Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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# Derive the Relationship Between the Peak and the Rms Value of Current in an A.C. Circuit. - Physics

ConceptAlternating Currents

#### Question

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

#### Solution

The instantaneous power dissipated in the resistor isp = 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 Irms.

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

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Solution Derive the Relationship Between the Peak and the Rms Value of Current in an A.C. Circuit. Concept: Alternating Currents.
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