---
title: "Wave-wound shunt generator, power absorbed by load"
subject: "Electrical Machines"
chapter: 3
key: "semester_4/electrical_machines:gen-shunt-378-conductors-load-power"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/gen-shunt-378-conductors-load-power
site: BCT Engineering
---

# Wave-wound shunt generator, power absorbed by load

Electrical Machines · Chapter 3: DC Generator

- Asked in: 2075 Baisakh (old course), 8 marks
- Video: [How to Calculate Power Absorbed by Load in a DC Shunt Generator (Step-by-Step Solved Problem)](https://www.youtube.com/watch?v=gDto9G4oY3E) — 5 Minute Electrical 
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/gen-shunt-378-conductors-load-power
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-3

## Question

A 4 pole d.c shunt generator with a field resistance of $100\ \Omega$ and an armature resistance of $1\ \Omega$ has 378 wave connected conductors in its armature. The flux per pole is 0.02 Wb. If a total resistance of $10\ \Omega$ is connected across the armature terminals and the generator is driven at 1000 rpm. Calculate the power absorbed by the load.

## Short answer

Generated emf E = (0.02×378×1000×4)/(60×2) = 252 V. Load and field in parallel with armature yield equivalent resistance 9.091 Ω. Armature current 24.973 A, terminal voltage 227.03 V. Load power = (227.03)²/10 = 5154 W.

## Answer

The load (10 Ω) and the field winding (100 Ω) are both across the armature terminals, so they are in parallel. The generated emf comes from the emf equation for a wave winding ($A = 2$).

#### Step 1: Generated emf

$$
E = \frac{\Phi Z N P}{60A} = \frac{0.02\times 378\times 1000\times 4}{60\times 2} = 252\ \text{V}
$$

#### Step 2: Terminal resistance and armature current

Load in parallel with field:

$$
R_p = \frac{100\times 10}{100 + 10} = 9.091\ \Omega
$$

$$
I_a = \frac{E}{R_a + R_p} = \frac{252}{1 + 9.091} = 24.973\ \text{A}
$$

#### Step 3: Terminal voltage

$$
V = I_aR_p = (24.973)(9.091) = 227.03\ \text{V}
$$

#### Step 4: Load current and power

$$
I_L = \frac{V}{10} = 22.703\ \text{A}, \qquad I_f = \frac{V}{100} = 2.270\ \text{A}
$$

(Check: $I_L + I_f = 24.973\ \text{A} = I_a$.)

$$
P_{load} = \frac{V^2}{R_L} = \frac{(227.03)^2}{10} = 5154\ \text{W}
$$

**Answer:** power absorbed by the load $\approx 5154\ \text{W}$ ($\approx 5.15$ kW) at a terminal voltage of $227.03\ \text{V}$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For wave winding with A = 2, the emf is:
   A. 126 V
   B. 252 V
   C. 504 V
   D. 1008 V
   Answer: B. E = 0.02x378x1000x4/(60x2) = 252 V. (Key point: Wave winding emf)

2. The 10 ohm load and the 100 ohm field are:
   A. In series, giving 110 ohm
   B. Only the load matters
   C. In parallel across the armature terminals, giving 9.091 ohm
   D. Field in series with Ra
   Answer: C. Both are across armature terminals. (Key point: Parallel combination)

3. Terminal voltage is:
   A. 252 V
   B. 25 V
   C. E / Ra
   D. Ia x Rp = 227 V
   Answer: D. Ia = 252/(1 + 9.091) = 24.97 A; V = 24.97x9.091 = 227.03 V. (Key point: Terminal voltage)

4. The power absorbed by the load is:
   A. V^2/RL = about 5154 W
   B. about 6350 W
   C. about 5750 W
   D. about 62,500 W
   Answer: A. 227.03^2/10 = 5154 W; field power not included. (Key point: Load power)

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