--- title: "Shunt motor speed with 0.05 Ω armature resistance added" subject: "Electrical Machines I" chapter: 4 key: "semester_4/electrical_machines_i:shunt-motor-220v-25a-0p05ohm-torque" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/shunt-motor-220v-25a-0p05ohm-torque site: BCT Engineering --- # Shunt motor speed with 0.05 Ω armature resistance added Electrical Machines I · Chapter 4: DC Motor - Asked in: 2071 Magh, 8 marks - Video: [Speed Control Of DC Motor | Armature resistance control | DC machines | Lec-60](https://www.youtube.com/watch?v=7lQaKwoz91Q) — Education 4u - 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-25a-0p05ohm-torque - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-4 ## Question A dc shunt motor supplied by 220 V dc draws a current of 25 A and runs at a speed of 1500 rpm. The armature and field winding resistances are 0.08 Ω and 110 Ω respectively. A resistance of 0.05 Ω is added in series with armature and load torque is increased by 20%, calculate the new speed. ## Short answer Field current 2 A, flux constant. Initial back emf 218.16 V; after adding 0.05 Ω, armature current 27.6 A, back emf 216.41 V. Speed = 1500 × 216.41/218.16 ≈ 1488 rpm. ## Answer Adding resistance in the armature circuit does not change the shunt field, so flux is constant; then $T \propto I_a$ and $N \propto E_b$. **Given:** $V = 220$ V, $I_{L1} = 25$ A, $N_1 = 1500$ rpm, $R_a = 0.08\ \Omega$, $R_{sh} = 110\ \Omega$; extra 0.05 Ω in armature; load torque +20%. #### Initial condition $$ \begin{aligned} I_{sh} &= \frac{220}{110} = 2\ \text{A}, \quad I_{a1} = 25 - 2 = 23\ \text{A} \\ E_{b1} &= 220 - 23 \times 0.08 = 218.16\ \text{V} \end{aligned} $$ #### New condition $$ \begin{aligned} I_{a2} &= 1.2 \times 23 = 27.6\ \text{A} \\ E_{b2} &= 220 - 27.6 \times (0.08 + 0.05) = 220 - 3.588 = 216.41\ \text{V} \end{aligned} $$ #### New speed $$ N_2 = 1500 \times \frac{216.41}{218.16} = 1488.0\ \text{rpm} $$ **Answer:** New speed $\approx 1488$ rpm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why can flux be taken constant when 0.05 ohm is added in series with the armature? A. The armature resistance cancels armature reaction B. Torque stays constant C. The shunt field is across the supply and is unaffected by armature-circuit resistance D. Back emf stays constant Answer: C. The field current V/Rsh is unchanged, so flux is constant. (Key point: Flux with armature resistance) 2. With constant flux and torque up 20%, what is Ia2 (Ia1 = 23 A)? A. 27.6 A B. 23 A C. 19.2 A D. 30 A Answer: A. With constant flux T is proportional to Ia, so Ia2 = 1.2 x 23 = 27.6 A. (Key point: New armature current) 3. Which expression gives the new back emf? A. 220 - 27.6 x 0.08 B. 220 - 27.6 x 0.05 C. 220 - 23 x 0.13 D. 220 - 27.6 x 0.13 Answer: D. The armature circuit now has 0.08 + 0.05 = 0.13 ohm, so Eb2 = 220 - 27.6 x 0.13 = 216.41 V. (Key point: Back emf after resistance added) 4. Using N proportional to Eb, how does the speed change? A. Rises to about 1512 rpm B. Falls slightly to about 1488 rpm C. Falls to about 1000 rpm D. Unchanged at 1500 rpm Answer: B. N2 = 1500 x 216.41/218.16 = 1488 rpm, a slight fall. 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