---
title: "Series motor resistance for 800 rpm"
subject: "Electrical Machine"
chapter: 4
key: "semester_4/electrical_machine:series-motor-armature-resistance-800rpm"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/series-motor-armature-resistance-800rpm
site: BCT Engineering
---

# Series motor resistance for 800 rpm

Electrical Machine · Chapter 4: DC Motor

- Asked in: 2073 Bhadra, 4+4 marks; 2069 Poush (asked 2 times)
- Video: [Problem in DC series motor](https://www.youtube.com/watch?v=sYf7Uo5WxSo) — BeTECH SIVA
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/series-motor-armature-resistance-800rpm
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-4

## Question

A 220 V, DC series motor draws 100 A current and runs at 1200 RPM. What is the value of armature resistance required to run the motor at 800 RPM keeping load torque constant. Given armature resistance = 0.2 ohm, field winding = 0.05 ohm.

## Short answer

Constant-torque load keeps armature current at 100 A. Back emf at 1200 rpm: 220 − 100×(0.2+0.05) = 195 V. At 800 rpm: 195 × (800/1200) = 130 V. Total circuit resistance needed: (220−130)/100 = 0.9 Ω. Extra resistance = 0.9 − 0.25 = 0.65 Ω.

## Answer

For a series motor the flux is proportional to the armature current (below saturation). At constant load torque, T ∝ Φ I_a ∝ I_a², so the current stays the same (100 A) after the speed is reduced, and the flux is unchanged too.

#### Equivalent circuit
```text
 +--[Ra=0.2]--[Rse=0.05]--[R extra]--(Eb)--+
 |            I = 100 A                     |
 +---------------- V = 220 V ---------------+
```

#### Step 1: back emf at 1200 rpm
E_b1 = V − I(R_a + R_se) = 220 − 100 × (0.2 + 0.05) = 220 − 25 = 195 V

#### Step 2: back emf at 800 rpm
E_b ∝ Φ N, and Φ is the same (current is same), so
E_b2 = E_b1 × (N₂/N₁) = 195 × (800/1200) = 130 V

#### Step 3: total resistance needed
V = E_b2 + I (R_a + R_se + R)

220 = 130 + 100 × (0.25 + R)

0.25 + R = 90/100 = 0.9 Ω

**R = 0.9 − 0.25 = 0.65 Ω**

| Condition | Speed (rpm) | E_b (V) | Total series R (Ω) |
|---|---|---|---|
| Normal | 1200 | 195 | 0.25 |
| Reduced | 800 | 130 | 0.90 |

**Answer: an extra resistance of 0.65 Ω must be connected in series with the armature.**

## Check yourself

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

1. Why does the current remain 100 A at constant load torque?
   A. Because voltage is constant
   B. Because speed is constant
   C. Because Ra is constant
   D. T ∝ Φ I_a ∝ I_a² in a series motor, so constant torque means constant I_a
   Answer: D. Same torque needs the same I_a, so flux is also the same. (Key point: Constant torque)

2. What is E_b at 800 rpm?
   A. 130 V
   B. 195 V
   C. 180 V
   D. 146 V
   Answer: A. E_b1 = 220 - 100 x 0.25 = 195 V; E_b2 = 195 x 800/1200. (Key point: Back emf at 800 rpm)

3. What extra resistance is needed?
   A. 0.9 ohm
   B. 0.65 ohm
   C. 0.25 ohm
   D. 0.4 ohm
   Answer: B. 220 = 130 + 100 x (0.25 + R) gives R = 0.65 ohm. (Key point: Extra resistance)

## Related in Electrical Machine

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## Related in other subjects

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- [Speed of 220 V series motor at reduced load current](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aseries-motor-220v-38a-speed) — Electrical Machines I, chapter 4
- [Shunt motor speed with 0.1 Ω armature resistance, less load](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-240v-5a-armature-0p1) — Electrical Machines I, chapter 4, asked 3×