Question

The average half-cycle value of a sine wave with a $40V$ peak is

• A. $25.48V$
  
• B. $6.37V$
  
• C. $14.14V$
  
• D. $50.96V$
  

Answer 1

Answer: (A)

$25.48V$

The source voltage $V=V_o\sin wt$

Average half-cycle value of sine wave $V_{avg}=\frac{{\int }_o^{T/2}{V}_o\sin \left(wt\right)dt}{T/2}$

Or $V_{avg}=\frac{V_o}{T/2}\times {\int }_o^{T/2}\sin wtdt$

Or $V_{avg}=\frac{V_o}{T/2}\times \frac{\cos wt}{w}{|}_o^{T/2}$

Or $V_{avg}=\frac{V_o}{T/2}\times \frac{\cos \frac{wT}{2}-1}{w}$

Or $V_{avg}=\frac{V_o}{T/2}\times \frac{-1-1}{w}$

Or $V_{avg}=\frac{V_o}{\pi }\times (-2)$

$⟹$ $|V_{avg}|=\frac{2V_o}{\pi }$

We get $|V_{avg}|=\frac{2\times 40}{\pi }=25.48$ volts