--- title: "Plugging of 200 V, 1000 rpm separately excited motor" subject: "Utilization of Electrical Energy" chapter: 3 key: "semester_7/utilization_of_electrical_energy:separately-excited-plugging-200v-1000rpm" source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/separately-excited-plugging-200v-1000rpm site: BCT Engineering --- # Plugging of 200 V, 1000 rpm separately excited motor Utilization of Electrical Energy · Chapter 3: Control of Electric Drive - Asked in: 2072 Chaitra, 6 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/separately-excited-plugging-200v-1000rpm - Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-3 ## Question A 200 V, 1000 rpm, 50 A separately excited dc motor is to be stopped to zero speed by plugging. Armature resistance is 0.2 ohm. Find (i) additional resistance to be connected in series with armature to limit braking current to twice the rated current (ii) braking torque (iii) torque when speed has decreased to zero. Assume that initial speed is 1000 rpm. ## Short answer For plugging, total armature circuit resistance is (200+190)/100 = 3.9 ohm, so additional resistance is 3.7 ohm. Initial braking torque is 181.4 N·m, and torque at zero speed is 93.0 N·m, so supply must be disconnected at standstill. ## Answer 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. At the instant of plugging, which voltage drives the braking current through Ra + R? A. V - Eb B. V + Eb, because supply and back emf add C. Eb only D. V only Answer: B. Reversing the supply makes supply voltage and back emf act in the same direction, so Ra + R = (V + Eb)/Ib. (Key point: Plugging circuit relation) 2. For the 200 V, 50 A motor with Ra = 0.2 Ω, what is the back emf at rated speed? A. 200 V B. 210 V C. 180 V D. 190 V Answer: D. Eb = V - Ia Ra = 200 - 50 x 0.2 = 190 V. (Key point: Back emf) 3. The braking current is limited to 100 A. What additional resistance is required? A. 3.7 Ω B. 3.9 Ω C. 2.0 Ω D. 1.9 Ω Answer: A. Ra + R = (200 + 190)/100 = 3.9 Ω, so R = 3.9 - 0.2 = 3.7 Ω. (Key point: Additional resistance) 4. After plugging brings the speed to zero, what must be done and why? A. Nothing; torque is zero at zero speed B. Disconnect the supply, because torque (about 93 N-m) is still present and would drive the motor in reverse C. Increase the series resistance to zero D. Connect the field in reverse Answer: B. At zero speed Eb = 0, I = 200/3.9 = 51.3 A and T = K I = 93 N-m, so the motor would accelerate in reverse unless the supply is removed. 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