---
title: "6-pole slip-ring IM no-load current"
subject: "Electrical Machine"
chapter: 5
key: "semester_4/electrical_machine:im-6pole-slipring-standstill-no-load-current"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/im-6pole-slipring-standstill-no-load-current
site: BCT Engineering
---

# 6-pole slip-ring IM no-load current

Electrical Machine · Chapter 5: Three Phase Induction Machines

- 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/im-6pole-slipring-standstill-no-load-current
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-5

## Question

A 6 pole, 50 Hz, 3-ph, slip ring induction motor has star connected stator and rotor windings. The rotor windings has impedance of 0.8 + j4 Ω/phase at standstill. The induced emf between slip rings at standstill is 400 V. The stator to rotor turn ratio is 4. The motor runs at 960 rpm at no load. Calculate current drawn by motor at standstill and no load.

## Short answer

At standstill, rotor current = 230.9/4.079 = 56.6 A, stator current = 14.15 A. At no-load (960 rpm, slip 0.04), rotor current = 9.24/0.816 = 11.3 A, stator current = 2.83 A.

## Answer

**Given:** 6-pole, 50 Hz, 3-phase slip-ring motor, star stator and rotor; Z₂ = 0.8 + j4 Ω/phase at standstill; emf between slip rings at standstill = 400 V; stator:rotor turns ratio = 4; no-load speed = 960 rpm.

**Assumption:** the magnetising (no-load) current of the stator is neglected, so the current drawn is the rotor current reflected to the stator by the turns ratio. Stator current I₁ = I₂ / 4.

N_s = 120 × 50 / 6 = 1000 rpm.

Rotor emf per phase E₂ = 400/√3 = 230.9 V. |Z₂| at standstill = √(0.8² + 4²) = 4.079 Ω.

#### At standstill (s = 1)
```text
 I2 = 230.9 / 4.079 = 56.6 A
 I1 = 56.6 / 4      = 14.15 A
```

#### At no load (960 rpm)
Slip s = (1000 − 960)/1000 = 0.04.

```text
 sE2  = 0.04 × 230.9 = 9.24 V
 sX2  = 0.04 × 4     = 0.16 ohm
 |Z2r| = √(0.8² + 0.16²) = 0.816 ohm
 I2 = 9.24 / 0.816 = 11.32 A
 I1 = 11.32 / 4    = 2.83 A
```
| Condition | Rotor current | Stator (line) current |
|---|---|---|
| Standstill | 56.6 A | 14.15 A |
| No load, 960 rpm | 11.32 A | 2.83 A |

**Answer: the motor draws about 14.2 A at standstill and about 2.83 A at no load (rotor current 56.6 A and 11.3 A respectively).**

## Check yourself

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

1. What is assumed so that the stator current equals reflected rotor current?
   A. Rotor resistance is zero
   B. Magnetising current is neglected
   C. Slip is zero at no load
   D. Stator impedance is large
   Answer: B. Neglecting I0, the stator current is just the rotor current divided by the turns ratio. (Key point: Assumption)

2. How is the standstill rotor current found?
   A. I2 = 230.9/4.079 = 56.6 A, then I1 = 56.6/4 = 14.15 A
   B. I2 = 400/4.079
   C. I2 = 230.9/0.8
   D. I1 = 56.6 x 4
   Answer: A. At s = 1 the full impedance 0.8 + j4 applies and the stator current is I2/4. (Key point: Standstill current)

3. The no-load speed is 960 rpm on a 6-pole 50 Hz motor. What is the slip?
   A. 0.04
   B. 0.96
   C. 0.4
   D. 0.004
   Answer: A. Ns = 1000, s = 40/1000 = 0.04. (Key point: No-load slip)

4. At s = 0.04 what happens to rotor emf and reactance?
   A. Both stay at standstill values
   B. Emf becomes sE2 = 9.24 V and reactance sX2 = 0.16 ohm
   C. Resistance scales to 0.032 ohm
   D. Emf becomes E2/s
   Answer: B. Both emf and reactance scale with s; resistance stays 0.8 ohm, so I2 = 11.32 A and I1 = 2.83 A. (Key point: No-load rotor impedance)

## Related in Electrical Machine

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- [Slip ring motor: starting, no-load, full-load currents](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aim-4pole-slipring-1490-1300-currents) — Electrical Machines, chapter 5
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- [Delta induction motor: slip, rotor emf, turns ratio](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aim-delta-440v-1440rpm-rotor-emf) — Electrical Machines, chapter 5, asked 2×
- [Slip-ring IM rotor current at 4% slip](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Aim-slipring-rotor-current-4-percent-slip) — Electrical Machine, chapter 5
- [Starting current and rotor resistance for 150 kW slip ring motor](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aim-150kw-3000v-starting-current-30a) — Electrical Machines I, chapter 5

## Related in other subjects

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- [8-pole motor: speed at max torque (50 N-m, j4)](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aim-8pole-50nm-starting-torque-j4) — Electrical Machines, chapter 5, asked 4×