00:01
Hello everyone, let us do the following question.
00:02
Here we have given a three -phase four -wire line with the phase voltage of 120.
00:07
Supplies a balance loaded, 260 -volt motor -volt -volt -volt -gear lamps are given.
00:13
We have to find the current in the neutral wire.
00:16
We have to find current in the neutral wire and we have to find voltage, average power observed.
00:22
We will find firstly a, we will say the power to the motor is.
00:29
Firstly we will say the power to the motor is p .t is equal to s cos theta, s is 260, cost theta is 0 .85.
00:45
This is 2 to 1 kilowatt.
00:48
Next the motor power.
00:50
Next we will say the motor power per face is.
00:57
The motor power is the motor power.
00:59
Per faces p p is written as p t by 3 this is 73 .67 kilowatt hence the watt meter readings are hence the watt meter hence the watt meter readings are hence the wattmeter readings are w .a., this is 73 .67 plus 24, this is 97 .67 kilowatt.
01:36
It will write wv, this is 73 .67 plus 15, this can be 88 .67 kilowatt...