---
title: "Induction motor current with load; stator connection"
subject: "Electrical Machines"
chapter: 5
key: "semester_4/electrical_machines:im-current-adjust-load-stator-connection"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/im-current-adjust-load-stator-connection
site: BCT Engineering
---

# Induction motor current with load; stator connection

Electrical Machines · Chapter 5: Three Phase Induction Machines

- 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/im-current-adjust-load-stator-connection
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-5

## Question

How does an induction motor adjust its current with the changes in shaft load? Explain the effect of type of connection of stator winding of slip ring induction motor.

## Short answer

As load increases, rotor slows, slip rises, boosting rotor current and torque; stator current automatically rises to balance rotor MMF and maintain flux. Stator winding connection: star reduces phase voltage to VL/√3, limiting starting current and torque to one-third of delta, used for star-delta starting.

## Answer

#### How the motor adjusts its current with load
An induction motor behaves like a transformer with a rotating secondary.

1. At no load, the rotor runs near $N_s$ and the slip is tiny, so the rotor emf $sE_2$ and the rotor current are very small. The stator takes only the no-load current $I_0$ (mostly magnetising, power factor 0.1 to 0.2).
2. When the shaft load increases, the load torque exceeds the electromagnetic torque and the rotor **slows down**.
3. The slip $s$ increases, so the relative speed between the field and the rotor increases. The rotor emf $sE_2$ and its frequency $sf$ increase, so the **rotor current** $I_2 = \frac{sE_2}{\sqrt{R_2^2+(sX_2)^2}}$ increases. The rotor power factor also improves.
4. The torque therefore rises until it matches the load, and the motor settles at a new speed with slightly higher slip.
5. The rotor mmf opposes the stator flux (Lenz's law) and tends to reduce it. To keep the flux constant ($\phi \propto V_1$), the stator draws an extra current $I_2'$:
$$
\vec I_1 = \vec I_0 + \vec I_2'
$$
So the **stator current rises automatically with load**, and the input power factor improves with load.

#### Effect of stator connection (slip-ring motor)
The stator can be connected in star or delta (the rotor is normally star). For the same line voltage $V_L$:

| Quantity | Star | Delta |
|---|---|---|
| Phase voltage | $V_L/\sqrt3$ | $V_L$ |
| Phase current (relative) | 1/√3 of delta | 1 |
| Line current | 1/3 of delta | 1 |
| Torque ($\propto V_{ph}^2$) | 1/3 of delta | 1 |

- A delta-connected winding is used for normal running; it takes more current but delivers full torque.
- Star connection at starting reduces the starting current to one third and also gives one third of the torque (**star-delta starter** for delta-rated motors).
- The rotor connection does not change the stator current directly. With a star-connected rotor, the phase emf is $E_{L}/\sqrt3$, where $E_L$ is the emf between slip rings; this sets the rotor current and the values of the external resistance.

## Check yourself

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

1. When shaft load increases, the rotor first:
   A. Speeds up
   B. Slows down, increasing slip
   C. Stays the same
   D. Stops
   Answer: B. Load exceeds torque so rotor slows. (Key point: Load change)

2. Why does rotor current rise with slip?
   A. Rotor emf sE_2 increases with slip
   B. R_2 falls
   C. The flux falls
   D. Voltage rises
   Answer: A. I_2 = sE_2/sqrt(R_2^2 + (sX_2)^2) increases. (Key point: Rotor current)

3. Stator current increases with load because:
   A. The rotor mmf opposes flux, so the stator draws extra I_2' (I_1 = I_0 + I_2')
   B. Supply voltage rises
   C. Flux increases
   D. Frequency rises
   Answer: A. Flux is held constant by the supply voltage. (Key point: Stator current)

4. For the same line voltage, star connection compared with delta gives line current of:
   A. 3 times
   B. 1/3
   C. Same
   D. 1/sqrt3
   Answer: B. Star line current is one third of delta. (Key point: Star vs delta)

## Related in Electrical Machines

- [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 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
- [Slip ring motor max-torque slip and full-load output](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aslip-ring-380v-180v-60nm-starting) — Electrical Machines I, chapter 5, asked 4×
- [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×
- [Effect of five-times rotor resistance on 40 kW slip ring motor](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aslip-ring-40kw-rotor-resistance-5x) — Electrical Machines I, chapter 5

## Related in other subjects

- [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
- [IM and sync motor current adjust with load](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Aim-current-adjust-with-load) — Electrical Machine, chapter 5
- [Slip power recovery for slip ring induction motor](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Aslip-power-recovery-slip-ring-motor) — Utilization of Electrical Energy, chapter 3, asked 2×
- [Slip-ring IM starting and running current](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Aim-4pole-slipring-starting-running-current) — Electrical Machine, chapter 5