--- title: "Shunt motor speed with 75 Ω field resistance added" subject: "Electrical Machines I" chapter: 4 key: "semester_4/electrical_machines_i:shunt-motor-250v-75ohm-torque-20pc" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/shunt-motor-250v-75ohm-torque-20pc site: BCT Engineering --- # Shunt motor speed with 75 Ω field resistance added Electrical Machines I · Chapter 4: DC Motor - Asked in: 2076 Bhadra, 8 marks; 2069 Bhadra, 8 marks (asked 2 times) - Video: [EE213 - 26a - Speed control of shunt DC motors - Changing field resistance](https://www.youtube.com/watch?v=qmsgMsE1xzo) — MAFarooqi - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/shunt-motor-250v-75ohm-torque-20pc - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-4 ## Question A 250 V d.c shunt motor has armature winding resistance of 0.2 ohm and field winding resistance of 125 ohm. It draws a current of 32 Amp. and runs at a speed of 1500 r.p.m. If a resistance of 75 ohm is connected in series with the field and the load torque on the shaft is increased by 20%, what will be the new speed of the motor? ## Short answer Initially If=2 A, Ia=30 A, Eb=244 V. After adding 75 Ω, If=1.25 A, flux ratio=0.625; with 20% more torque Ia=57.6 A, Eb=238.48 V. New speed N2=1500×(238.48/244)×(1/0.625)≈2346 rpm. ## Answer **Given:** $V = 250$ V, $R_a = 0.2\ \Omega$, $R_{sh} = 125\ \Omega$, $I_{L1} = 32$ A, $N_1 = 1500$ rpm. Extra 75 Ω in field circuit, load torque raised by 20%. For a shunt motor, take flux proportional to field current (no saturation). Then $$ T \propto \phi I_a, \qquad N \propto \frac{E_b}{\phi}, \qquad E_b = V - I_a R_a $$ #### Initial condition $$ \begin{aligned} I_{sh1} &= \frac{250}{125} = 2\ \text{A} \\ I_{a1} &= 32 - 2 = 30\ \text{A} \\ E_{b1} &= 250 - 30 \times 0.2 = 244\ \text{V} \end{aligned} $$ #### New field current and flux $$ I_{sh2} = \frac{250}{125 + 75} = 1.25\ \text{A}, \qquad \frac{\phi_2}{\phi_1} = \frac{1.25}{2} = 0.625 $$ #### New armature current $T_2 = 1.2\, T_1$ and $T \propto \phi I_a$: $$ \begin{aligned} \phi_2 I_{a2} &= 1.2\, \phi_1 I_{a1} \\ I_{a2} &= \frac{1.2 \times 30}{0.625} = 57.6\ \text{A} \end{aligned} $$ #### New speed $$ \begin{aligned} E_{b2} &= 250 - 57.6 \times 0.2 = 238.48\ \text{V} \\ \frac{N_2}{N_1} &= \frac{E_{b2}}{E_{b1}} \times \frac{\phi_1}{\phi_2} \\ N_2 &= 1500 \times \frac{238.48}{244} \times \frac{1}{0.625} = 2345.7\ \text{rpm} \end{aligned} $$ **Answer:** New speed $N_2 \approx 2346$ rpm (armature current rises to 57.6 A). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the flux ratio φ2/φ1 after adding 75 Ω to the 125 Ω field? A. 0.625 B. 0.6 C. 0.375 D. 1.6 Answer: A. Ish2 = 250/200 = 1.25 A vs 2 A, so the ratio is 0.625 assuming flux ∝ field current. (Key point: New flux) 2. How is the new armature current found when load torque rises by 20%? A. From Ia2 = 1.2·Ia1 because flux is unchanged B. From φ2·Ia2 = 1.2·φ1·Ia1 C. From Ia2 = Ia1/1.2 D. From Ia2 = Ia1·φ2/φ1 Answer: B. T ∝ φIa, so Ia2 = 1.2×30/0.625 = 57.6 A. (Key point: Torque relation) 3. Which relation gives the new speed? A. N2/N1 = (Eb1/Eb2)·(φ1/φ2) B. N2/N1 = (Eb2/Eb1)·(φ2/φ1) C. N2/N1 = (Eb2/Eb1)·(φ1/φ2) D. N2/N1 = Ia2/Ia1 Answer: C. N ∝ Eb/φ, so a weaker field raises speed. (Key point: Speed relation) 4. What is the new speed approximately? A. 1500 rpm B. 938 rpm C. 2400 rpm D. 2346 rpm Answer: D. Eb2 = 250 - 57.6×0.2 = 238.48 V; N2 = 1500×(238.48/244)/0.625 = 2346 rpm. 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