Who Is Correct in Calculating Sinusoidal Waveform Values?

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SUMMARY

The discussion centers on calculating sinusoidal waveform values for an AC circuit described by the equation v = 100 sin(50πt - 0.523) V. Key findings include the correct peak-to-peak voltage of 200 V, a periodic time of 0.04 s, a frequency of 25 Hz, and a phase angle of 29.97° lagging. The user reported discrepancies in their calculations for the times when the voltage is 60V and -40V, specifically 7.43 ms and 12.57 ms for 60V, and 25.95 ms and 34.05 ms for -40V, which were inconsistent with the established answers of 7.426 ms, 19.23 ms, and 25.95 ms, 40.71 ms respectively.

PREREQUISITES
  • Understanding of sinusoidal functions in AC circuits
  • Knowledge of phase angles in waveform analysis
  • Ability to perform calculations involving trigonometric functions
  • Familiarity with periodicity and frequency concepts
NEXT STEPS
  • Review the calculation methods for sinusoidal waveform values in AC circuits
  • Study the implications of phase angles on voltage calculations
  • Learn about the graphical representation of sinusoidal functions
  • Explore the mathematical properties of sine functions in electrical engineering
USEFUL FOR

Electrical engineers, students in engineering mathematics, and anyone involved in analyzing AC circuit behavior will benefit from this discussion.

Utech
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I was routinely checking a question in an Engineering Maths book as listed below:

The instantaneous value of voltage in an a.c.
circuit at any time t seconds is given by
v = 100 sin(50[tex]\pi[/tex]t - 0.523) V. Find:
(a) the peak-to-peak voltage, the periodic
time, the frequency and the phase angle
(b) the voltage when t = 0
(c) the voltage when t = 8ms
(d) the times in the first cycle when the voltage
is 60V
(e) the times in the first cycle when the voltage
is =-40V
(f) the first time when the voltage is a maximum.
Sketch the curve for one cycle showing
relevant points. [(a) 200 V, 0.04 s, 25 Hz,
29.97° lagging (b) -49.95V (c) 66.96V
(d) 7.426 ms, 19.23 ms (e) 25.95 ms, 40.71 ms
(f) 13.33 ms]

Questions (d) and (e) are the one of interest to me, I worked those two questions and my answers were as follows:

(d) 7.43 ms, 12.57 ms
(e) 25.95 ms, 34.05 ms

I have highlighted the discrepancies between my answers and the authors answers. Who is correct ? A cursory glance at second answer of part (e) would show an issue as the periodic cycle is 40 ms so how could it fall out of the first cycle. Is there a different way of calculating negative voltages also. Thanks, any detailed explanations / corrections would be appreciated
 
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Utech said:
I was routinely checking a question in an Engineering Maths book as listed below:

The instantaneous value of voltage in an a.c.
circuit at any time t seconds is given by
v = 100 sin(50[tex]\pi[/tex]t - 0.523) V. Find:
(a) the peak-to-peak voltage, the periodic
time, the frequency and the phase angle
(b) the voltage when t = 0
(c) the voltage when t = 8ms
(d) the times in the first cycle when the voltage
is 60V
(e) the times in the first cycle when the voltage
is =-40V
(f) the first time when the voltage is a maximum.
Sketch the curve for one cycle showing
relevant points. [(a) 200 V, 0.04 s, 25 Hz,
29.97° lagging (b) -49.95V (c) 66.96V
(d) 7.426 ms, 19.23 ms (e) 25.95 ms, 40.71 ms
(f) 13.33 ms]

Questions (d) and (e) are the one of interest to me, I worked those two questions and my answers were as follows:

(d) 7.43 ms, 12.57 ms
(e) 25.95 ms, 34.05 ms

I have highlighted the discrepancies between my answers and the authors answers. Who is correct ? A cursory glance at second answer of part (e) would show an issue as the periodic cycle is 40 ms so how could it fall out of the first cycle. Is there a different way of calculating negative voltages also. Thanks, any detailed explanations / corrections would be appreciated

Welcome to the PF. Could you please show the way you solved the equation for d) and e)? That would be the quickest way for us to check your math.
 

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