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//Chapter 5:Dc Motor Drives
//Example 12
clc;
//Variable Initialization
//Ratings of the 2-pole separately excited DC motor with the fields coils connected in parallel
V=220 // rated voltage in V
N=750 // rated speed in rpm
Ia1=100 // rated current in A
Ra=0.1 // armature resistance in ohms
//Solution
E1=V-Ia1*Ra //rated back emf at rated operation
Wm1=2*%pi*N/60 //angular speed
Trated=E1*Ia1/Wm1 //rated torque
Ke_phi1=E1/Wm1
//(i) When the armature voltage is reduced to 110V
Wm2=2*%pi*N/60 //angular speed
E2=Ke_phi1*Wm2
//Now there are two linear equations...that we have to solve
//They are given by 0.3*N2+2.674*Ia2=500 and 0.28*N2+0.1*Ia2=110
a = [0.3,2.674;0.28,0.1]
b = [500;110]
x = inv(a)*b
N2=x(1) //let the motor speed be N2
Ia2=x(2) //let the motor current be Ia2
//(ii)When the field coils are connected in series
K=Ke_phi1/2
Wm3=2*%pi*N/60 //angular speed
E3=K*Wm3
//Now there are two linear equations...that we have to solve"
//They are given by 0.3*N3+1.337*Ia3=500 and 0.14*N3+0.1*Ia3=220"
a = [0.3,1.337;0.14,0.1]
b = [500;220]
x = inv(a)*b
N3=x(1) //let the motor speed be N3
Ia3=x(2) //let the motor current be Ia3
//Results
mprintf("(i)Hence the motor armature current is Ia2 :%.1f A",Ia2)
mprintf("\nAnd the required speed is N2 :%.1f rpm",N2)
mprintf("\n(ii)Hence the motor armature current is Ia3 :%.1f A",Ia3)
mprintf("\nAnd the required speed is N3 :%.1f rpm",N3)
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