--- title: "Power to 10 Ω load from shunt generator" subject: "Electrical Machines I" chapter: 3 key: "semester_4/electrical_machines_i:shunt-gen-378-conductors-10ohm-load" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/shunt-gen-378-conductors-10ohm-load site: BCT Engineering --- # Power to 10 Ω load from shunt generator Electrical Machines I · Chapter 3: DC Generator - Asked in: 2075 Baisakh, 8 marks - Video: [DC. Shunt generator : problem & solution](https://www.youtube.com/watch?v=lU9cq1u_hLE) — Sumit Kumar Maitra - 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-378-conductors-10ohm-load - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-3 ## Question A 4 pole d.c shunt generator with a field resistance of 100 Ω and an armature resistance of 1 Ω has 378 wave connected conductors in its armature. The flux per pole is 0.02 Wb. If a total resistance of 10 Ω is connected across the armature terminals and the generator is driven at 1000 rpm, calculate the power absorbed by the load. ## Short answer Use E_g = P φ Z N / (60 A) with wave winding A=2 to get 252 V. Load and shunt field in parallel give 9.091 Ω, so armature current is 24.97 A and terminal voltage 227.03 V. Load power is V²/10 = 5154 W ≈ 5.15 kW. ## Answer First find the generated emf from the emf equation, then solve the circuit (armature resistance in series with the load and shunt field in parallel) to get the terminal voltage. **Given:** $P = 4$, wave winding so $A = 2$, $Z = 378$, $\phi = 0.02$ Wb, $N = 1000$ rpm, $R_a = 1\ \Omega$, $R_{sh} = 100\ \Omega$, $R_L = 10\ \Omega$. #### Step 1: Generated emf $$ \begin{aligned} E_g &= \frac{P \phi Z N}{60 A} \\ &= \frac{4 \times 0.02 \times 378 \times 1000}{60 \times 2} = 252\ \text{V} \end{aligned} $$ #### Step 2: Equivalent resistance across the terminals Load and shunt field are in parallel: $$ R_p = \frac{R_L R_{sh}}{R_L + R_{sh}} = \frac{10 \times 100}{110} = 9.091\ \Omega $$ ```text +---Ra=1---+-------+ | | | (252 V) Rsh=100 RL=10 | | | +----------+-------+ ``` #### Step 3: Armature current and terminal voltage $$ \begin{aligned} I_a &= \frac{E_g}{R_a + R_p} = \frac{252}{1 + 9.091} = 24.97\ \text{A} \\ V &= I_a R_p = 24.97 \times 9.091 = 227.03\ \text{V} \end{aligned} $$ Check: $V = E_g - I_a R_a = 252 - 24.97 = 227.03$ V. #### Step 4: Power absorbed by load $$ \begin{aligned} I_L &= \frac{227.03}{10} = 22.70\ \text{A} \\ P_L &= \frac{V^2}{R_L} = \frac{227.03^2}{10} = 5154\ \text{W} \end{aligned} $$ **Answer:** Power absorbed by the load $\approx 5.15$ kW (5154 W), with $E_g = 252$ V and $V = 227.03$ V. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How many parallel paths A are used in the emf equation of a wave winding? A. A = P = 4 B. A = 2 C. A = 1 D. A = 8 Answer: B. Wave winding always has A = 2 regardless of the number of poles. (Key point: Parallel paths in wave winding) 2. What is E_g? A. 4 x 0.02 x 378 x 1000/(60 x 4) = 126 V B. 4 x 0.02 x 378 x 1000/(60 x 2) = 252 V C. 0.02 x 378 x 1000/60 = 126 V D. 4 x 0.02 x 378 x 1000/60 = 504 V Answer: B. E_g = P phi Z N/(60 A) = 252 V. (Key point: Generated emf) 3. How is the terminal voltage found? A. Divide E_g between R_a and the load only B. Take V = E_g C. Load and shunt field are in parallel (9.091 ohm), in series with R_a; V = E_g x R_p/(R_a + R_p) D. Add R_sh and R_L in series Answer: C. Load and field are across the terminals in parallel; with R_a they form a series circuit, giving I_a = 24.97 A and V = 227.03 V. (Key point: Circuit solution) 4. What is the power absorbed by the load? A. E_g x I_a = 6.3 kW B. I_a^2 R_a C. V^2/R_L = 227.03^2/10 = 5154 W D. V x I_a = 5669 W Answer: C. Only the load resistor's power counts: V^2/R_L, not the power that includes the field. 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