AC circuit, voltage and frequency problem

In summary, an ac voltage V is given as a function of time t by V = Vo sin 2ft, where Vo is the peak voltage and f is the frequency (in hertz). For a frequency of 64.7 Hz, the smallest value of the time at which the voltage equals one-half of the peak-value is .0012880664 s.
  • #1
36
2

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
 

Attachments

  • lower_pi.gif
    lower_pi.gif
    84 bytes · Views: 736
Physics news on Phys.org
  • #2
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
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.
 
  • #4
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?
 
  • #5
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
 
  • #6
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
lower_pi-gif.gif
ft
part is in radians. The presence of the "
lower_pi-gif.gif
" just about clinches it.
 

Attachments

  • lower_pi-gif.gif
    lower_pi-gif.gif
    84 bytes · Views: 286
  • lower_pi-gif.gif
    lower_pi-gif.gif
    84 bytes · Views: 314
  • lower_pi-gif.gif
    lower_pi-gif.gif
    84 bytes · Views: 305
  • #7
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
 
  • #8
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.
 
  • #9
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 ##.
 
Last edited:

1. What is an AC circuit?

An AC circuit is a type of electrical circuit that uses alternating current (AC) to transfer electricity. In an AC circuit, the current and voltage alternate between positive and negative values, unlike in a direct current (DC) circuit where the current flows in one direction.

2. What is voltage in an AC circuit?

Voltage in an AC circuit refers to the potential difference between two points in the circuit. It is measured in volts and determines the amount of energy that can be transferred from the power source to the load. In an AC circuit, the voltage constantly changes direction and magnitude, making it more complex than in a DC circuit.

3. How does frequency affect an AC circuit?

Frequency is the number of cycles per second in an AC circuit. It determines the speed at which the current alternates and is measured in Hertz (Hz). In an AC circuit, the frequency is directly proportional to the voltage and inversely proportional to the wavelength. A higher frequency means a faster alternating current and vice versa.

4. What is the difference between series and parallel AC circuits?

In a series AC circuit, the components are connected in a single loop, with the same current flowing through each component. In a parallel AC circuit, the components are connected in multiple branches, with the same voltage applied to each component. The total current in a parallel circuit is equal to the sum of the currents in each branch, while in a series circuit, the current is the same throughout.

5. How do you calculate the impedance in an AC circuit?

Impedance in an AC circuit is the total opposition to the flow of current. It is a combination of resistance and reactance (resistance to the changing current) and is measured in ohms. To calculate impedance, you can use the formula Z = √(R^2 + X^2), where R is the resistance and X is the reactance.

Suggested for: AC circuit, voltage and frequency problem

Back
Top