00:01
So solving part a is a focus on a slip calculation.
00:30
The slip s is an induction motor, and this is given by, so we have n sub s minus n subr divided into n sub s.
01:24
So n sub as is the synchronous speed, and n subr is the rotor speed.
01:32
So for a full 4 -pole 60 -hertz motor, the ns, n -sub -s, is equal to 120 times 60, then divided into 4.
01:51
So this is 1 ,800 rpm.
01:59
So if the rotor runs at 1746 rpm, this will translate to an s value equal to 1800 minus 1746.
02:24
Divided into 1800 times 100%.
02:32
So this is 3 % is the value for a.
02:40
And then b, we have the rotor input power.
02:57
So the rotor input power is the portion of the power transferred to the rotor from the stator.
03:06
It is related to the rotor copper loss, the r -c -l.
03:10
So that is an equation, equal to the p input rotor that is equal to the rotor copper loss divided into s.
03:43
So if we solve this based on the information available, we have p input rotor is equal to 2 % divided into 3 % times input power.
04:04
So this is equal to 66 .67 % of input power.
04:27
So i'll put that in the next line...