AC power vs DC power

[iminbglr]

Hi ,

Given a half waveform , the Average DC value is given by I/Pie and DC power generated at load = (I/pie)^2 * R
AC power at load = (I/2)^2 * R

so AC power is greater than DC power .

I am not able to understand , since the waveform is same in both case ..how can its power be different?

Please throw some light on this ..may be its basic question ..but I am completely confused how can the power generated be different ?

Regards
carol

 

[albbg]

Why do you consider the mean power of an half waveform ? It's not correct.

In AC your voltage is Vac=A*sin(wt), then the istantaneous power delivered to a load R will be:

Pac=Vac²/R=A²*sin²(wt)/R, thus the mean power is given by:

<Pac>=1/T*∫A²*sin²(wt)/R*dt (integral over a period T)

the term √(1/T*∫A²*sin²(wt)*dt) is known ad the RMS voltage. For a sinusoid (like this case) its value is Vrms=A/√2

then Pac=Vrms²/R=A²/(2*R)

 

[iminbglr]

Why do you consider the mean power of an half waveform ? It's not correct.

In AC your voltage is Vac=A*sin(wt), then the istantaneous power delivered to a load R will be:

Pac=Vac²/R=A²*sin²(wt)/R, thus the mean power is given by:

<Pac>=1/T*∫A²*sin²(wt)/R*dt (integral over a period T)

the term √(1/T*∫A²*sin²(wt)*dt) is known ad the RMS voltage. For a sinusoid (like this case) its value is Vrms=A/√2

then Pac=Vrms²/R=A²/(2*R)

Thanks albbg for making me understand the concept right . the half wave rectifier output is unidirectional and thus has half wave form , it cant be considered as AC since unidirectional , so calculating AC power of this wrong .

Thanks again !