--- title: "Shunt generator feeding six lamps" subject: "Electrical Machines I" chapter: 3 key: "semester_4/electrical_machines_i:shunt-gen-lap-six-lamps" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/shunt-gen-lap-six-lamps site: BCT Engineering --- # Shunt generator feeding six lamps Electrical Machines I · Chapter 3: DC Generator - Asked in: 2078 Chaitra, 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-gen-lap-six-lamps - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-3 ## Question A 4 pole shunt generator with a lap wound armature has armature resistance of 0.1 ohm and field circuit resistance of 50 ohm. The generator supplies power to six filament lamps of rated 200 V 600 Watt each. The brush contact drop is 1 V per brush. If generator terminal voltage is 220 V after lamps are connected then find armature current and generated emf. ## Short answer Each lamp's resistance is 66.67 Ω, drawing 3.3 A at 220 V; six lamps give IL=19.8 A. Field current Ish=4.4 A, so armature current Ia=24.2 A. Generated emf Eg = V + IaRa + brush drop = 220 + 2.42 + 2 = 224.42 V. ## Answer In a shunt generator the armature current supplies both the load (lamps) and the shunt field. The lamps are treated as fixed resistances found from their rating, and they now work at 220 V. **Given:** $V = 220$ V, $R_a = 0.1\ \Omega$, $R_{sh} = 50\ \Omega$, six lamps of 200 V, 600 W, brush drop 1 V per brush. #### Lamp current Resistance of one lamp (from its rating): $$ R_{lamp} = \frac{V_{rated}^2}{P_{rated}} = \frac{200^2}{600} = 66.67\ \Omega $$ At 220 V, current through one lamp $= \dfrac{220}{66.67} = 3.3$ A. Six lamps are in parallel, so $$ I_L = 6 \times 3.3 = 19.8\ \text{A} $$ #### Field and armature current $$ \begin{aligned} I_{sh} &= \frac{V}{R_{sh}} = \frac{220}{50} = 4.4\ \text{A} \\ I_a &= I_L + I_{sh} = 19.8 + 4.4 = 24.2\ \text{A} \end{aligned} $$ #### Generated emf In a lap winding the brushes of the same polarity are in parallel, so the current path passes through one positive and one negative brush set. Total brush drop $= 2 \times 1 = 2$ V. $$ \begin{aligned} E_g &= V + I_a R_a + V_{brush} \\ &= 220 + 24.2 \times 0.1 + 2 \\ &= 220 + 2.42 + 2 = 224.42\ \text{V} \end{aligned} $$ ```text +----Ra---+------+--------+ | | | | (Eg) Rsh lamps x6 V=220 V | | | | +---------+------+--------+ ``` **Answer:** Armature current $I_a = 24.2$ A, generated emf $E_g = 224.42$ V. (If the lamps were assumed to draw their rated current of 3 A each, $I_L = 18$ A, $I_a = 22.4$ A and $E_g = 224.24$ V. The resistance method above is more exact because the lamps are not at rated voltage.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the resistance of each lamp found? A. R = P/V = 600/200 = 3 ohm B. R = V^2/P = 200^2/600 = 66.67 ohm C. R = V/P = 200/600 = 0.33 ohm D. R = P^2/V = 600^2/200 = 1800 ohm Answer: B. Lamp resistance comes from its rating as R = V_rated^2/P_rated, giving 66.67 ohm. (Key point: Lamp resistance from rating) 2. Why is the load current 19.8 A and not 6 x 3 = 18 A? A. Because the lamps are connected in series B. Because the shunt field adds to the lamp current C. Because the lamps run at 220 V, not 200 V, so each draws 3.3 A D. Because the brush drop increases the lamp current Answer: C. Lamps are fixed resistances; at 220 V each draws 220/66.67 = 3.3 A, so six draw 19.8 A. (Key point: Lamps at non-rated voltage) 3. How is the armature current obtained? A. I_a = I_L - I_sh = 15.4 A B. I_a = I_L = 19.8 A C. I_a = I_L + I_sh = 19.8 + 4.4 = 24.2 A D. I_a = V/R_a = 2200 A Answer: C. The armature supplies both the load and the shunt field, so I_a = I_L + I_sh with I_sh = 220/50 = 4.4 A. (Key point: Armature current of shunt generator) 4. What is the generated emf? A. 220 + 2.42 + 2 = 224.42 V B. 220 - 2.42 - 2 = 215.58 V C. 220 + 2.42 + 1 = 223.42 V D. 220 + 2.42 = 222.42 V Answer: A. E_g = V + I_a R_a + total brush drop; the current passes one positive and one negative brush, so the drop is 2 V. 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