0

726views

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

written 3.6 years ago by teamques10 ★ 68k

modified 2.8 years ago by pedsangini276 • 4.8k

basic electrical engineering

ADD COMMENT FOLLOW SHARE EDIT

1 Answer

0

17views

written 3.6 years ago by teamques10 ★ 68k

$\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 SHARE EDIT