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//Chapter 5:Dc Motor Drives
//Example 11
clc;
//Variable Initialization
//Ratings of the separately excited motor
V=220 // rated voltage in V
N=600 // rated speed in rpm
Ia=500 // rated current in A
Ra=0.02 // armature resistance in ohms
Rf=10 // field resistance in ohms
//Solution
Ia1=2*Ia
E1=V-Ia*Ra //rated back emf at rated operation
Wm1=2*%pi*N/60 //angular speed
Trated=E1*Ia1/Wm1 //rated torque
//(i) When the speed of the motor is 450rpm
N1=450 //given speed in rpm
Tl=2000-2*N1 //load torque is a function of the speed as given
Ia2=Tl/Trated*Ia1 //for a torque of Tl as a function of current
E2=N1/N*E1 //for a given speed of 450rpm
V2=E2+Ia2*Ra //terminal voltage for a given speed of 450 rpm
//(ii) when the speed of the motor is 750rpm
N1=750 //given speed in rpm
Tl=2000-2*N1 //load torque is a function of the speed as given
Wm_=2*%pi*N1/60
Ke_phi1=E1/Wm1
//Since we know that V=Ke*phi*Wm+Ia*Ra by solving we get that 0.02*(Ia_)**2 -220*Ia_ + 39270 = 0"
a = 0.02
b = -220
c = 39270
//Discriminant
d = (b**2) - (4*a*c)
Ia_1 = (-b-sqrt(d))/(2*a)
Ia_2 = (-b+sqrt(d))/(2*a)
Ke_phi=Tl/abs(Ia_1)
V1=V*Ke_phi/Ke_phi1 //required field voltage
//Results
mprintf("(i)Hence motor terminal voltage is :%.1f V",V2)
mprintf("\nAnd the armature current is :%.1f A",Ia2)
mprintf("\n(ii)The solutions for Ia_ are %.1f A and %.1f A",abs(Ia_1),abs(Ia_2))
mprintf("\nWe ignore %d A since it is infeasible,\n Hence armature current is :%.1f A",abs(Ia_2),abs(Ia_1))
mprintf("\nHence the required field voltage is :%.1f V",V1)
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