---
title: "Shunt motor 480 V, 740 conductors"
subject: "Electrical Machine"
chapter: 4
key: "semester_4/electrical_machine:shunt-motor-480v-740-conductors"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/shunt-motor-480v-740-conductors
site: BCT Engineering
---

# Shunt motor 480 V, 740 conductors

Electrical Machine · Chapter 4: DC Motor

- Asked in: 2069 Poush
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/shunt-motor-480v-740-conductors
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-4

## Question

A dc shunt motor drains a current of 120 A from 480 V dc source, the armature winding and field winding resistances are [?] Ω and [?] Ω respectively. The motor has 6 poles and the armature winding has 740 conductors. The flux per pole is 0.08 Wb. Calculate: (i) Armature current (ii) Speed (iii) Torque developed by armature.

## Short answer

With assumed Ra=0.2 Ω, Rf=240 Ω, and lap winding: armature current = 118 A, speed ≈ 463 rpm, torque ≈ 1112 N·m.

## Answer

The armature and field resistances are not readable in the question. **Assumed values: R_a = 0.2 Ω and R_f = 240 Ω** (so the field current is 2 A). Winding type is also not stated, so a **lap winding (A = P = 6)** is assumed.

#### Equivalent circuit
```text
 I_L = 120 A --+--> Ia --[Ra=0.2]--(Eb)--+
               |                          |
               +--> If --[Rf=240]---------+--- V = 480 V
```

#### (i) Armature current
- I_f = 480 / 240 = 2 A
- **I_a = 120 − 2 = 118 A**

#### Back emf
E_b = V − I_aR_a = 480 − 118 × 0.2 = 480 − 23.6 = 456.4 V

#### (ii) Speed
E_b = PΦZN / (60A), so N = 60 A E_b / (P Φ Z)

N = (60 × 6 × 456.4) / (6 × 0.08 × 740) = 164 304 / 355.2

**N = 462.6 rpm**

#### (iii) Torque developed by the armature
ω = 2πN/60 = 2π × 462.6 / 60 = 48.44 rad/s

T = E_b I_a / ω = (456.4 × 118) / 48.44 = 53 855 / 48.44

**T = 1112 N·m**

Check: T = PΦZ I_a / (2πA) = (6 × 0.08 × 740 × 118) / (2π × 6) = 1112 N·m ✓

| Quantity | Value |
|---|---|
| I_a | 118 A |
| E_b | 456.4 V |
| N | 462.6 rpm |
| T | 1112 N·m |

If the armature is wave wound (A = 2), the speed would be 154.2 rpm and the torque 3335 N·m.

**Answer: I_a = 118 A, N ≈ 463 rpm, T ≈ 1112 N·m (with the assumed R_a = 0.2 Ω, R_f = 240 Ω, lap winding).**

## Check yourself

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

1. How is armature current found?
   A. Line current minus field current
   B. Line current plus field current
   C. V/R_a
   D. Equal to line current
   Answer: A. I_a = 120 - 2 = 118 A. (Key point: Armature current)

2. Which formula gives speed?
   A. N = PΦZ/(60 A E_b)
   B. N = E_b/I_a
   C. N = 60 A E_b/(PΦZ)
   D. N = 120f/P
   Answer: C. From E_b = PΦZN/(60A). (Key point: Speed formula)

3. For a lap winding with 6 poles, A equals?
   A. 2
   B. 1
   C. 12
   D. 6
   Answer: D. Lap winding has A = P = 6; a wave winding would have A = 2. (Key point: Winding assumption)

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