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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+clc;
+v=240; // supply voltage
+n=1000; // speed of motor
+i=40; // current drawn from supply
+rf=0.2; // field resistance
+ra=0.25; // armture resistance
+rd=0.3; // diverter resistance
+// torque is constant for different speeds
+// when diverter is put in parallel with series resistance then some fraction of armature current flows through series circuit this current for constant torque is given by
+ia2=sqrt(i^2/(rd/(rf+rd)));
+Ea1=v-i*(ra+rf); // counter EMf at n=1000 rpm
+Ea2=v-ia2*(ra+((rf*rd)/(rf+rd))); // counter EMF at new speed
+n2=(Ea2*n*i)/(Ea1*(rd/(rf+rd))*ia2);
+printf('Motor speed after diverter is put in parallel with series field winding is %f rpm',ceil(n2));