---
title: "Efficiency and engine BHP of 48.75 kW shunt generator"
subject: "Electrical Machines I"
chapter: 3
key: "semester_4/electrical_machines_i:shunt-gen-48p75kw-250v-efficiency-bhp"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/shunt-gen-48p75kw-250v-efficiency-bhp
site: BCT Engineering
---

# Efficiency and engine BHP of 48.75 kW shunt generator

Electrical Machines I · Chapter 3: DC Generator

- Asked in: 2075 Bhadra, 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-48p75kw-250v-efficiency-bhp
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-3

## Question

A dc shunt generator gives an output of 48.75 kW at 250 V across the load. The armature winding resistance and field winding resistance are 0.02 Ω and 50 Ω respectively. No load power loss is 950 watt. Calculate the efficiency of the generator and B.H.P of the driving engine.

## Short answer

Load current 195 A, field current 5 A, armature current 200 A. Losses: armature Cu 800 W, field Cu 1250 W, stray 950 W, total 3000 W. Input 51750 W, efficiency 94.2%. BHP = 51750/746 ≈ 69.4 hp.

## Answer

Efficiency is output divided by input, where input = output + all losses. The driving engine must supply this input, so its B.H.P. is the input power in horsepower.

**Given:** $P_{out} = 48.75$ kW, $V = 250$ V, $R_a = 0.02\ \Omega$, $R_{sh} = 50\ \Omega$, no-load (iron + friction) loss = 950 W.

**Assumption:** the 950 W no-load loss is the iron and mechanical (stray) loss; shunt field copper loss is added separately.

#### Step 1: Currents
$$
\begin{aligned}
I_L &= \frac{48750}{250} = 195\ \text{A} \\
I_{sh} &= \frac{250}{50} = 5\ \text{A} \\
I_a &= I_L + I_{sh} = 200\ \text{A}
\end{aligned}
$$

#### Step 2: Losses
$$
\begin{aligned}
P_{cu,a} &= I_a^2 R_a = 200^2 \times 0.02 = 800\ \text{W} \\
P_{cu,sh} &= V I_{sh} = 250 \times 5 = 1250\ \text{W} \\
P_{stray} &= 950\ \text{W} \\
P_{loss} &= 800 + 1250 + 950 = 3000\ \text{W}
\end{aligned}
$$

| Loss | Value (W) |
|---|---|
| Armature copper | 800 |
| Shunt field copper | 1250 |
| Iron + friction | 950 |
| Total | 3000 |

#### Step 3: Efficiency
$$
\begin{aligned}
P_{in} &= 48750 + 3000 = 51750\ \text{W} \\
\eta &= \frac{48750}{51750} \times 100 = 94.20\%
\end{aligned}
$$

#### Step 4: B.H.P. of the engine
Taking 1 hp = 746 W:

$$
\text{B.H.P.} = \frac{51750}{746} = 69.37\ \text{hp}
$$

(With metric hp, 735.5 W, this is 70.36 hp.)

**Answer:** Efficiency $\eta = 94.2\%$; B.H.P. of driving engine $\approx 69.37$ hp.

## Check yourself

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

1. How is I_a obtained?
   A. I_a = I_L - I_sh = 190 A
   B. I_a = I_L + I_sh = 195 + 5 = 200 A
   C. I_a = I_L = 195 A
   D. I_a = V/R_a = 12500 A
   Answer: B. Output power gives I_L = 195 A, shunt field takes 5 A, so the armature carries 200 A. (Key point: Armature current)

2. Which set lists the generator's total losses correctly?
   A. 800 W armature Cu + 950 W stray, 1750 W
   B. 800 W + 1250 W shunt Cu + 950 W = 3000 W
   C. 1250 W + 950 W = 2200 W
   D. 195^2 x 0.02 + 950 W
   Answer: B. Armature Cu = 200^2 x 0.02 = 800 W, shunt field Cu = 250 x 5 = 1250 W, plus 950 W iron and friction. (Key point: Losses)

3. What is the efficiency?
   A. 48750/51750 = 94.2%
   B. 48750/48750+950 = 98%
   C. 51750/48750 = 106%
   D. 48750/3000 = 16.25
   Answer: A. Input = output + losses = 51750 W, so efficiency = 48750/51750 = 94.2%. (Key point: Efficiency)

4. How is the engine B.H.P. found?
   A. Output power / 746 = 65.35 hp
   B. Input power / 746 = 51750/746 = 69.37 hp
   C. Loss / 746 = 4.02 hp
   D. Output x 746
   Answer: B. The engine must supply the generator's input power, so B.H.P. = input in watts divided by 746. (Key point: B.H.P. of engine)

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