AC circuit, voltage and frequency problem

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  • #1
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Homework Statement


According to Equation 20.7, an ac voltage V is given as a function of time t by V = Vo sin 2
lower_pi.gif
ft
, where Vo is the peak voltage and f is the frequency (in hertz). For a frequency of 64.7 Hz, what is the smallest value of the time at which the voltage equals one-half of the peak-value?

Homework Equations


V = Vo sin2(3.14)ft

The Attempt at a Solution


1/2Vo = Vo sin(2*3.14*64.7*t)

I know I set this equation wrong with the 1/2Vo but I don't understand how to do that part, the answer I got this way is .0738 V
 

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  • #2
Charles Link
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You need to know the angle ## \theta ## that has ##\sin(\theta)=\frac{1}{2} ##. And here, with this voltage function as a function of time, we are working in radians, so you need to write that angle ## \theta ## in radians. ## \\ ## For starters, you need to know the angle in degrees that has ## \sin(\theta)=\frac{1}{2} ##, and then convert that angle to radians.
 
  • #3
haruspex
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I know I set this equation wrong
Looks right to me.
the answer I got this way is .0738 V
Please show your working. The answer should be a period of time, not a voltage.
 
  • #4
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You need to know the angle ## \theta ## that has ##\sin(\theta)=\frac{1}{2} ##. And here, with this voltage function as a function of time, we are working in radians, so you need to write that angle ## \theta ## in radians. ## \\ ## For starters, you need to know the angle in degrees that has ## \sin(\theta)=\frac{1}{2} ##, and then convert that angle to radians.
How do you know its in radians?
 
  • #5
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You need to know the angle ## \theta ## that has ##\sin(\theta)=\frac{1}{2} ##. And here, with this voltage function as a function of time, we are working in radians, so you need to write that angle ## \theta ## in radians. ## \\ ## For starters, you need to know the angle in degrees that has ## \sin(\theta)=\frac{1}{2} ##, and then convert that angle to radians.
Never mind, I angular velocity=2pif
 
  • #6
haruspex
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How do you know its in radians?
At this level, radians would be standard as input to a trig function. A calculator would generally assume radians unless you tell it otherwise.
In the present case, you have to assume that in the given sin(2
lower_pi-gif.gif
ft)
the 2
lower_pi-gif.gif
ft
part is in radians. The presence of the "
lower_pi-gif.gif
" just about clinches it.
 

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  • #7
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Looks right to me.

Please show your working. The answer should be a period of time, not a voltage.
*Vo cancel out*
1/2= sin(406.5t)
1/2= sin(.523599)
t=406.5/.523599 = .0012880664 s
 
  • #8
haruspex
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*Vo cancel out*
1/2= sin(406.5t)
1/2= sin(.523599)
t=406.5/.523599 = .0012880664 s
Correct, but don't quote so many digits. The frequency is only specified to three sig figs.
 
  • #9
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2
I entered .001 s and I still have it wrong
 
  • #10
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I got it, the answer was supposed to be in scientific notation
 
  • #11
Charles Link
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Instead of using ## .523599=\sin^{-1}(\frac{1}{2})=\theta ## (radians) from a calculator, could you recognize that ## \theta=30^o=\frac{\pi}{6} ## radians? (and then compute ## \frac{\pi}{6} ##?) Note: ## \pi \,(radians)=180^o ##. Otherwise, it is correct. ##\\ ## ## \sin(30^o)=\frac{1}{2} ##. The angle of ## 30^o ## is well-known as having the ## \sin(30^o)=\frac{1}{2} ##. The ## \sin(45^o)=\frac{\sqrt{2}}{2} ##. Both of these trigonometric values come up time and time again. And also ## \sin(90^o)=1 ##.
 
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