--- title: "Armature resistance to reduce shunt motor speed to 1250 rpm" subject: "Electrical Machines I" chapter: 4 key: "semester_4/electrical_machines_i:shunt-motor-250v-30a-1250rpm-armature" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/shunt-motor-250v-30a-1250rpm-armature site: BCT Engineering --- # Armature resistance to reduce shunt motor speed to 1250 rpm Electrical Machines I · Chapter 4: DC Motor - Asked in: 2069 Poush, 8 marks - 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-30a-1250rpm-armature - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-4 ## Question A 250 V dc shunt motor has armature winding resistance of 0.2 ohm and field winding resistance of 125 ohms. It draws a current of 30 amp and runs at a speed of 1500 rpm. Calculate the value of resistance to be connected in series with armature in order to reduce the speed to 1250 rpm keeping the load torque constant. ## Short answer With constant flux and load torque, armature current remains 28 A. Back emf at 1500 rpm is 244.4 V. For 1250 rpm, back emf must be 203.67 V. Required total armature circuit resistance = (250−203.67)/28 = 1.655 Ω, so added series resistance ≈ 1.455 Ω. ## Answer This is armature-resistance speed control. For a shunt motor with constant flux, $N \propto E_b$, and with constant load torque the armature current also stays constant ($T \propto I_a$). **Given:** $V = 250$ V, $R_a = 0.2\ \Omega$, $R_{sh} = 125\ \Omega$, $I_L = 30$ A, $N_1 = 1500$ rpm, $N_2 = 1250$ rpm, torque constant. #### Step 1: Armature current and back emf at 1500 rpm $$ \begin{aligned} I_{sh} &= \frac{250}{125} = 2\ \text{A} \\ I_a &= 30 - 2 = 28\ \text{A} \\ E_{b1} &= 250 - 28 \times 0.2 = 244.4\ \text{V} \end{aligned} $$ #### Step 2: Back emf needed at 1250 rpm Torque constant and flux constant, so $I_{a2} = I_{a1} = 28$ A. $$ \begin{aligned} E_{b2} &= E_{b1} \times \frac{N_2}{N_1} \\ &= 244.4 \times \frac{1250}{1500} = 203.67\ \text{V} \end{aligned} $$ #### Step 3: Series resistance $$ E_{b2} = V - I_a (R_a + R) $$ $$ \begin{aligned} R_a + R &= \frac{V - E_{b2}}{I_a} = \frac{250 - 203.67}{28} = 1.655\ \Omega \\ R &= 1.655 - 0.2 = 1.455\ \Omega \end{aligned} $$ Check: $E_{b2} = 250 - 28 \times 1.655 = 203.67$ V, and $1500 \times 203.67/244.4 = 1250$ rpm. Power wasted in the added resistor $= 28^2 \times 1.455 \approx 1141$ W, which is why this method is inefficient. **Answer: A resistance of about $1.455\ \Omega$ must be connected in series with the armature.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does Ia stay at 28 A after the resistance is added? A. Load torque is constant and flux is constant, so T is proportional to Ia stays fixed B. Supply voltage is constant C. Back emf stays fixed D. Speed is constant Answer: A. With constant flux and torque, Ia is unchanged. (Key point: Armature current constant) 2. What back emf is needed at 1250 rpm? A. 244.4 x 1250/1500 = 203.67 V B. 244.4 x 1500/1250 = 293 V C. 250 x 1250/1500 = 208 V D. 244.4 V Answer: A. At constant flux Eb is proportional to N, so Eb2 = 203.67 V. (Key point: Required back emf) 3. Which equation gives the required series resistance R? A. R = (V - Eb2)/Ia B. R = (V - Eb2)/Ia - Ra C. R = Eb2/Ia D. R = Ra x N2/N1 Answer: B. Ra + R = (250 - 203.67)/28 = 1.655 ohm, so R = 1.455 ohm. (Key point: Series resistance) 4. Why is this method of speed reduction inefficient? A. Flux is reduced B. Power is lost as I^2 R in the added resistor (about 1.1 kW here) C. It raises armature reaction D. It lowers supply voltage Answer: B. The external resistor carries full armature current and dissipates 28^2 x 1.455 = 1141 W. 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