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RLC Circuit - AC Power Supply

  1. Jul 12, 2008 #1
    1. The problem statement, all variables and given/known data

    The figure below shows the time variation of the current through an electrical heater when it is plugged into a 240 V, 50 Hz outlet.

    1) What is the peak voltage?
    2) What is the rms value of the current drawn by the heater?
    3) What is the resistance of the heater?
    4) If t1 = 0.06 s, what is the value of t2? Note: The graph does not start at t = 0 s.

    2. Relevant equations

    1) RMS voltage value = [tex]\frac{Vo}{\sqrt{2}}[/tex] , the hint given for the problem is Use the relation between the peak voltage and the rms voltage.

    2) RMS current value = [tex]\frac{Io}{\sqrt{2}}[/tex]

    3) R = V/I

    4) Couldn't find an equation in my book. HINT: Remember the relation between frequency and the time to complete a cycle. How many cycles are between t1 and t2?


    3. The attempt at a solution

    1) 240 V/[tex]\sqrt{2}[/tex] = 169.7 Volts
    2)
    3) R = 240 V/I
    4) I know that there is 2 cycles betweem t1 and t2.
     

    Attached Files:

  2. jcsd
  3. Jul 12, 2008 #2

    alphysicist

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    Hi purduegirl,



    I can't see the attachment right now, but comparing your equation for #1 and your work for #1, what numerical value does the rms voltage have? What does the variable [itex]V_0[/itex] stand for? If you think about why it is called peak voltage, you'll see why your answer can't be true.
     
  4. Jul 12, 2008 #3
    I am completely confused about RMS values. My book doesn't have a clear explanation and our prof. didn't lecture over it. I have been trying to look up information about RMS.

    I found this equation: e = emaxsin(wt) But as in my LR circuit post, I don't know what to do about the time here.
     
  5. Jul 12, 2008 #4

    alphysicist

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    Well, I was referring to your equation. You have

    [tex]
    V_{\rm rms} = \frac{V_0}{\sqrt{2}}
    [/itex]

    but when you did the calculation you switched the two voltage values. (If it helps, I've always thought of the equation as:

    [tex]
    V_{\rm rms} = \frac{V_{\rm max}}{\sqrt{2}}
    [/itex]




    But here is one practical way of looking at why the rms value is important. For a normal US outlet, the voltage varies from about +170 V to -170 V; which means it's always going back and forth between those two values.

    But when it comes to calculating the power used by whatever you plug in there, do you just use P = I V, with V=170 volts? The power companies would like that, since you pay for the amount of power used!

    But the point is, the voltage only reaches 170V for an instant, so that's really overcalculating the power--most of the time you are not near 170V. What is needed is a sort of "averaged" voltage--and by averaged I mean that a constant voltage of [itex]V_{\rm rms}[/itex] connected to a resistor uses the same amount of power as a varying voltage of [itex]V_{\rm max}[/itex]. And that's why the average power formula for a pure resistor AC circuit is [itex]P=I_{\rm rms}V_{\rm rms}[/itex].

    So for a normal outlet, the peak voltage is about 170V, and the rms voltage is about 120V. Then the power used by plugging in a 1ohm resistor to the outlet with a peak voltage of 170V, is the same power used by connecting the 1ohm resistor to a 120V battery.
     
    Last edited: Jul 12, 2008
  6. Jul 12, 2008 #5
    Well, looking at this problem again, I am given the Vrms = 240 , the frequency = 50Hz, and I calculated w = 314.16 Hz.

    For part one, I am supposed to use the picture (which is now available) I know that the Irms = Amplitude/1.41 , 1.41 being the sqrt root of 2.
     
  7. Jul 12, 2008 #6

    alphysicist

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    Right, so you can calculate [itex]V_{\rm max}[/itex] now that you have identified [itex]V_{\rm rms}[/itex]. Did you get the right answe?


    You mean part two, right? But that looks right. Just find the amplitude from the graph and you're done.
     
  8. Jul 12, 2008 #7
    Okay, got those, but I'm having problems with the last problem. I looked at this again and can see only 1 cycle has occured between t1 and t2. But, from there I don't know what to do.
     
  9. Jul 12, 2008 #8

    alphysicist

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    No, it's not 1 cycle. Think of cycles corresponding to wavelengths. How many wavelengths of the sine function are between t1 and t2?
     
  10. Jul 12, 2008 #9
    There are two cyles.
     
  11. Jul 12, 2008 #10

    alphysicist

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    No; picture what a sine curve looks like (since t1 starts at zero), and what one wavelength looks like (going from 0 to 360 degrees). How many wavelengths would fit from t1 to t2? It's not an integer.
     
  12. Jul 13, 2008 #11
    after comparing it to a figure in my book, I believe there is 3/2 of a cycle.
     
  13. Jul 13, 2008 #12

    alphysicist

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    That's right. So you know that there is 3/2 of a cycle between the times t1 and t2. So can you find the time of one cycle with what they've given you in the problem? (What is the term that means the time to complete one oscillation?)
     
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