Plugging of 200 V, 1000 rpm separately excited motor

Data: $V = 200$ V, $N = 1000$ rpm, $I = 50$ A, $R_a = 0.2\ \Omega$. Braking current limit $= 2\times50 = 100$ A.

Back emf and motor constant

$$\begin{aligned} E_b &= 200 - 50\times0.2 = 190\ \text{V} \\ \omega &= \frac{2\pi\times1000}{60} = 104.72\ \text{rad/s} \\ K &= \frac{190}{104.72} = 1.8144\ \text{V s/rad} \end{aligned}$$

(i) Additional resistance

At plugging, supply voltage and back emf add:

$$\begin{aligned} R_a + R &= \frac{V + E_b}{I_b} = \frac{200 + 190}{100} = 3.9\ \Omega \\ R &= 3.9 - 0.2 = 3.7\ \Omega \end{aligned}$$

(ii) Initial braking torque

$$T_b = K I_b = 1.8144 \times 100 = 181.4\ \text{N·m}$$

(iii) Torque at zero speed

At zero speed $E_b = 0$:

$$\begin{aligned} I &= \frac{200}{3.9} = 51.28\ \text{A} \\ T &= 1.8144 \times 51.28 = 93.04\ \text{N·m} \end{aligned}$$

Answer: (i) 3.7 Ω, (ii) 181.4 N·m, (iii) 93.0 N·m.

Since torque is not zero at standstill, the supply must be disconnected at zero speed; otherwise the motor will accelerate in reverse.