Find the effective value of current if its mean value is `I_(0)` and its time dependence is
`(a)` shown in figure,
` (b)` `I~| sin omegat|.`

Explanation

(a) `I(t)=I_(1)(t)/(T)0 le t lt T`
`I(t+-T)=I(t)`
Now mean current
`lt I gt =(1)/(T)int_(0)^(T)I_(1)(t)/(T)dt=I_(1)(T_(1)^(2)//2)/(T^(2))=I_(2)//2`
Then , `=I_(1)=2I_(0)` since `lt I gt = I_(0)`
Now mean square current `lt I^(2) gt`
`=4I_(0)^(2)(1)/(T) int _(0)^(T)(t^(2))/( t^(2))dt=(4I_(0)^(2))/( 3)`
so effective current `=(2I_(0))/( sqrt(3))`.
`(b)` In this case `I=I_(1)|sin omegat|`
and `I_(0)=(1)/(T)int_(0)^(T)I_(1)|sin omegat|dt`
`=(1)/(2pi) I_(1)int _(0)^(2pi) | sin theta| d theta=(I_(1))/( pi)int _(0)^(pi)sin theta d theta=(2I_(1))/( pi)`
So `I_(1)=(pi l_(0))/( 2)`
Then , mean square current`= lt I^(2) gt =(pi^(2) I_(0)^(2))/( 4T)int _(0)^(T) sin ^(2) omegat dt`
`=(pi^(2)I_(0)^(2))/(4)xx(1)/(2pi)int_(0)^(2pi)sin^(2) theta d theta=(pi^(2) I_(0)^(2))/( 8)`
so effective current `=(pi I_(0))/( sqrt(8))`