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