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+//Problem 21.30: A series motor runs at 800 rev/min when the voltag is 400 V and the current is 25 A. The armature resistance is 0.4 ohm and the series field resistance is 0.2 ohm. Determine the resistance to be connected in series to reduce the speed to 600 rev/min with the same current.
+
+//initializing the variables:
+Ia1 = 25; // in Amperes
+Ra = 0.4; // in ohm
+Rse = 0.2; // in ohm
+n1 = 800/60; // in rev/sec
+n2 = 600/60; // in rev/sec
+V = 400; // in Volts
+
+//calculation:
+//e.m.f. E1
+E1 = V - Ia1*(Ra + Rse)
+//At n2, since the current is unchanged, the flux is unchanged.
+//E1/E2 = n1/n2
+E2 = E1*n2/n1
+//and E2 = V - Ia1(Ra + Rse + R)
+R = (V - E2)/Ia1 - Ra - Rse
+
+printf("\n\n Result \n\n")
+printf("\n Resistance is %.2f ohm", R)
+\ No newline at end of file