Understanding AC Waveforms: How Does Phase Shift Affect Voltage?

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Discussion Overview

The discussion revolves around understanding the effects of phase shift in AC waveforms, specifically analyzing the voltage of a given waveform equation at a certain time. Participants explore the implications of the phase shift and the calculations involved in determining the voltage output.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a waveform equation and seeks clarification on the impact of the phase shift of +60º on the voltage calculation.
  • Another participant explains that the phase shift simply shifts the angle in the sine function but does not provide a definitive method for calculating the voltage.
  • A participant shares their calculated values and expresses confusion over discrepancies between their results and theoretical expectations.
  • There is a discussion on the peak-to-peak voltage and its relevance to the calculations, with some participants questioning the interpretation of this value.
  • Multiple participants attempt to verify calculations, with one noting a potential error in the use of radians versus degrees in their calculations.
  • Some participants arrive at the same voltage result of -11.66V, while others express uncertainty about the correctness of this value based on the original problem's parameters.
  • There is a suggestion that if the problem was indeed in degrees, it would be indicated in the angular frequency notation.

Areas of Agreement / Disagreement

Participants express differing views on the correctness of the calculated voltage, with some agreeing on the value of -11.66V while others remain uncertain about the calculations and the interpretation of the phase shift. The discussion does not reach a consensus on the method or the final answer.

Contextual Notes

There are unresolved issues regarding the use of degrees versus radians in calculations, and participants note potential errors in their computational steps without reaching a definitive conclusion on the correct approach.

newbie991
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Homework Statement



v = 60sin(458t + 60º)

Find the voltage of the waveform when t = 5ms

It may be a silly question but I've just started studying ac, just wondering how that + 60º affects the equation?

any help would be appreciated as I've looked everywhere for the method...


Homework Equations



V = VmSin(2πft)

The Attempt at a Solution



v = 60sin(458t + 60º)

=> v=60sin(458(5x10-3) + 60º)

=> v=60(0.039957 + sin60º) ? -----> not sure if this is correct


I would appreciate any reply :smile:




~Newbie
 
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Welcome to PF, newbie991!

The 60º only sifts the angle, well 60 degrees. You've probably done trigonometry, where sin(∏+∏)=sin(2∏). You just got to be careful, though. Here the voltage is in the form v=60sin(ωt+60º).
 
Thanks, very interesting forum indeed! :)

Thanks for the reply, I've roughly plotted out this function and I am getting a different answer to the theoretical value.

what would you estimate it to be?

from the forumula I've extracted:

vmax=60 p-p = 120
f=72.892Hz
T=13.71ms

assuming all that is correct, i found v after 5ms to be between -40 and -20, although I am getting -11.66.

60 Sin (458 (5x10-3) + 1/3 ∏) = -11.66

I know there's something right in front of me I am not seeing!

would appreciate any further help!

~Newbie
 
newbie991 said:
vmax=60 p-p = 120

I'm not quite following you here. What does the last part, p-p = 120, mean?

When I graphically checked the answer, I got the same as plugging in the values to the given signal.
 
ah sorry that's kind of irrelevant that's the peak to peak voltage,

could you tell me what your answer was so i can see if I am on the right track?

or could you tell me where I am going wrong here:
60 Sin (458 (5x10-3) + 1/3 ∏)
 
newbie991 said:
ah sorry that's kind of irrelevant that's the peak to peak voltage,

could you tell me what your answer was so i can see if I am on the right track?

or could you tell me where I am going wrong here:
60 Sin (458 (5x10-3) + 1/3 ∏)

The way the problem was originally listed, it looks to all be in degrees. You just made an error with your calculator in multiplying 5ms * 458:

=> v=60sin(458(5x10-3) + 60º)

=> v=60(0.039957 + sin60º) ? -----> not sure if this is correct

Re-check 458 * 0.005 = ______
 
berkeman said:
The way the problem was originally listed, it looks to all be in degrees. You just made an error with your calculator in multiplying 5ms * 458:

=> v=60sin(458(5x10-3) + 60º)

=> v=60(0.039957 + sin60º) ?

Re-check 458 * 0.005 = ______
__________________

thanks for the reply :smile:

458 * 0.005 = 2.29 :-p

ok this is exactly what I am entering:

rad mode:
60 sin (458(0.005) + 1/3 ∏)

and I am getting: -11.66159...
 
newbie991 said:
thanks for the reply :smile:

458 * 0.005 = 2.29 :-p

ok this is exactly what I am entering:

rad mode:
60 sin (458(0.005) + 1/3 ∏)

and I am getting: -11.66159...

sin(458 * 0.005) = 0.752

sin(PI/3) = 0.866

sin(458 * 0.005 + PI/3) = sin(2.29 + 1.047) = sin(3.337) = sin(3.337 - 3.142) = sin(0.195)

So the answer cannot be negative. Try doing the calculation in a different order, and also do the individual pieces to check your work.
 
Maybe your 1/3 PI isn't turning out to be PI/3...
 
  • #10
berkeman said:
sin(458 * 0.005) = 0.752

sin(PI/3) = 0.866

sin(458 * 0.005 + PI/3) = sin(2.29 + 1.047) = sin(3.337) = sin(3.337 - 3.142) = sin(0.195)

So the answer cannot be negative. Try doing the calculation in a different order, and also do the individual pieces to check your work.

Oops, my bad. I subtracted only one PI inside the sine... that's not right. Give me a sec...
 
  • #11
sin(3.337) = -0.194

60 * sin(3.337) = -11.65

Are you sure that's not the right answer? If it's not, then maybe the whole original problem was in degrees after all.

Is the answer 60sin(2.29 + 60) = 53.1 ?
 
  • #12
I get the same V=-11.66V as well. Is this an online assignment, or how do you know it's not the right answer?

If the problem was in degrees, I would assume the angular frequency would have the degree symbol added. After all, rad/s is the more commonly used unit.
 
Last edited:
  • #13
Kruum said:
I get the same V=-11.66V as well. Is this an online assignment, or how do you know it's not the right answer?

If the problem was in degrees, I would assume the angular frequency would have the degree symbol added. After all, rad/s is the more commonly used unit.

that is the correct answer :wink:

went over everything again and -11.66 is correct :biggrin:
was just a silly mistake on my part!

thanks a lot for your help Kruum and berkeman, much appreciated :wink:
 

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