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

Energy dissipated ` = int_(0)^(t) (E^(2))/(R ) dt`
Where `t = 1 ms = 10^(-3) s`
`= int_(0)^(t) ([(12) sin 250 pi t]^(2) dt)/(100)`
`= (144)/(100) int_(0)^(t) ((1 - cos 2 xx 250 pi t) dt)/(2)`
`= (144)/(200) [1 - (sin 500 pi t)/(500 pi)]_(0)^(t)`
As `t = 1 ms = 10^(-3) s,
`:. Energy dissipated ` = (144)/(200) [10^(-3) - (1)/(500 pi) ]`
`= 2.61 xx 10^(-4) J`