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anonymous

  • one year ago

in a sine wave: v=100sin(200π*t+π/4) 10.7) Find the value of the greatest voltage rate of changee – you have already found when this is and this value of t can be used in the expression for the rate of change that you have also found (t=-0.00125 or 0.00125) 10.8) Using integral calculus, calculate the RMS value of the voltage

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  1. anonymous
    • one year ago
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    equation edited...big hands and little keyboards don't mix well :-(

  2. amoodarya
    • one year ago
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    hint 200(.00125)=.25=1/4 \[-.00125\leq t \leq 0.00125\\-.25\leq 200t \leq .25\\-\frac{\pi}{4}\leq 200t\pi \leq \frac{\pi}{4}\\0\leq 200t\pi +\frac{\pi}{4}\leq \frac{\pi}{2}\]

  3. amoodarya
    • one year ago
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    |dw:1437478833253:dw|

  4. amoodarya
    • one year ago
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    can you go on ?

  5. anonymous
    • one year ago
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    I thought that the greatest voltage rate of change is simply where the line crosses the x-axis and is found by 2π*frequency*amplitude = The bit that's throwing me is having to use differential calculus to find it. Derivative of 100sin(200πt+4/π) = 20,000πcos(200πt+π/4). 2π*frequency*amplitude = 62831.8 but 20,000πcos(200πt+π/4) = 62808.2 surely these should equal the same? Also just realised that my original question didn't mention using differential calculus. Sorry about that.

  6. Mrhoola
    • one year ago
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    10. 7) If you already know at what 't' the greatest rate of change occurs , then what I would do is take the derivative of the function and plug in +t. 10.8) Equation for RMS using an integral X_rms = \[\sqrt{1/T \int\limits_{0}^{T}(x)^2dt}\]

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