Question

Find the effective value or rms value (in ampere) of an alternating current in one time period that changes according to the given law: (All quantities are in S.I. unit and symbols have their usual meaning)
$I=10$, when $0<t<T/8;I=0$, when $\frac{T}{8}<t<\frac{T}{2}$
$I=-10$, when $\frac{T}{2}<t<\frac{5}{8}T$
$I=0$, when $\frac{5}{8}T<t<T$
$I=10$, when$T<t<\frac{9}{8}T$

Answer 1

Here, one time period is $0$ to $T$.
Thus,$I_{rms}^2=\frac{{\int }_0^T{I}^{2}dt}{{\int }_0^{T}dt}$
$I_{rms}^2=\frac{1}{T}[{\int }_0^{T/8}{I}^{2}dt+{\int }_{T/8}^{T/2}{I}^{2}dt+{\int }_{T/2}^{5T/8}{I}^{2}dt+{\int }_{5T/8}^T{I}^{2}dt]$
$=\frac{1}{T}\left[\right(10{)}^2(T/8-0)+0+(-10{)}^{(}5T/8-T/2)+0]$
$=100(1/8+5/8-1/2)=25$
$\therefore I_{rms}=5A$