Question

 Derive the relationship between the peak and the rms value of current in an a.c. circuit.

Answer 1

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)`

Then,

`<sin^2 omegat > =1/2 (1- <cos2omegat>)`

The average value of cos 2 ωt is zero.

We have:

`< sin^2 omegat  > =1/2 (1-0)`

`< sin^2 omegat> =1/2`

Thus,

`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 I_rms.

Peak current is `i_m`

Therefore,

`I = (i_m)/sqrt2 =0.707 i_m`

`I^2 R = (i_m^2)/2 R`

`I = i_m/sqrt2`