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After rectifying an AC sine wave (full-wave or half-wave), we get Vdc, Idc, V Rms and I Rms.
However, if we want to calculate the power dissipated in a load, we use the RMS values, so why we used the RMS values instead of the DC values?? and what the difference between AC and DC power??
And the definition of the efficiency is not clear, what's the meaning of the ratio between DC power and AC power??

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You are talking about two different powers and you are neglecting the losses.

The Power Out (or load), for a rectifier, is DC, and would be given by the DC equation \$P_{dc} = V_{dc} * I_{dc}\$ The Power In for a rectifier, is AC, and would be given by the AC equation \$P = V_{rms} * I_{rms} * pf\$

The difference between the two numbers are your losses: \$P_{out} = P_{in} - P_{losses}\$ or \$P_{losses} = P_{in} - P_{out}\$

Your efficiency, in any system, DC, AC or rectifier is always the ratio of your output to input. \$\eta = P_{out} / P_{in}\$

Your load power for the rectifier should be using the DC values, not AC RMS, that should be your input power.

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if we want to calculate the power dissipated in a load, we use the RMS values, so why we used the RMS values instead of the DC values?

If you had a lightbulb fed from AC mains taking 100 watts but then fed it via a perfect bridge rectifier, the lightbulb would still take 100 watts. If you don't grasp this yet you will probably want to think about it until you do.

This is substantiated by the fact that the RMS value of an AC waveform is numerically the same as the RMS value of said waveform when fed through a perfect bridge rectifier.

Another fact: the average value of the output of a bridge rectifier when fed a sinewave is NOT the same as the RMS value: -

enter image description here

And the definition of the efficiency is not clear, what's the meaning of the ratio between DC power and AC power?

I have no idea what the ratio is - maybe you can give a better indication of what you mean?

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