin62.8t

Standard sinusoidal equation is:

  v(t)=Vmsin(ωt±θ)

Comparing, we get

  ω=62.8rad/sec

Value of frequency will be

  ω=2πff=ω2πSubstitute thevalueofω,f=62.82πf=10Hz

Voltage at t=0

  v(t)=4sin62.8tv(0)=4sin62.8(0)v(0)=0V

Voltage at t=0.02

  v(t)=4sin62.8tv(0.02)=4sin62.8(0.02)v(0.02)=3.80V

Voltage at t=0.04

  v(t)=4sin62.8tv(0.04)=4sin62.8(0.04)v(0.04)=2.355V

Voltage at t=0.06

  v(t)=4sin62.8tv(0.06)=4sin62.8(0.06)v(0.06)=2.35V

Voltage at t=0.08 :

  v(t)=4sin62.8tv(0.08)=4sin62.8(0.08)v(0.08)=3.80V

Voltage at t=0.1

  v(t)=4sin62.8tv(0.1)=4sin62.8(0.1)v(0.1)=0V

Graph showing voltage versus time

Laboratory Manual for Introductory Circuit Analysis, Chapter 14, Problem 1P , additional homework tip  1

Derivative of standard sinusoidal equation is

  ddtv(t)=ωVmcos(ωt±θ)

Value of derivative of voltage at t=0

  ddtv(t)=ωVmcos(ωt±θ)ddtv(0)=(62.8)(4)cos(62.8×0)ddtv(0)=251.2V

Value of derivative of voltage at t=0.02

  ddtv(t)=ωVmcos(ωt±θ)ddtv(0.02)=(62.8)(4)cos(62.8×0.02)ddtv(0.02)=77.77V

Value of derivative of voltage at t=0.04

  ddtv(t)=ωVmcos(ωt±θ)ddtv(0.04)=(62.8)(4)cos(62.8×0.04)ddtv(0.04)=203.04V

Value of derivative of voltage at t=0.06

  ddtv(t)=ωVmcos(ωt±θ)ddtv(0.06)=(62.8)(4)cos(62.8×0.06)ddtv(0.06)=203.50V

Value of derivative of voltage at t=0.08

  ddtv(t)=ωVmcos(ωt±θ)ddtv(0.08)=(62.8)(4)cos(62.8×0.08)ddtv(0.08)=77.01V

Value of derivative of voltage at t=0.1

  ddtv(t)=ωVmcos(ωt±θ)ddtv(0.1)=(62.8)(4)cos(62.8×0.1)ddtv(0.1)=251.2V

Graph of derivative of voltage directly under the original waveform:

Laboratory Manual for Introductory Circuit Analysis, Chapter 14, Problem 1P , additional homework tip  2

Laboratory Manual for Introductory Circuit Analysis, Chapter 14, Problem 1P , additional homework tip  3

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Chapter 14 Solutions

Laboratory Manual for Introductory Circuit Analysis

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