00:01
Hello, in the question we have given a 500 volt dc shunt motor runs at its normal speed of 250 rpm when armature current is 200 ampere, the resistance of armature is 0 .1 to om.
00:14
So, see, what is given.
00:17
So here we are applying some voltage which is given 500 volt.
00:22
Here we have the armature and your device.
00:31
So now see, it is shunt.
00:35
It is mentioned in the question it is sent.
00:37
So it is parallel.
00:38
So this has the resistance of 0 .12 oms and the current from this is 200 ampere.
00:46
So it is 200 ampere.
00:51
So now what is given? so let us write.
00:54
So v is equal to 500 volt.
00:57
Then r, a, that is the armature radio resistance is 0 .12 oms.
01:05
Then i a is 200 mpheres so now in this in this case this is the dc motor so we have this dc motor so what it will do it will produce some back emf so there will be a back emf back emf of dc motor so what this is it will be eb1 so for this case i'm representing it as ab1 so it will be v minus i a r a so this will be 500 minus this is 200 times 0 .12 so this will turn out to be 4178 volts so this is the back emf in the first case when this was the amateur now in the second case they are they as they have said that we are going to put one resistance.
02:11
So when resistance is inserted in field for reducing the shund field to 80 % of the normal value...