--- title: "Firing angles for 200 V, 875 rpm DC motor" subject: "Utilization of Electrical Energy" chapter: 3 key: "semester_7/utilization_of_electrical_energy:full-converter-firing-angles-875rpm" source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/full-converter-firing-angles-875rpm site: BCT Engineering --- # Firing angles for 200 V, 875 rpm DC motor Utilization of Electrical Energy · Chapter 3: Control of Electric Drive - Asked in: 2080 Bhadra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/full-converter-firing-angles-875rpm - Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-3 ## Question A 200V, 875-rpm, 150A separately excited dc motor has an armature resistance of 0.06Ω. It is fed from a single phase fully-controlled rectifier with an AC source voltage of 220V, 50Hz. Assuming continuous conduction, calculate: i) Firing angle for rated motor torque and 750rpm. ii) Firing angle for rated motor torque and (−500)rpm. iii) Motor speed for α = 160° and rated torque. ## Short answer Motor constant K = 0.2183 V/rpm. (i) For 750 rpm, α = 29.3°. (ii) For −500 rpm, α = 120.4°. (iii) At α = 160°, speed = −893.9 rpm. ## Answer Data: $V = 200$ V, $N = 875$ rpm, $I_a = 150$ A, $R_a = 0.06\ \Omega$; supply 220 V, 50 Hz, $V_m = 311.13$ V. For a single-phase full converter (continuous conduction): $$ V_a = \frac{2V_m}{\pi}\cos\alpha = 198.07\cos\alpha $$ #### Motor constant At rated condition: $$ \begin{aligned} E &= V - I_a R_a = 200 - 150\times0.06 = 191\ \text{V} \\ K &= \frac{E}{N} = \frac{191}{875} = 0.21829\ \text{V/rpm} \end{aligned} $$ Rated torque means rated current, so $I_a R_a = 9$ V in every part. #### i) Rated torque, 750 rpm $$ \begin{aligned} E &= 0.21829 \times 750 = 163.71\ \text{V} \\ V_a &= 163.71 + 9 = 172.71\ \text{V} \\ \cos\alpha &= \frac{172.71}{198.07} = 0.8720 \Rightarrow \alpha = 29.31^\circ \end{aligned} $$ #### ii) Rated torque, −500 rpm Speed reversed, torque positive: regenerative braking in quadrant IV (e.g. an overhauling load). $$ \begin{aligned} E &= 0.21829 \times (-500) = -109.14\ \text{V} \\ V_a &= -109.14 + 9 = -100.14\ \text{V} \\ \cos\alpha &= \frac{-100.14}{198.07} = -0.5056 \Rightarrow \alpha = 120.37^\circ \end{aligned} $$ #### iii) Speed at $\alpha = 160^\circ$ and rated torque $$ \begin{aligned} V_a &= 198.07\cos160^\circ = -186.12\ \text{V} \\ E &= V_a - I_a R_a = -186.12 - 9 = -195.12\ \text{V} \\ N &= \frac{-195.12}{0.21829} = -893.9\ \text{rpm} \end{aligned} $$ | Case | $V_a$ (V) | $E$ (V) | Result | |---|---|---|---| | i | 172.71 | 163.71 | $\alpha$ = 29.3° | | ii | −100.14 | −109.14 | $\alpha$ = 120.4° | | iii | −186.12 | −195.12 | $N$ = −893.9 rpm | **Answer: (i) 29.3°, (ii) 120.4°, (iii) −894 rpm (reverse direction, regenerative braking).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the motor constant K found? A. K = V/N = 200/875 B. K = (V - Ia Ra)/N = 191/875 V/rpm C. K = Ia/N D. K = (V + Ia Ra)/N Answer: B. Rated back emf E = 200 - 150 x 0.06 = 191 V, so K = 0.2183 V/rpm. (Key point: Motor constant) 2. Why is Ia Ra = 9 V in every part? A. Because speed is fixed B. Because rated torque means rated current of 150 A C. Because Va is fixed D. Because alpha is fixed Answer: B. T is proportional to Ia, so rated torque means Ia = 150 A and Ia Ra = 9 V. (Key point: Rated torque) 3. For rated torque at -500 rpm, which gives the firing angle? A. alpha < 90° since Va positive B. alpha = 120.4° since Va = -109.14 + 9 = -100.14 V is negative (regenerative) C. alpha = 29.3° D. alpha = 160° Answer: B. E is negative, so Va is negative; cos alpha = -0.5056, alpha = 120.4°. (Key point: Negative speed) 4. At alpha = 160° with rated torque, the motor speed is: A. +894 rpm B. 0 rpm C. -894 rpm (reverse, regenerative) D. -500 rpm Answer: C. Va = 198.07 cos160° = -186.12 V, E = -195.12 V, N = -893.9 rpm. 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