Why I avg. (Current average) multiplied by Vavg (voltage average) and Time period does not give me heat loss in full cycle ?

As I know I avg gives me dc component for ac ,I can take + I avg & -I avg for for +half cycle and -half cycle respectively,then net current will be zero in full cycle but both the current are responsible for heat production in ac so why can'take and do 

I avg ×v avg×T= Heat produced in ac current .

(I know that vrms ×Irms×T will give me heat loss )

But my quetion it is why it is wrong which I had first .

thank you

Asked by Jaydeepkalal2003 | 16th Aug, 2021, 09:32: PM

Expert Answer:

Alternating voltage is assumed to be sinusoidal waveform of period T .
Since waveform repeats after every time period T , we need to average the voltage variation for a time period T
Let V = Vm sin(ωt) 
If we average over a time period T , then we have 
begin mathsize 14px style V with bar on top space equals space 1 over T integral subscript 0 superscript T V subscript m space sin left parenthesis space omega t space right parenthesis space d t space equals space 0 end style
Hence average square of instantaneous voltage is detrmined and square-root of this average is considered
as RMS ( Root-mean-square ) value of voltage
begin mathsize 14px style V subscript r m s end subscript space equals space square root of 1 over T integral subscript o superscript T V subscript o superscript 2 sin squared left parenthesis omega t right parenthesis d t end root space equals space fraction numerator V subscript o over denominator square root of 2 end fraction end style
Similarly RMS current over time period T of full cycle is determined .
Product of RMS voltage and RMS current gives average power
If average power is multiplied by time t , we get heat energy generated for time t .
Basically, answer to your question is averaging has to be done for full cycle , not for half cycle,
because waveform repeats after every full cycle.

Answered by Thiyagarajan K | 16th Aug, 2021, 10:18: PM

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