---
title: "Slip-ring IM starting and running current"
subject: "Electrical Machine"
chapter: 5
key: "semester_4/electrical_machine:im-4pole-slipring-starting-running-current"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/im-4pole-slipring-starting-running-current
site: BCT Engineering
---

# Slip-ring IM starting and running current

Electrical Machine · Chapter 5: Three Phase Induction Machines

- Asked in: 2073 Magh, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/im-4pole-slipring-starting-running-current
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-5

## Question

A 4 pole, 50 Hz, 3-phase slip ring induction motor has star connected stator and rotor windings. The rotor winding has impedance of (1+j4) Ω per phase at stand still. The stator to rotor turn ratio is 2. Given that emf induced in rotor circuit is 400 V between two slip rings at stand still. Calculate starting current and running current at 1400 rpm.

## Short answer

At standstill, rotor current = 230.9/4.123 = 56.0 A, stator current = 28.0 A. At 1400 rpm (slip 0.0667), rotor current = 15.40/1.035 = 14.9 A, stator current = 7.44 A.

## Answer

**Given:** 4-pole, 50 Hz, star-connected stator and rotor; rotor impedance at standstill Z₂ = 1 + j4 Ω/phase; stator:rotor turns ratio = 2; rotor open-circuit emf between slip rings = 400 V (standstill).

N_s = 120 × 50 / 4 = 1500 rpm.

Rotor emf per phase (star) E₂ = 400 / √3 = 230.9 V.
|Z₂| = √(1² + 4²) = 4.123 Ω.

#### Starting current (s = 1)
```text
 Rotor current  I2 = E2 / |Z2| = 230.9 / 4.123 = 56.0 A
 Stator current I1 = I2 / K    = 56.0 / 2     = 28.0 A
```
(K = stator turns/rotor turns = 2; the magnetising current is neglected.)

#### Running current at 1400 rpm
Slip s = (1500 − 1400) / 1500 = 0.0667.

```text
 Rotor emf        sE2 = 0.0667 × 230.9   = 15.40 V
 Rotor reactance  sX2 = 0.0667 × 4       = 0.267 ohm
 |Z2r| = √(1² + 0.267²) = 1.035 ohm
 I2 = 15.40 / 1.035 = 14.88 A
 I1 = I2 / 2        = 7.44 A
```
| Condition | Slip | Rotor current | Stator current |
|---|---|---|---|
| Starting | 1 | 56.0 A | 28.0 A |
| 1400 rpm | 0.0667 | 14.88 A | 7.44 A |

**Answer: starting current is 56.0 A in the rotor (28.0 A in the stator); running current at 1400 rpm is 14.9 A in the rotor (7.44 A in the stator).**

## Check yourself

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

1. The emf between slip rings is 400 V for a star rotor. What is E2 per phase?
   A. 400 V
   B. 400 x sqrt(3) V
   C. 400/sqrt(3) = 230.9 V
   D. 400/3 V
   Answer: C. In a star connection phase emf = line emf / sqrt(3). (Key point: Rotor emf per phase)

2. Which gives rotor current at start?
   A. I2 = E2/|Z2| = 230.9/4.123 = 56 A
   B. I2 = E2/R2 = 230.9 A
   C. I2 = sE2/|Z2| with s = 0
   D. I2 = 400/4.123
   Answer: A. At s = 1, the full standstill impedance 1 + j4 applies. (Key point: Starting rotor current)

3. How is the stator current obtained from the rotor current?
   A. Multiply by turns ratio 2
   B. Divide by turns ratio 2, giving 28 A
   C. Add the magnetising current
   D. Divide by sqrt(3)
   Answer: B. With stator:rotor turns ratio 2, I1 = I2/2 = 28 A (magnetising current neglected). (Key point: Reflecting to the stator)

4. What is the slip at 1400 rpm and how does rotor impedance change?
   A. s = 0.0667; reactance becomes sX2 = 0.267 ohm, R stays 1 ohm
   B. s = 0.0667; resistance becomes sR2
   C. s = 0.933; reactance 3.73 ohm
   D. s = 0.0667; impedance stays 4.123 ohm
   Answer: A. Ns = 1500, s = 100/1500 = 0.0667; reactance scales with s, resistance does not, so I2 = 15.4/1.035 = 14.88 A. (Key point: Running at 1400 rpm)

## Related in Electrical Machine

- [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
- [6-pole slip-ring IM no-load current](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Aim-6pole-slipring-standstill-no-load-current) — Electrical Machine, chapter 5
- [Starting, no-load and full-load current of slip ring motor](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aslip-ring-0p8-4ohm-400v-ratio4) — Electrical Machines I, chapter 5
- [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
- [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
- [Slip ring motor: rotor current at 4% slip](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aim-slipring-120v-4-slip-current) — Electrical Machines, chapter 5

## Related in other subjects

- [200 kW 1420 rpm slip ring motor (star)](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Aslip-ring-200kw-star) — Electric Machine Design, chapter 4, asked 5×
- [Slip for maximum torque and full-load output of slip ring motor](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aslip-ring-400v-delta-star-190v) — Electrical Machines I, chapter 5, asked 2×
- [Split-phase motor winding currents at starting](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Asplit-phase-starting-currents) — Electric Machines II, chapter 3
- [6-pole induction motor: slip, losses, efficiency](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aim-4pole-975rpm-slip-losses) — Electrical Machines, chapter 5