The waveform below is from a 70-Vac source.- 50 ms180 VDetermine:a. Vpk (ans: 98.9949 V)b. Vpk-pk (ans: 197.9899 V)c. Vrms (ans: 70 V)d. Period(ans: 100 ms)e. Frequency(ans: 10 Hz)f. Angular velocity(ans:20n rad/sec)g. Phase angle(ans: 125.0876°)h. equation of the waveform(in time domain)i. the instantaenous voltage when t = 120ms(ans: -29.0880 V)j. the angle (1st occurence)after t= 0 when the voltage is + 80 V(ans: 0.99959 °)k. the time (2nd occurence) after t= 0 when the voltage is – 10 V(ans: 63.643 ms)(ans: v(t) = 98.9949sin(20nt – 125.0876°))

As per the honor code I will do only first three parts . Kindly post the other questions again.…

Answered: The waveform below is from a 70-Vac…

The waveform below is from a 70-Vac source. - 50 ms 180 V Determine: a. Vpk (ans: 98.9949 V) b. Vpk-pk (ans: 197.9899 V) c. Vrms (ans:70 V)

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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Question

The waveform below is from a 70-Vac source.
- 50 ms
180 V
Determine:
a. Vpk (ans: 98.9949 V)
b. Vpk-pk (ans: 197.9899 V)
c. Vrms (ans: 70 V)
d. Period(ans: 100 ms)
e. Frequency(ans: 10 Hz)
f. Angular velocity(ans:20n rad/sec)
g. Phase angle(ans: 125.0876°)
h. equation of the waveform(in time domain)
i. the instantaenous voltage when t = 120ms(ans: -29.0880 V)
j. the angle (1st occurence)after t
= 0 when the voltage is + 80 V(ans: 0.99959 °)
k. the time (2nd occurence) after t
= 0 when the voltage is – 10 V(ans: 63.643 ms)
(ans: v(t) = 98.9949sin(20nt – 125.0876°))

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