How does the presence of a diode affect the average power in a circuit?

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The presence of a diode in a circuit with alternating current alters the average power delivered to a resistor. In the discussed scenario, with a maximum voltage of 6.0 V and a resistance of 100 ohms, the average power calculated is 0.18 W, while the maximum power reaches 0.36 W. When a diode is connected, the power graph shows a sinusoidal pattern, with the first half-cycle exhibiting a peak of 3.6 W and the second half-cycle at zero. The average power for a half-rectified wave is determined to be half that of the full wave, leading to a formula where the mean power of the half-rectified wave is one-fourth of the maximum power. Thus, the average power in the circuit is effectively reduced due to the diode's presence.
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[SOLVED] Average power and diodes.

Hi,

I have this question. If a diode were connected to a circuit with an alternating current and there was a resistor was connected to it, will the average power change? For example, if we take the following values,

V max = 6.0 V
Resistance = 100 ohms,

the Average power will be 0.18 W and the max power 0.36W. When we plot the graph with respect to power dissipated by the resistor with a diode connected, we get a sinusoidal graph with it's second half cycle at a null value and it's first half at a peak value of 3.6W (assuming that the diode is forward biased at the start).

I'm not very sure how to calculate max power using graphs; but is the average power still 0.18W?

I thank you for your help in advance.
 
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Anyway, I think I solved it. Thanking the Half Rectified wave to be H and the original unrectified full wave to be F, the mean power of the half rectified wave should be half that of the full wave.

Therefore,

<P_{H}> = 1/2<P_{F}>
= 1/2(P_{o}/2)
= 1/4 P_{o}

where P_{o} is the max power.
 
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