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AC Waveform question

  • Thread starter newbie991
  • Start date
  • #1
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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 ive just started studying ac, just wondering how that + 60º affects the equation?

any help would be appreciated as ive 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
 

Answers and Replies

  • #2
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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).
 
  • #3
5
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Thanks, very interesting forum indeed! :)

Thanks for the reply, ive roughly plotted out this function and im getting a different answer to the theoretical value.

what would you estimate it to be?

from the forumula ive 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 im getting -11.66.

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

I know theres something right in front of me im not seeing!

would appreciate any further help!

~Newbie
 
  • #4
220
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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.
 
  • #5
5
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ah sorry thats kind of irrelevant thats the peak to peak voltage,

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

or could you tell me where im going wrong here:
60 Sin (458 (5x10-3) + 1/3 ∏)
 
  • #6
berkeman
Mentor
56,623
6,525
ah sorry thats kind of irrelevant thats the peak to peak voltage,

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

or could you tell me where im 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 = ______
 
  • #7
5
0
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 :tongue:

ok this is exactly what im entering:

rad mode:
60 sin (458(0.005) + 1/3 ∏)

and im getting: -11.66159....
 
  • #8
berkeman
Mentor
56,623
6,525
thanks for the reply :smile:

458 * 0.005 = 2.29 :tongue:

ok this is exactly what im entering:

rad mode:
60 sin (458(0.005) + 1/3 ∏)

and im 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.
 
  • #9
berkeman
Mentor
56,623
6,525
Maybe your 1/3 PI isn't turning out to be PI/3....
 
  • #10
berkeman
Mentor
56,623
6,525
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
berkeman
Mentor
56,623
6,525
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
220
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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
5
0
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 alot for your help Kruum and berkeman, much appreciated :wink:
 

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