The waveform below is from a 70-Vac source. 180 V 50 ms Determine: a. Vpk (ans: 98.9949 V) b. Vpr-pk(ans: 197.9899 V) c.Vyms (ans:70 V) i. j.
The waveform below is from a 70-Vac source. 180 V 50 ms Determine: a. Vpk (ans: 98.9949 V) b. Vpr-pk(ans: 197.9899 V) c.Vyms (ans:70 V) i. j.
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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![The waveform below is from a 70-Vac source.
180 V
50 ms
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: 54.9124°)
h. equation of the waveform(in time domain)
i. the instantaemous voltage when t = 120ms(ans: 29.0880 V)
j. the angle (1st occurence)after t
= 0 when the voltage is + 80 V(ans:108.8253 V)
k. the time (2nd occurence) after t
= 0 when the voltage is – 10 V(ans: 66.8639 ms)
(ans: v(t) = 98.9949sin(20nt – 54.9124°))](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F35c74e5e-a302-4aa3-a502-0c0c0abba88b%2F843ba2bd-d4c0-409b-8aad-e439ef26511a%2Fwlqzkb_processed.png&w=3840&q=75)
Transcribed Image Text:The waveform below is from a 70-Vac source.
180 V
50 ms
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: 54.9124°)
h. equation of the waveform(in time domain)
i. the instantaemous voltage when t = 120ms(ans: 29.0880 V)
j. the angle (1st occurence)after t
= 0 when the voltage is + 80 V(ans:108.8253 V)
k. the time (2nd occurence) after t
= 0 when the voltage is – 10 V(ans: 66.8639 ms)
(ans: v(t) = 98.9949sin(20nt – 54.9124°))
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