00:01
Hello students, to solve this question, first we have given for load 1 that absorbs 200 w for pure resistive and load 2 absorbs 400 w at 0 .8 leading resistive and capacitance.
00:21
So voltage across load is 300 by root 3 vrms.
00:33
Now take the voltage as reference.
00:36
So vs is 300 by root 3, 0 degree vrms.
00:42
Then vs i1 is equal to 200 for load 1.
00:48
So i1 is equal to 200 by vs.
00:51
Now substituting the value, so 200 divided by 300 by root 3 that gives 200 root 3 by 300 that is equal to 1 .155 a for rms.
01:06
Therefore i1 is equal to i1 for 0 degree.
01:11
Now this is for resistive load.
01:13
I1 is in the phase vs.
01:15
So i1 is equal to 1 .155, 0 degree for arms.
01:22
Now for load 2, so cos of phi 2 is equal to 0 .8.
01:29
So phi 2 will be equal to cos inverse of 0 .8 that gives value 36 .87 degree.
01:38
Now we can write vs i2 cos of phi 2 will be equal to 400.
01:48
So i2 will be equal to 400 divided by vs cos of phi 2 that is equal to 400 divided by 300 by root 3 into 0 .8.
02:02
So after solving this, we get 400 root 3 by 300 into 0 .8 that gives 2 .887 for a rms...