--- title: "Shunt motor speed with 30 Ω field resistance added" subject: "Electrical Machines I" chapter: 4 key: "semester_4/electrical_machines_i:shunt-motor-220v-30ohm-field-torque" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/shunt-motor-220v-30ohm-field-torque site: BCT Engineering --- # Shunt motor speed with 30 Ω field resistance added Electrical Machines I · Chapter 4: DC Motor - Asked in: 2078 Baisakh, 8 marks - 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-220v-30ohm-field-torque - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-4 ## Question A 220 V dc shunt motor has armature winding resistance of 0.08 ohm and field winding resistance of 110 ohms. It runs at a speed of 1400 rpm at half load. If a resistance of 30 ohms is connected in series with the field winding and load torque is increased by 10%, calculate the new speed of the motor. ## Short answer Field current changes from 2 A to 220/(110+30) ≈ 1.571 A; flux ratio φ2/φ1 = 0.7857. Neglecting armature drop, speed = 1400 × (2/1.571) ≈ 1782 rpm. If half-load armature current is 21 A, corrected E_b gives ~1776 rpm. ## Answer **Given:** $V = 220$ V, $R_a = 0.08\ \Omega$, $R_{sh} = 110\ \Omega$, $N_1 = 1400$ rpm at half load, extra 30 Ω in field, load torque +10%. **Note:** the current at half load is not given. Since $R_a$ is very small, the $I_a R_a$ drop (about 1-2 V) hardly affects the speed, so the main answer neglects it. A check with an assumed current follows. For a shunt motor, take flux proportional to field current (no saturation). Then $$ T \propto \phi I_a, \qquad N \propto \frac{E_b}{\phi}, \qquad E_b = V - I_a R_a $$ #### Flux ratio $$ \begin{aligned} I_{sh1} &= \frac{220}{110} = 2\ \text{A}, \quad I_{sh2} = \frac{220}{110 + 30} = 1.5714\ \text{A} \\ \frac{\phi_2}{\phi_1} &= \frac{110}{140} = 0.7857 \end{aligned} $$ #### New speed (neglecting $I_a R_a$, so $E_{b1} \approx E_{b2}$) $$ N_2 = N_1 \times \frac{\phi_1}{\phi_2} = 1400 \times \frac{140}{110} = 1781.8\ \text{rpm} $$ #### Check with an assumed half-load line current of 23 A $$ \begin{aligned} I_{a1} &= 23 - 2 = 21\ \text{A}, \quad E_{b1} = 220 - 21 \times 0.08 = 218.32\ \text{V} \\ I_{a2} &= \frac{1.1 \times 21}{0.7857} = 29.4\ \text{A}, \quad E_{b2} = 220 - 29.4 \times 0.08 = 217.65\ \text{V} \\ N_2 &= 1400 \times \frac{217.65}{218.32} \times \frac{1}{0.7857} = 1776.3\ \text{rpm} \end{aligned} $$ The 10% torque increase only raises $I_a$; its effect on speed is under 0.5%. **Answer:** New speed $\approx 1782$ rpm (about 1776 rpm if a half-load current of 23 A is assumed). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is φ2/φ1 with 30 Ω added to a 110 Ω field? A. 110/140 = 0.7857 B. 140/110 C. 0.5 D. 1 Answer: A. Flux ∝ field current ∝ 1/R. (Key point: Flux ratio) 2. Why can the speed be found as N1·φ1/φ2 roughly? A. Torque is constant B. Ra is very small so Eb1 ≈ Eb2 C. Flux is constant D. Ia is unchanged Answer: B. IaRa drop is only 1-2 V. (Key point: Neglecting IaRa) 3. What is the new speed (main answer)? A. About 1400 rpm B. About 1100 rpm C. About 1782 rpm D. About 1540 rpm Answer: C. 1400×140/110. (Key point: Result) 4. How much does the 10% torque increase affect the speed? A. It raises speed by 10% B. It lowers speed by 10% C. It doubles Ia D. Very little (under 0.5%); it mostly raises Ia Answer: D. Only the small IaRa drop changes. 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