AC circuit, voltage and frequency problem

AI Thread Summary
The discussion focuses on solving an AC circuit problem involving voltage as a function of time, specifically determining the time at which the voltage equals one-half of the peak voltage for a frequency of 64.7 Hz. The equation V = Vo sin(2πft) is used, and participants emphasize the need to convert the angle where sin(θ) = 1/2 into radians. The correct approach involves recognizing that θ corresponds to 30 degrees or π/6 radians, leading to the calculation of time using t = θ/(2πf). There is a clarification that the final answer should be expressed in scientific notation and with appropriate significant figures. The conversation highlights the importance of understanding trigonometric functions and their applications in AC circuit analysis.
Alice7979
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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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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.
 
Alice7979 said:
I know I set this equation wrong
Looks right to me.
Alice7979 said:
the answer I got this way is .0738 V
Please show your working. The answer should be a period of time, not a voltage.
 
Charles Link said:
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?
 
Charles Link said:
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
 
Alice7979 said:
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
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ft
part is in radians. The presence of the "
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" just about clinches it.
 

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haruspex said:
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
 
Alice7979 said:
*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.
 
I entered .001 s and I still have it wrong
 
  • #10
I got it, the answer was supposed to be in scientific notation
 
  • #11
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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