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In the book the average value is given but as it is the dc component i think it should be the rms value? Input is sine wave of amplitude 10V.half wave diode rectifier

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The output waveform \$v_0\$ of a half wave rectifier consists of the positive pulses of the input signal, less a bit for the diode drop.

The DC component of that is defined as its average. That's the DC value that would appear after a while on the output of a lowpass filter connected to \$v_0\$.

The RMS value of \$v_0\$ is defined as the DC value that gives the equivalent heating in the load resistor.

The difference is small but significant for modestly rippling waveforms like this. There is no difference if \$v_0\$ is DC. The difference increases as the waveform becomes more spiky.

One way to see what's happening, which I understand not everybody will grok, is that the RMS value of the waveform measures the total power that's available in all its components, AC and DC. The average value measures the power of only the DC component. The difference between the DC value and the RMS value will therefore be a measure of the power available in the non-DC components, that is the AC power of the waveform. This means that we can say that \$ RMS \ge averageDC\$, always.

If you were to connect a capacitor across the output resistor, it would hold the peak voltage of \$v_0\$, which is larger than the RMS value of the input voltage \$v_I\$, a fact that often confuses the uninitiated.

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For a sinewave input: -

A perfect half wave rectifer provides only half the power to the load (compared to a straight load connection) so, square the RMS value of the sinewave, divide it by 2 and then take the square root. RMS(half wave) = \$\dfrac{V_{PEAK}}{2}\$.

For the average you need to integrate like this: -

enter image description here

As you can see the RMS and the average do not mathematically coincide for a sinewave.

For other input waveforms there can be equality between RMS and average or great inequalities.

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