00:01
In this question, let us solve a part a.
00:03
So in dc shunt motor, d .c.
00:09
Shunt motor, e2 divided by e1, is equal to i of f2 and 2 divided by i of f1 n1.
00:25
So here given and loaded, so terminal voltage will be equal to, that is, vt, will be equal to the induce, voltage so e is given that the same terminal of the voltage so it is given that there is same terminal of the voltage this means that e1 will be equal to e2 without series resistance 3 the current of i of f2 will become v1 or let us write it as vt divided by rf so vt value is 125 divided by rf now this is basically for the out series resistance field current.
01:10
Now with out series resistant field current, the if2 will be equal to this can be written as vt divided by rf plus rae.
01:22
So this will be equal to 125 divided by rf plus 8.
01:28
So let us multiply this 125 divided by 125 divided by 125.
01:34
So this will be equal to 125 divided by rf plus 8 which is multiplied by 1560 divided by 1560 divided by 1560 divided by 1420.
01:50
This will be 1420 is multiplied by 1420 is divided by 1420 divided by 1560.
01:58
So this will be equal to rf divided by rf plus 8 which is multiplied by 0 .91.
02:06
Which will be equal to rf divided by rf plus 8.
02:11
This is the field resistance.
02:13
So from this, field resistance, that is, rf will be equal to 81 .14 oom.
02:19
So this is the required answer for the a part...