Question #187524

For the function v = 40sin calculate the mean and rms over a range of


1
Expert's answer
2021-05-07T10:24:38-0400

Let function is v=40sinωtv = 40sin\omega t


Mean value over interval 0 to T where ω=2πT\omega = \frac{2 \pi}{T}


Vmean=0Tvdt0Tdt=0T40sinωtdt0Tdt=40ω[cosωt]0Tt0TV_{mean} = \frac{\int_0^T vdt}{\int_0^T dt} = \frac{\int_0^T 40sin\omega t dt}{\int_0^T dt} = \frac{\frac{40}{\omega}[-cos\omega t]_0^T}{t_0^T}

Vmean=40[cosωTcos0]ωT=40ωT[cos2πcos0]=40ωT[11]=0V_{mean} = \frac{ -40[cos\omega T - cos0]}{\omega T} = -\frac{40}{\omega T}[cos2\pi - cos0] = -\frac{40}{\omega T}[1-1] = 0



RMS value in interval 0 to T,

Vrms=0Tv2dt0TdtV_{rms} = \sqrt{\frac{\int_0^T v^2 dt}{\int_0^T dt}}

Vrms=0T1600sin2ωtdt0Tdt=0T1600(1cos2ωt)2dt0TdtV_{rms} = \sqrt{\frac{\int_0^T 1600sin^2\omega t dt}{\int_0^T dt}} = \sqrt{\frac{\int_0^T 1600 \frac{(1-cos2\omega t)}{2}dt}{\int_0^T dt} }

Vrms=800(t12ωsin2ωt)0Tt0T=800(T12ω(sin2ωTsin0))TV_{rms} = \sqrt{\frac{ 800(t - \frac{1}{2\omega}sin2\omega t)_0^T }{t_0^T}} = \sqrt{\frac{ 800(T - \frac{1}{2\omega}(sin2\omega T - sin0)) }{T} }

Vrms=800TT=800=28.284V_{rms} = \sqrt{\frac{800T}{T}} = \sqrt{800} = 28.284



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