SSE JE Electrical Engineering 10 Dec 2020 Papers

Let the full load ${\mathrm{L}}_1$ = 100 kW operated for time ${\mathrm{t}}_1$ = 10 min
Half-full load ${\mathrm{L}}_1=\frac{{\mathrm{L}}_1}{2}$ = 50 kW operated for time ${\mathrm{t}}_2$ = 20 min
No-load ${\mathrm{L}}_3$ = 0 KW operated for ${\mathrm{t}}_3$ = 20 min
∴ Rating of the motor $=\sqrt{\frac{{100}^2\times 10+{50}^2\times 20}{10+20+20}}$
∴ Rating of the suitable motor is $=\sqrt{3000}\mathrm{kW}$