(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A sinusoidal voltage waveform is rectified so that it becomes a half-wave waveform (i.e. the bottom half has been "chopped off"). Determine:

i) The average p.d.

ii) The RMS p.d.

iii) The heat dissipated in a 200ohm resistor in 100s

2. Relevant equations

[itex]P_{avg} = \frac{(V_{rms})^2}{R}[/itex]

3. The attempt at a solution

I'm not sure, but I would guess that the average p.d. is the line that cuts the rectified waveform in such a way that the area above and below is the same. I have a vague recollection about it being [itex]\frac{2}{\pi}[/itex] times the peak voltage.

I am also quite uncertain about the RMS, how would the voltage being zero half the time affect this value? I'm guessing it will no longer be [itex]\frac{\sqrt2}{2} V_{peak}[/itex].

Please could anyone give me some clues on how to work out the RMS and average values?

EDIT:Oops sorry I did not mean to post the same thread twice...just ignore the other one!

**Physics Forums - The Fusion of Science and Community**

# Rectified Voltage Waveform

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: Rectified Voltage Waveform

Loading...

**Physics Forums - The Fusion of Science and Community**