Average Value and RMS Value : Problem 1 – AC Circuits – Basic Electrical Engineering

– hi friends in this video we are going to see how to calculate average and RMS value of AC quantity so here I have taken a waveform which is changing like this or a triangular manner but it is a AC waveform because it is changing its magnitude with respect to time so for this waveform we are supposed to calculate average value and RMS value we will use analytical method now two things we have to remember one tracing the cycle and second writing equation of a waveform so this particular waveform having this cycle which will repeat itself so once we decided a cycle we can get a time period of that waveform so cycle exists from 0 to 1 so time period is nothing but capital T equal to 1 second because cycle exists for this much of time secondly we have to write an equation of wave form during a time period so it is a simple straight line given like this so coordinates of these two points is 0 comma 0 and for this point it is 1 comma 10 so I can use equation of a straight line it says y 2 minus y 1 upon X 2 minus X 1 equal to Y 1 minus y divided by X 1 minus X so here I will consider this as x1 y1 x2 y2 and x-axis is T y-axis is I of T so if I substitute what I will get Y 2 is 10 y 1 is 0 divided by X 2 X 2 is 1 X 1 is 0 equal to Y 1 so Y 1 is 0 again minus Y is I of t upon X 1 H 1 is 0 again minus X x axis is time axis is T so if I simplify what I will get n upon 1 equal to minus I of T upon minus T so if I solve I will get I of T equal to 10 times T so finally what I get for this waveform is equation I of T equal to 10 T for T greater than 0 and less than 1 so what I have done I have decided what is a time period secondly I have got equation of the wave form now the problem will become very easy because we know average value equal to 1 upon time period we have to integrate that particular waveform for the time period t so in this case what I will get 1 upon 1 and I am having a waveform only from 0 to 1 so 0 to 1 and equation of this I is 10 T so our average equals 10 I will take out of a integration integration of T is T Square by 2 and lets apply the limits lower limit is 0 upper limit is 1 so if I solve I will get average value as 10 divided by 2 in bracket 1 square minus 0 square so ultimately it is 5 only so it is a 5 ampere so average value I am getting 5 ampere same way I can get RMS value so to get an RMS value the formula is I RMS equals root of 1 by T integration over a time period t but this time it is the integration of I square so lets substitute the value 1 upon 1 waveform exist from 0 to 1 I is 10 T so 10 T Square into DT so it is route integral 0 to 1 100 t-square 100 is a constant I will take it out of integration so finally I will get integral 0 to 1 T Square DT and we know integration of T Square is T cube divided by 3 and limits are 0 and 1 so if I apply the limits what I will get I RMS equal to root 100 divided by 3 1 cube minus 0 cube so ultimately it is root 100 divided by 3 and if I solve what I will get I RMS as 5.7733 ampere so what I have done over here a simple triangular wave I am consider and Illustrated how to calculate average and RMS value by considering two thing one we have to define what is the cycle and second in the cycle how the time period is changing as per that we have to write equation of wave form thank you

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