What is root mean square value of alternating current? Derive a relation between peak value and virtual value of alternating current.

Explanation

Root mean square value of alternating current : It is that steady value of current which produces same amount of heat in a given resistance as is alone by the a.c. when a.c. passes through the same resistance for same time.
Let at any time, alternating current flowing through the circuit is given by
`I=I_0 sin omegat " "..(i)`
Let this current flows through a conductor of resistance R for time dt.
Then amount of heat produced in the conductor is given by
`dH =I^(2)Rdt`
`dH = (I_0 sin omegat)^(2) R dt `[Using eq.(i)]
`dH = I_0^2 R sin^(2) omegat dt `
Total heat produced in one complete cycle i.e., in T is given by
` intdH=int_0^(T)I_0^2Rsin^2omegat dt`
`H=I_0^2Rint_0^(T) sin^2 omegat dt `
But `sin^2 omegat = (1-cos2omegat)/(2)`
`implies H = I_0^2 Rint_0^T ((1-cos2omegat)/2)dt`
or `H = (I_0^2 R)/2[int_0^T1dt -int _0^T cos 2omegat dt ]`
`(I_0^2 R)/2[|t|_0^T-|(sin2omegat)/(2omega)|_0^T]`
`(I_0^2 R)/2[(T-0)-((sin2omegaT)/(2omega)-sin0^@)]`
`(I_0^2 R)/2[(T-0)-(sin2xx2pi)/(2omega)]" " [:. omega T = 2pi]`
`H=(I_0^2RT)/(2) " "...(ii)`
If `I_(r.m.s)` the root mean square value of the current, flows through the conductor or resistance R for time T, then heat produced in the conductor.
`H =I_(r.m.s)^(2)RT " "..(iii)`
(iii) Comparing eqns. (ii) and (iii), we get
`I_(r.m.s)^(2)RT= (I_0^2RT)/(2)`
or `I_(r.m.s)=I_0/(sqrt2)=0.707I_0`
Hence, the c.m.s. value or effective value or virtual value of a.c. is 0.707 times the peak value of a.c. i.e., 70.7% of the peak value of a.c.