00:01
Hi, here in this given problem peak value of the triangular wave, peak value of the potential of the triangular wave that is given to be v naught and its time period is t.
00:37
So, for the first one quarter wave means from 0 to infinity, so this is the first one quarter t by 4 equation of the line straight line will be using coordinate geometry, we can say equation of straight line that will be v is equal to 4 v naught by t multiplied by t instantaneous value of voltage.
01:28
So, using the definition of root mean square voltage r m s voltage that will be given by square root of 4 by t for the first one quarter motion that is 4 by t integration of v square with respect to time in the limits from 0 to t by 4.
01:55
So, we can say this is 4 by t multiplied by integration of having the limits from 0 to t by 4 for this v we can use 4 v naught by t having a square of 8 and d t and it will be having t also sorry.
02:21
So, multiplied by t and having a square of 8 and then d t, so 4 v naught by t and having a square of 8 it will come out as it is a constant we take it out of the integration.
02:34
So, we get 4 by t multiplied by square of 4 v naught which will become 16 v naught square divided by t square and again the integration this time integration will be of t square with respect to d t from 0 to t by 4...