Plugging of 220 V separately excited motor (100 A)

The armature resistance is not given; assume $R_a = 0.1\ \Omega$ (a typical value for a motor of this size). Braking current limit $= 2\times100 = 200$ A.

Back emf and motor constant

$$\begin{aligned} E_b &= 220 - 100\times0.1 = 210\ \text{V} \\ \omega &= \frac{2\pi\times1000}{60} = 104.72\ \text{rad/s} \\ K &= \frac{210}{104.72} = 2.0054\ \text{V s/rad} \end{aligned}$$

(i) Resistance to be inserted

$$\begin{aligned} R_a + R &= \frac{V + E_b}{I_b} = \frac{220 + 210}{200} = 2.15\ \Omega \\ R &= 2.15 - 0.1 = 2.05\ \Omega \end{aligned}$$

(ii) Initial braking torque

$$T_b = K I_b = 2.0054 \times 200 = 401.1\ \text{N·m}$$

(iii) Braking torque at 500 rpm

$$\begin{aligned} E &= 210\times\frac{500}{1000} = 105\ \text{V} \\ I &= \frac{220 + 105}{2.15} = 151.16\ \text{A} \\ T &= 2.0054 \times 151.16 = 303.1\ \text{N·m} \end{aligned}$$

Answer (with $R_a$ = 0.1 Ω): (i) 2.05 Ω, (ii) 401.1 N·m, (iii) 303.1 N·m.

If $R_a$ is neglected: $E_b = 220$ V, $R = 440/200 = 2.2\ \Omega$, $K = 2.1008$, $T_b = 420.2$ N·m and at 500 rpm $I = 330/2.2 = 150$ A, $T = 315.1$ N·m.