--- title: "Shunt motor: field resistance to raise speed to 1100 rpm" subject: "Electrical Machines" chapter: 4 key: "semester_4/electrical_machines:dcm-shunt-field-resistance-1100rpm" source: https://www.bctengineer.com/q/semester_4/electrical_machines/dcm-shunt-field-resistance-1100rpm site: BCT Engineering --- # Shunt motor: field resistance to raise speed to 1100 rpm Electrical Machines · Chapter 4: DC Motor - Asked in: 2078 Baisakh, 8 marks - Video: [Numerical on speed control of DC shunt motors #electricmotors #electricalengineering](https://www.youtube.com/watch?v=T7KmAY7FYA4) — EEE, SJBIT - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines/dcm-shunt-field-resistance-1100rpm - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-4 ## Question A 250 V dc shunt motor has armature winding resistance of 0.5 ohm and field winding resistance of 125 ohms. It draws a current of 25 A at a speed of 900 rpm. It is required to increase the speed to 1100 rpm keeping the load torque constant. Calculate the value of additional resistance to be connected in series with the field winding to achieve this speed. ## Short answer To raise speed from 900 to 1100 rpm at constant load torque, add about 29.54 Ω in series with the field winding. The field current falls to 1.618 A, requiring total field resistance of 154.54 Ω; the additional resistance is 154.54 – 125 = 29.54 Ω. ## Answer Assume the flux is proportional to the field current (unsaturated). The load torque is constant, so $T\propto\Phi I_a$ is constant. Let $R_x$ be the added field resistance. #### Initial condition $$ I_{f1} = \frac{250}{125} = 2\ \text{A}, \qquad I_{a1} = 25 - 2 = 23\ \text{A}, \qquad E_{b1} = 250 - 23(0.5) = 238.5\ \text{V} $$ #### New condition (1100 rpm) Let $I_{f2} = s$ (A). **Constant torque:** $\Phi_1I_{a1} = \Phi_2I_{a2}$, so $2\times 23 = s\,I_{a2}$: $$ I_{a2} = \frac{46}{s} $$ **Back emf:** $$ E_{b2} = 250 - 0.5I_{a2} = 250 - \frac{23}{s} $$ **Speed ratio:** $E_b\propto\Phi N$: $$ E_{b2} = E_{b1}\times\frac{\Phi_2}{\Phi_1}\times\frac{N_2}{N_1} = 238.5\times\frac{s}{2}\times\frac{1100}{900} = 145.750\,s $$ Equating: $$ 250 - \frac{23}{s} = 145.750\,s \;\Rightarrow\; 145.750\,s^2 - 250s + 23 = 0 $$ $$ s = \frac{250 + \sqrt{250^2 - 4(145.750)(23)}}{2(145.750)} = 1.6177\ \text{A} $$ (The larger root is taken; the smaller one gives an unrealistically weak field.) #### Added resistance $$ R_f + R_x = \frac{250}{s} = \frac{250}{1.6177} = 154.54\ \Omega $$ $$ R_x = 154.54 - 125 = 29.54\ \Omega $$ Check: $I_{a2} = 46/1.6177 = 28.44$ A, $E_{b2} = 250 - 0.5(28.44) = 235.78$ V, and $145.750(1.6177) = 235.78$ V. **Answer:** additional field resistance $\approx 29.54\ \Omega$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. When the field is weakened at constant load torque, the armature current: A. Falls B. Stays the same C. Rises so that Phi Ia stays constant D. Becomes zero Answer: C. T proportional to Phi Ia is fixed, so lower Phi needs higher Ia. (Key point: Constant torque with field weakening) 2. Initially If, Ia, Eb1 are: A. 2 A, 23 A, 238.5 V B. 2 A, 25 A, 237.5 V C. 1 A, 24 A, 238 V D. 2 A, 23 A, 250 V Answer: A. If = 250/125 = 2; Ia = 25 - 2 = 23; Eb1 = 250 - 11.5 = 238.5 V. (Key point: Initial values) 3. Which root of the quadratic in s (field current) is taken? A. The smaller root B. The larger root, as the smaller gives unrealistically weak field C. Both D. Neither Answer: B. The larger root s = 1.6177 A is practical. (Key point: Root choice) 4. The additional field resistance is about: A. 154.5 ohm B. 125 ohm C. 29.5 ohm D. 1.6 ohm Answer: C. 250/1.6177 = 154.54 ohm total; minus 125 gives 29.54 ohm. 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