// FUNDAMENTALS OF ELECTICAL MACHINES // M.A.SALAM // NAROSA PUBLISHING HOUSE // SECOND EDITION // Chapter 5 : DIRECT CURRENT MOTORS // Example : 5.6 clc;clear; // clears the console and command history // Given data I_L1 = 5 // dc shunt motor current V_t = 230 // terminal voltage in V N_1 = 1000 // speed in rpm R_a = 0.2 // armature resistance in ohm R_F = 230 // field resistance in ohm I_L2 = 30 // dc shunt motor current // caclulations // at noload condition I_sh = V_t/R_F // shunt field current in A I_a1 = I_L1-I_sh // armature current in A E_b1 = V_t-I_a1*R_a // back emf in V // under load condition I_a2 = I_L2-I_sh // armature current in A E_b2 = V_t-I_a2*R_a // back emf in V N_2 = (E_b2/E_b1)*N_1 // motor speed under load condtion in rpm // display the result disp("Example 5.6 solution"); printf(" \n Speed under load condition \n N_2 = %.1f rpm \n", N_2 );