Consider a 500 volt d.c shunt motor, when on load, running at 750 rpm with armature current of 100 A where armature resistance is 0.5 Ω. If the flux is reduced by 50% without changing the load torque then the new speed of motor is

Correct option is D

​$$E_{b1} = 500 - I_{a1} R_a \\= 500 - 100(0.5) \\= 500 - 50 \\= 450\ \text{V} \\\text{Since } \phi_1 I_1 = \phi_2 I_2 \\1 \times 100 = 0.5 \times I_2 \\I_2 = 200\ \text{A} = I_{a2} \\E_{b2} = 500 - I_{a2} R_a \\E_{b2} = 500 - 200(0.5) \\= 400\ \text{V} \\\frac{E_{b1}}{E_{b2}} = \frac{N_1}{N_2} \times \frac{\phi_1}{\phi_2} \\\phi_2 = 0.5 \times \phi_1 \\\frac{450}{400} = \frac{750}{N_2} \times \frac{1}{0.5} \\N_2 = 1333.33\ \text{RPM}$$​​