--- title: "Short shunt to long shunt compound generator (10 kW, 220 V)" subject: "Electrical Machines I" chapter: 3 key: "semester_4/electrical_machines_i:short-shunt-cumulative-10kw-220v-long" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/short-shunt-cumulative-10kw-220v-long site: BCT Engineering --- # Short shunt to long shunt compound generator (10 kW, 220 V) Electrical Machines I · Chapter 3: DC Generator - Asked in: 2070 Bhadra, 8 marks - Video: [Compound Generator 1. Short shunt compound generator 2. Long shunt compound generator](https://www.youtube.com/watch?v=GpP5E92egzQ) — Electrical learning & practice - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/short-shunt-cumulative-10kw-220v-long - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-3 ## Question Case-I: A short shunt cumulative compound dc generator supplies 10 kW at 220 V. The shunt field, series field and armature resistance are 100, 0.3 and 0.4 ohms respectively. Calculate the induced emf. Case-2: If the same generator is re-connected in long shunt compound mode and same value of emf and armature current is generated, calculate the terminal voltage across the load. ## Short answer In short-shunt, induced emf is 252.75 V. Reconnected as long-shunt with same emf and armature current, terminal voltage becomes 219.3 V. ## Answer **Given:** $P = 10$ kW, $V = 220$ V, $R_{sh} = 100\ \Omega$, $R_{se} = 0.3\ \Omega$, $R_a = 0.4\ \Omega$. Brush drop neglected. #### Case 1: Short-shunt connection ```text +--Ra--+-------Rse-------+ | | | (Eg) Rsh Load 220 V | | | +------+-----------------+ ``` $$ \begin{aligned} I_L &= \frac{10000}{220} = 45.455\ \text{A} \\ V_{sh} &= V + I_L R_{se} = 220 + 45.455 \times 0.3 = 233.64\ \text{V} \\ I_{sh} &= \frac{233.64}{100} = 2.336\ \text{A} \\ I_a &= I_L + I_{sh} = 47.791\ \text{A} \\ E_g &= V_{sh} + I_a R_a = 233.64 + 47.791 \times 0.4 \\ &= 233.64 + 19.12 = 252.75\ \text{V} \end{aligned} $$ #### Case 2: Reconnected as long shunt, same $E_g$ and $I_a$ ```text +--Ra--Rse--+-----------+ | | | (Eg) Rsh Load V | | | +-----------+-----------+ ``` Now the armature current flows through both $R_a$ and $R_{se}$: $$ \begin{aligned} V &= E_g - I_a (R_a + R_{se}) \\ &= 252.75 - 47.791 \times (0.4 + 0.3) \\ &= 252.75 - 33.45 = 219.30\ \text{V} \end{aligned} $$ Check: $I_{sh} = 219.30/100 = 2.193$ A, $I_L = 47.791 - 2.193 = 45.60$ A, which is consistent. **Answer:** Case 1: induced emf $E_g \approx 252.75$ V. Case 2: terminal voltage in long shunt $V \approx 219.3$ V (slightly lower, because the series field now carries the larger armature current). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is E_g in Case 1? A. 220 + 47.791 x 0.4 = 239.1 V B. 233.64 + 19.12 = 252.75 V C. 220 + 45.455 x 0.7 D. 233.64 - 19.12 Answer: B. V_sh = 220 + 45.455 x 0.3 = 233.64 V; E_g = V_sh + I_a R_a. (Key point: Short shunt emf) 2. In Case 2, which resistance carries I_a? A. R_a only B. R_a + R_se C. R_a + R_sh D. R_se only Answer: B. In long shunt the series field is in series with the armature. (Key point: Long shunt armature drop) 3. What is V in Case 2? A. 252.75 - 47.791 x 0.7 = 219.30 V B. 252.75 - 47.791 x 0.4 = 233.6 V C. 252.75 + 33.45 D. 252.75 - 45.455 x 0.7 Answer: A. V = E_g - I_a(R_a + R_se) = 252.75 - 33.45. (Key point: Terminal voltage) 4. Why is the long-shunt terminal voltage slightly lower? A. Shunt resistance rises B. Series field now carries the larger armature current, so the drop is larger C. Brush drop appears D. Flux changes sign Answer: B. Same E_g and I_a but the drop in R_se is now taken from E_g, not from the shunt branch. 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