--- title: "Field diverter to raise series motor speed" subject: "Electrical Machines I" chapter: 4 key: "semester_4/electrical_machines_i:series-motor-500v-diverter-600rpm" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/series-motor-500v-diverter-600rpm site: BCT Engineering --- # Field diverter to raise series motor speed Electrical Machines I · Chapter 4: DC Motor - Asked in: 2074 Bhadra, 8 marks - Video: [(English)EM || (Ch-9)(4ed)|| Series DC Motor with Diverter || Q3a (Assignment Question) ||](https://www.youtube.com/watch?v=xE1GJTQrSq8) — Electrical Engineering Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/series-motor-500v-diverter-600rpm - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-4 ## Question A 500 V dc series motor runs at 500 rpm and takes 60 A. The resistance of the field and the armature are 0.3 Ω and 0.2 Ω respectively. Calculate the value of the resistance to be shunted with the series field in order that speed be increased to 600 rpm, if the load torque is assumed to be constant. Saturation may be neglected. ## Short answer Armature current becomes 71.9 A, series field current 50.1 A. Diverter resistance is R_d = I_se2 R_se / (I_a2 - I_se2) = 50.06 × 0.3 / 21.84 ≈ 0.69 Ω. ## Answer A **diverter** is a resistance $R_d$ in parallel with the series field. It takes part of the armature current, so the field current and flux fall and the speed rises. **Given:** $V = 500$ V, $N_1 = 500$ rpm, $I_1 = 60$ A, $R_{se} = 0.3\ \Omega$, $R_a = 0.2\ \Omega$, $N_2 = 600$ rpm, torque constant, no saturation ($\phi \propto I_{se}$). ```text +--Rse--+ + o-----| |----Ra---(M)----o - +--Rd---+ Ia2 splits: Ise2 in Rse, Ia2-Ise2 in Rd ``` #### Step 1: Initial back emf $$ E_{b1} = 500 - 60 \times (0.3 + 0.2) = 470\ \text{V} $$ #### Step 2: Torque condition $T \propto \phi I_a \propto I_{se} I_a$: $$ I_{se2} I_{a2} = 60 \times 60 = 3600 $$ #### Step 3: Speed condition $$ \frac{N_2}{N_1} = \frac{E_{b2}}{E_{b1}} \times \frac{I_{se1}}{I_{se2}} = 1.2 $$ with $E_{b2} = 500 - 0.2 I_{a2} - 0.3 I_{se2}$ (drop across field and diverter $= 0.3 I_{se2}$). Substituting $I_{se2} = 3600/I_{a2}$: $$ \begin{aligned} E_{b2} \times \frac{60\, I_{a2}}{3600} &= 1.2 \times 470 = 564 \\ E_{b2} I_{a2} &= 33\,840 \\ 500 I_{a2} - 0.2 I_{a2}^2 - 1080 &= 33\,840 \\ 0.2 I_{a2}^2 - 500 I_{a2} + 34\,920 &= 0 \\ I_{a2} &= 71.91\ \text{A} \end{aligned} $$ (The other root, about 2428 A, is not practical.) $$ I_{se2} = \frac{3600}{71.91} = 50.06\ \text{A}, \quad E_{b2} = 470.60\ \text{V} $$ Check: $N_2 = 500 \times \dfrac{470.60}{470} \times \dfrac{60}{50.06} = 600$ rpm. #### Step 4: Diverter resistance $$ \begin{aligned} I_d &= I_{a2} - I_{se2} = 71.91 - 50.06 = 21.84\ \text{A} \\ R_d &= \frac{I_{se2} R_{se}}{I_d} = \frac{50.06 \times 0.3}{21.84} = 0.688\ \Omega \end{aligned} $$ **Answer:** Diverter resistance $R_d \approx 0.69\ \Omega$ (armature current becomes 71.9 A, field current 50.1 A). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the diverter do? A. Shunts part of the armature current away from the series field, weakening flux B. Strengthens the flux C. Reduces armature resistance D. Adds series resistance to the armature Answer: A. Weaker flux raises speed. (Key point: Diverter) 2. With constant torque, what holds? A. Ia2 = 60 B. Ise2·Ia2 = 60×60 = 3600 C. Ise2 = 60 D. Ia2/Ise2 = 1 Answer: B. T ∝ Ise Ia. (Key point: Torque) 3. What is the armature current? A. 60 A B. 50 A C. About 71.9 A D. 2428 A Answer: C. Positive practical root. (Key point: Ia2) 4. What is the diverter resistance? A. 0.3 Ω B. 1.4 Ω C. 0.2 Ω D. About 0.69 Ω Answer: D. Rd = 50.06×0.3/21.84. 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