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I have a DC source where DC Voltage = Vdd and the current supplied by this source is rectified sine wave to a circuit. Half wave rectified current

Now I would like to find the average power delivered by DC source. There are two ways of doing it $$ P_\mathrm{avg1} = \frac{1}{T} \int_{0}^{T} v(t) i(t) \, \mathrm{d}t $$ $$ P_\mathrm{avg2} = V_\mathrm{rms}\times I_\mathrm{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} v^2(t) \, \mathrm{d}t} \times \sqrt{\frac{1}{T} \int_{0}^{T} i^2(t) \, \mathrm{d}t} $$

Where 'T' is one complete period of \$ I(t) = I_p sin(t) \$ and \$ V(t) = vdd \$ Now the average power delivered in both cases will be \$ P_{avg1} = vdd \times \frac{I_p} {T/2} \$ and in the second case \$ P_{avg2} = vdd \times \frac{I_p} {2} \$

Which one is correct and why?

According to me the average way is correct, because the average of instantaneous voltage times current should give you the average power. But then I cannot understand why RMS way is wrong! There is plenty of literature on RMS and average power but they generally deal with either sinusoidal or square waves voltages and currents.

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  • \$\begingroup\$ With your calculation, Pavg1 is 0. \$\endgroup\$ – anhnha Sep 8 '17 at 18:19
  • \$\begingroup\$ As you use instantaneous (time-dependent) variables, you should lower-case them: \$i(t)\$ and \$v(t)\$. As stated in the answers, average power equals the product of rms variables when the power factor is unity. Integrating the instantaneous power is the way to go and works for all waveforms. \$\endgroup\$ – Verbal Kint Sep 9 '17 at 7:23
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The source of the confusion here is from the second equation you list:

$$P_\mathrm{avg2} = V_\mathrm{rms}\times I_\mathrm{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} V^2(t) \, \mathrm{d}t} \times \sqrt{\frac{1}{T} \int_{0}^{T} I^2(t) \, \mathrm{d}t}$$

This equation is true under the assumption that both the voltage and currents are sinusoidal and in phase. The derivation works out that way, so it is a simple way to calculate average power under that specific circumstance. The instantaneous method you used is the general case, and should work under all circumstances.

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  • \$\begingroup\$ Hi, Thank you for the answer, I thought the same, but i needed verification. Now its all clear to me. \$\endgroup\$ – RAN Sep 11 '17 at 8:31
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Power only equals \$V_{RMS} \times I_{RMS}\$ when both quantities are in phase or both quantities are DC. Put a DC ammeter in series with any load on a DC voltage supply and you can, with complete assurance, say \$V_{DC} \times I_{DC}\$ = power taken by the load.

So, if you have a 10 volt DC power source and you have a load that takes 2 amp DC from it, the power is 20 watts. If the same power source is connected to a load that also takes 2 amps (average or DC) but, on top of that current there is an alternating content of 2 amps p-p, the power taken is still 20 watts.

You can examine this in your head using superposition.

The DC parts are easy and come to 20 watts as explained in my first paragraph. The AC analysis implies is a 10 volt source that is having a positive half cycle of current taken from it that has a peak of 1 amp followed by a negative half cycle of current with a peak of -1 amps.

So, on positive half cycles, the AC power is peaking at 10 watts whilst on the negative part of the cycle the AC power is peaking at -10 watts.

The net effect of those AC powers is zero.

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  • \$\begingroup\$ Hi, than you for the detailed explanation. \$\endgroup\$ – RAN Sep 11 '17 at 8:32

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