Derive the relationship between the peak and the rms value of current in an a.c. circuit.
The instantaneous power dissipated in the resistor is`p = i^2 R = i_m^2 sin^2 omegat`
The average value of p over a cycle is:
`p = <i^2R> =<i_m^2 R sin^2 omegat>`
`i_m^2 and R` are constants. Therefore,
`p = i_m^2 R<sin^2 omegat>`
By trigonometric identity,
`sin^2 omegat = 1/2 (1-cos2 omegat)`
`<sin^2 omegat > =1/2 (1- <cos2omegat>)`
The average value of cos 2 ωt is zero.
`< sin^2 omegat > =1/2 (1-0)`
`< sin^2 omegat> =1/2`
`P = 1/2 i_m^2`
The rms value in the ac power is expressed in the same form as dc power root mean square or effective current and is denoted by Irms.
Peak current is `i_m`
`I = (i_m)/sqrt2 =0.707 i_m`
`I^2 R = (i_m^2)/2 R`
`I = i_m/sqrt2`