Question

How here sin²ωt is 1/2 ?

Answer 1

$\Large\mathfrak{\underline{\underline{Answer:-}}}$

$\Large\bold{\Rightarrow sin^2\omega\:t= \frac{1}{2} }$

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$\small\mathfrak{\underline{\underline{Step-By-Step\:Explanation:-}}}$

As we know from the graph of Sin²x

$\Large\bold{\lim\limits_{0\to\pi}sin^2\omega\:t = [0 → 1]}$

➛Since we always consider from $\Large\bold{0-\pi}$for sine curve

➛ Average value is taken always in one time period is 0—T which is between $\Large\bold{0-\pi}$

$\Large\boxed{Avg= \frac{max - min}{2}}$

$\Large\bold{\therefore sin^2\omega\:t= \frac{1-0}{2} }$

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Hence, Average value of

$\Large\bold{\Rightarrow sin^2\omega\:t= \frac{1}{2} }$

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Note- u can also find the Irms by Integration method.

Answer 2

$\huge{♡Answer♡}$

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