Q1 we have a separately excited dc machine with if 1 laf 2...

Q1. We have a separately excited dc machine with If = 1, Laf = 2, La = 0.01, Ra = 0.5, J = 1, B = 3. Design a current controller and set Zeta = 0.7. Q2. We have a series excited DC machine with J = 2, В = 3, Laf = 0.1, La = 0.1, Ra = 0.1, Tl = 10, and we connect this machine to 50V. Draw the linearized block diagram near the operating point and write in the values for all parameters.

Transcript

00:01 Hi, in this question we have two parts.

00:02 So, for first part we have so we have ir is given by so i is given by p by v which is equal to 6000 divided by 120 that is equal to 50 amperes.

00:15 Then we have the emf eg is given by so it is equal to it is u plus ia into ra which is equal to so it is 120 plus 50 into 0 .2 which is equal to 130 volts.

00:30 Then we have ifo as so we have this is ia so this is ia.

00:39 So, we have the value of ifo as 0.

00:41 So, we have ifo as 0 .8 ampere e naught is given as 114 volts.

00:51 So, 114 volts energy gap is given by 130 volts and n naught is given as 1200 rpm and n is given as 1500 rpm.

01:06 Now, we have the ratio for energy gap to the ground state energy is ng by n naught into ifg by if naught.

01:16 So, now we have to find ifg.

01:18 So, ifg will be equal to it is eg by e naught into ng by n naught into ifo.

01:26 So, this is equal to it is 130 divided by 114 into 1500 divided by 1200 into 0 .8.

01:36 So, we have the calculation.

01:38 So, calculation for the current ifg as 1 .14 amperes.

01:43 Now, we have to calculate.

01:45 So, we have to calculate rfg.

01:49 So, we have formula ifg is equal to vf divided by rfw plus rf.

01:57 So, rfg.

01:58 So, we have rfg will be equal to it is vf divided by ifg minus rfw.

02:09 So, now we have it is equal to so this is equal to 120 divided by 1 .14 minus 100.

02:17 So, we have the value of the resistance as 5 .26 ohm.

02:24 So, this is the solution for the first part...

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