A resistor of resistance `100 Omega` is connected to an AC source `epsilon = (12 V) sin (250 pi s^(-1))t`. Find the energy dissipated as heat during `t = 0` to `t = 1.0 ms`.

Explanation

Given `H=V^2/RT`,
`E_0 = 12 V`,
`omega = 250 pi`,
`R= 100 Omega`
`H=int_(0)^RdH`
`=int(E_(0)^2 sin^2 omegat)/(R)dt`
`=144/100intsin^2omegat dt`
`=1.44 int((1-cos2omegat)/(2))dt`
`=1.44/2[int_(0)^(10^-3)dt + int_(0)^(10^-3)cos2 omegatdt]`
`=0.72[10^(-3) - {(sin^2 omegat)/(2omega)}_0^(10^-3]]`
`=0.72 [(1)/(1000) - (1)/(500pi)]`
`=0.72[(1)/(1000) - (2)/(1000pi)]`
`=((pi-2)/(1000pi)) xx 0.72`
`=0.0002614 = 2.61 xx 10^-4 J`