Question

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

Answer 1

equation edited...big hands and little keyboards don't mix well :-(

Answer 2

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

Answer 3

can you go on ?

Answer 4

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.

Answer 5

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 = \[\sqrt{1/T \int\limits_{0}^{T}(x)^2dt}\]