00:01
In this question, there are two loads are connected in parallel with an input signal.
00:16
And let this is load 1 and this is load 2 and the input signal is 208 watt.
00:30
Here given the apparent power by load 1 is equal to 12 kv ampere.
00:44
Therefore, the real power drawn by load 1 is equal to p1 is equal to 12 kv ampere multiplied by pf1.
00:56
Where pf1 is the power factor of load 1 that is equal to 0 .7.
01:04
Therefore, p1 is equal to 12 multiplied by 0 .7 which is equal to 8 .4 kw.
01:14
Watt and the apparent power of load 2 is equal to 18 kilo volt ampere.
01:26
Therefore, the real power drawn by load 2 is p2 is equal to 18 kilo volt ampere multiplied by pf2.
01:37
Here pf2 p2 is equal to 0 .9 therefore p2 is equal to 18 multiplied by 0 .9 that is equal to 16 .2 kilowatt.
01:51
Therefore the total power drawn by load p is equal to p1 plus p2 that is equal to 8 .4 plus 16 .2 which is equal to 24 .6 kw and the source power p is equal to the total power drawn by the load and p is given by the expression root 3 vl il cos 5.
02:33
Therefore, root 3 vl il cos phi is equal to 24 .6 kw.
02:43
This is equation number 1.
02:45
Next, the kw ampere reactive drawn by load 1 is equal to kw ampere multiplied by sin phi 1.
03:01
Here given pf1 is equal to cos phi1 which is equal to 0 .7.
03:10
Therefore phi1 is equal to cos inverse 0 .7 which is equal to 45 .57 degree...