Show that `I_(rms)=(I_(0))/(sqrt(2))`, where `I_(rms)` is the root mean square value of alternating current and `I_(0)` the peak value.

Explanation

Let the alternating current is `I=I_(0)sin omegat`
Mean square value of current,
`I^(2)=(1)/(T)int_(0)^(T)I^(2)dt` for a complete cycle
`=(1)/(T)int_(0)^(T)I_(0)^(2)sin^(2)omegatdt=(I_(0)^(2))/(T)int_(0)^(T)((1-cos2omegat)/(2))dt`
`=(I_(0)^(2))/(2T)[t-(sin2omegat)/(2omega)]_(0)^(T)=(I_(0)^(2))/(2T)[T-0]=(I_(0)^(2))/(2)`
Now by taking roots of mean square value of current
i.e, `I_("run")=sqrt(I^(2))=sqrt(I_(0)^(2)//2)=(I_(0))/(sqrt(2))`
where, `I_(rms)` = root of mean square value,
`I_(0)` = peak value of current.