---
title: "Diverter resistance for flat compound generator"
subject: "Electrical Machines"
chapter: 3
key: "semester_4/electrical_machines:gen-flat-compound-diverter"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/gen-flat-compound-diverter
site: BCT Engineering
---

# Diverter resistance for flat compound generator

Electrical Machines · Chapter 3: DC Generator

- Asked in: 2067 Mangsir (old course), 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/gen-flat-compound-diverter
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-3

## Question

A long shunt compound generator has a shunt field winding of 1000 turns per pole, series field winding of 4 turns per pole and resistance of $0.05\ \Omega$. In order to obtain the rated voltage both at no load and full load for operation as shunt generator, it is necessary to increase field current by 0.2 A. The full load armature current of compound generator is 80 A. Calculate the diverter resistance connected in parallel with series field to obtain flat compound operation.

## Short answer

Extra field ampere-turns needed = 0.2×1000 = 200 AT. Series field current required 200/4 = 50 A. Diverter carries 80−50 = 30 A. Diverter resistance = (50×0.05)/30 = 0.0833 Ω.

## Answer

For flat compounding, the series field must supply the extra field ampere-turns that a shunt generator needs to keep the voltage constant from no load to full load. A diverter in parallel with the series field passes the surplus armature current.

```text
        Ia      +----[Rd]----+
   ---->--------+            +----+---> to load
                +--[ Rse ]---+    |
                   Ise            Rsh (across load)
```

#### Step 1: Extra ampere-turns needed

$$
AT_{extra} = \Delta I_{sh}\times N_{sh} = 0.2 \times 1000 = 200\ \text{AT per pole}
$$

#### Step 2: Series field current required

$$
I_{se} = \frac{AT_{extra}}{N_{se}} = \frac{200}{4} = 50\ \text{A}
$$

#### Step 3: Diverter current

In a long shunt machine the series field (with its diverter) is in the armature circuit, so the full-load armature current of 80 A divides between the series field and the diverter:

$$
I_d = I_a - I_{se} = 80 - 50 = 30\ \text{A}
$$

#### Step 4: Diverter resistance

The diverter and series field are in parallel, so their voltage drops are equal:

$$
I_dR_d = I_{se}R_{se} \;\Rightarrow\; R_d = \frac{I_{se}R_{se}}{I_d} = \frac{50\times 0.05}{30} = 0.0833\ \Omega
$$

**Answer:** diverter resistance $R_d = 0.0833\ \Omega$.

## Check yourself

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

1. Flat compounding means the series field must:
   A. Supply the extra ampere-turns needed to keep voltage constant from no load to full load
   B. Oppose the shunt field
   C. Carry zero current
   D. Supply all the field flux
   Answer: A. The series field replaces the extra shunt field current. (Key point: Flat compounding)

2. The extra ampere-turns per pole are:
   A. 0.2 x 4 = 0.8 AT
   B. 0.2 x 1000 = 200 AT
   C. 200 x 4 = 800 AT
   D. 1000 / 0.2 = 5000 AT
   Answer: B. AT = change in shunt current x shunt turns. (Key point: Extra ampere-turns)

3. The required series field current is:
   A. 200 x 4 = 800 A
   B. 80 A
   C. 200 / 4 = 50 A
   D. 30 A
   Answer: C. Series AT = Ise x 4 = 200. (Key point: Series current)

4. The diverter resistance is:
   A. 0.05 ohm
   B. 0.03 ohm
   C. 0.133 ohm
   D. 50x0.05/30 = 0.0833 ohm
   Answer: D. The diverter takes 80 - 50 = 30 A at the same voltage drop as the series field. (Key point: Diverter)

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