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//Chapter 5:Dc Motor Drives
//Example 23
clc;
//Variable Initialization
//The motor is operated using dynamic braking method
N=600 //speed in rpm
Vs=220 //source voltage in v
Ra=0.12 // armature resistance in ohms
delta_min=0.1 //manimum value of duty ratio
delta_max=0.9 //maximum value of duty ratio
//Magnetisation curve at N
If=[10, 20,30, 40, 50, 60, 70, 80] //field current in A
E =[64,118,150,170,184,194,202,210] //terminal voltage in V
//Solution
//(i) Maximum braking speed is 800rpm with armature current of 70 A
N1=800 //maximum braking speed in rpm
Ia=70 //armature current in A
E1=N1/N*E(7) //at 70A motor back emf is 202V
Rbe=E1/Ia-Ra //effective value of braking resistance
Rb=Rbe/(1-delta_min) //required braking resistance
//(ii)When the speed of the motor is 87 rpm
//now torque is maximum when the duty ratio is maximum
N1=87 //speed in rpm
R=Rb*(1-delta_max)+Ra
Ia=If(5) //value of armature current for the given value of E=184V
Ke_phi=E(5)/(2*%pi*N)*60
T=Ke_phi*Ia //required torque
//Results
mprintf("(i)Hence braking resistance is:%.2f ohm",Rb)
mprintf("\n(ii)Hence the required torque is :%.1f N-m",T)
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