GATE 2018 EE Q.24: solution and answer

Official answer: 600.0Source: Official answer key, GATE 2018 EE (IIT Guwahati)

1. At zero speed the back-emf is zero ($E=K\varphi \omega =0$), so the entire applied voltage drops across the armature resistance: $$I_a=\frac{V_a}{R_a}=\frac{100}{0.05}=2000\text{A}$$

2. Using the stall torque to find the torque constant (which also relates back-emf to speed, since the field excitation is unchanged): $$T=K\varphi I_a\Rightarrow K\varphi =\frac{500}{2000}=0.25\text{Nm/A}$$

3. At no load, with all mechanical losses neglected, the developed torque must be zero. Since $K\varphi \ne 0$, this forces the no-load armature current to be zero: $$T=K\varphi I_a=0\Rightarrow I_a=0$$

4. With $I_a=0$ there is no drop across $R_a$, so the back-emf equals the applied voltage: $$E=V_a-I_a{R}_a=150-0=150\text{V}$$

5. The no-load speed follows from $E=K\varphi \omega$: $$\omega =\frac{E}{K\varphi }=\frac{150}{0.25}=600\text{rad/s}$$