Derive an expression for the average value of a sinusoidally varying current in terms of Peak Value.

527views

written 2.9 years ago by

teamques10 ★ 64k

modified 2.1 years ago by

pedsangini276 • 4.7k

basic electrical engineering

ADD COMMENT EDIT

1 Answer

10views

written 2.9 years ago by

teamques10 ★ 64k

$\displaystyle I \ avg = \dfrac{1}{\Pi} \int_{0}^{\Pi} Im \ Sin \ \theta \ d \theta \Rightarrow \dfrac{Im}{\Pi} \left [ -cos \theta \right ]_0^\Pi\[2ex]

I \ avg = \dfrac{Im}{\Pi} [1-(-1)] \Rightarrow I\ avg =0.637 Im$

ADD COMMENT EDIT

Please log in to add an answer.