00:03
A circuit has an impedance, z equals 3 plus j4 and a source potential of 50 angle 30 degree volt at a frequency 1 .5 kilohertz.
00:31
Then determine the supply current.
00:38
Next part is to find the active.
00:44
Apparent and reactive power.
00:52
Third part is to find the rating of capacitor to be connected in parallel with the impedance z to improve the power factor of a circuit to 0 .96 lagging.
01:21
D part is the value of capacitance to improve the power factor by 0 .966 lagging.
01:37
The supply current denoted by is equals the supply voltage divided by impedance which is equal to 50 angle 30 degree divided by 5 angle 53 .1301 degrees converted into the angular domain your is equals 10 23 .131 degrees ampere.
02:09
Finding the active power for the next part, active power has the expression p equals vasis cos 5 and 5 can be written as angle vs minus angle is which is equal to 30 degree minus 23 .131 degrees this will be 53 .1301 degrees and the power will be equals to 50 into 10 into cost of 53 .1301 gives the active power as 300 watt now finding the reactive power, reactive power equals vs .i .s.
03:09
Sin 5.
03:10
Substitute the values 50 into 10 into sign of 53 .1301 degrees.
03:19
The reactor power is 400 var.
03:23
The apparent power is given by the apparent power is s equals p plus j theta which is equals to 300 plus j 400 volt ampere...