--- title: "Plugging of 220 V separately excited motor (100 A)" subject: "Utilization of Electrical Energy" chapter: 3 key: "semester_7/utilization_of_electrical_energy:separately-excited-plugging-220v-100a" source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/separately-excited-plugging-220v-100a site: BCT Engineering --- # Plugging of 220 V separately excited motor (100 A) Utilization of Electrical Energy · Chapter 3: Control of Electric Drive - Asked in: 2071 Chaitra - Video: [Breaking of dc motor](https://www.youtube.com/watch?v=0S9HiKro8Gc) — Parmod Saharan - 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-220v-100a - Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-3 ## Question A 220 V separately excited motor running at 1000 rpm draws 100 A from the source. The motor is braked by plugging. Calculate (i) Resistance to be inserted in the armature circuit to limit the braking current to twice the full load current, (ii) Initial braking torque, (iii) The braking torque when speed has reduced to 500 rpm. ## Short answer Assuming Ra = 0.1 ohm, Eb = 210 V and total resistance is (220+210)/200 = 2.15 ohm, so inserted resistance is 2.05 ohm. Initial braking torque is 401.1 N·m, and braking torque at 500 rpm is 303.1 N·m. ## Answer 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Ra is not given. How does the answer proceed? A. Assume a typical value Ra = 0.1 Ω to find Eb = 210 V B. The problem cannot be solved C. Assume Eb = 0 D. Assume Ra equals the inserted resistance Answer: A. A typical Ra = 0.1 Ω is assumed, so Eb = 220 - 100 x 0.1 = 210 V (neglecting Ra gives 220 V and slightly different results). (Key point: Handling missing Ra) 2. With Ra = 0.1 Ω and a braking current limit of 200 A, what resistance is inserted? A. 2.15 Ω B. 2.2 Ω C. 2.05 Ω D. 1.05 Ω Answer: C. Ra + R = (220 + 210)/200 = 2.15 Ω, so R = 2.15 - 0.1 = 2.05 Ω. (Key point: Series resistance for plugging) 3. At 500 rpm the braking torque is 303.1 N-m, lower than the initial 401.1 N-m. Why? A. The flux has halved B. Back emf halves to 105 V, so the current falls to (220 + 105)/2.15 = 151 A and T = K I falls C. The supply voltage has halved D. The resistance has doubled Answer: B. Flux (K) is constant for a separately excited motor; the back emf falls with speed, which reduces the current and torque. (Key point: Torque at lower speed) ## Related in Utilization of Electrical Energy - [Plugging of 200 V, 1000 rpm separately excited motor](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Aseparately-excited-plugging-200v-1000rpm) — Utilization of Electrical Energy, chapter 3 - [Reverse current (plugging) braking of 20 kW DC shunt motor](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Adc-shunt-reverse-current-braking-20kw) — Utilization of Electrical Energy, chapter 3 - [Induction motor plugging: braking current and torque](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Ainduction-motor-plugging-braking-current) — Utilization of Electrical Energy, chapter 3 - [Plugging resistance for 30 kW DC shunt motor](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Adc-shunt-30kw-plugging-resistance) — Utilization of Electrical Energy, chapter 3 - [Chopper ON time for separately excited DC motor](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Achopper-on-time-separately-excited-motor) — Utilization of Electrical Energy, chapter 3 - [Speed of separately excited DC motor (given J, B)](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Aseparately-excited-motor-speed-load-100nm) — Utilization of Electrical Energy, chapter 3 ## Related in other subjects - [Armature size, 7.5 kW 1000 rpm motor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adc-motor-7-5kw-1000rpm) — Electric Machine Design, chapter 5, asked 5× - [Shunt motor: resistance for 1200 rpm](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Ashunt-motor-40a-1400-to-1200) — Electrical Machine, chapter 4 - [Series motor: resistance to limit speed to 2400 rpm](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Adcm-back-emf-importance-series-resistance) — Electrical Machines, chapter 4 - [Armature resistance to reduce shunt motor speed to 1250 rpm](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-250v-30a-1250rpm-armature) — Electrical Machines I, chapter 4