---
title: "Split-phase motor winding currents at starting"
subject: "Electric Machines II"
chapter: 3
key: "semester_5/electric_machines_ii:split-phase-starting-currents"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/split-phase-starting-currents
site: BCT Engineering
---

# Split-phase motor winding currents at starting

Electric Machines II · Chapter 3: Fractional Kilowatt Motors

- Asked in: 2082 Bhadra (new course), 3 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/split-phase-starting-currents
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-3

## Question

At starting, the windings of a 230 V, 50 Hz, split-phase induction motor have the following parameters:
Main winding: R = 4 Ω; X1 = 7.5 Ω
Starting winding: R = 7.5 Ω; X1 = 4 Ω
Find
a) Current in the main winding
b) Current in the starting winding
c) Phase angle between Is and Im
d) Line current and
e) Power factor of the motor.

## Short answer

Both windings are across 230 V. Impedances: 8.5 angle 61.93 deg and 8.5 angle 28.07 deg. Currents: Im=27.06 A, Is=27.06 A, phase angle 33.86 deg (Is leads Im). Phasor sum gives line current 51.77 A at power factor 0.707 lagging.

## Answer

Both windings are connected directly across the 230 V supply at starting, so each current is $V/Z$ and the line current is their phasor sum.

**Given:** $Z_m = 4 + j7.5\ \Omega$, $Z_s = 7.5 + j4\ \Omega$, $V = 230\angle 0^\circ$ V.

$$
\begin{aligned}
Z_m &= \sqrt{4^2 + 7.5^2}\angle\tan^{-1}\tfrac{7.5}{4} = 8.5\angle 61.93^\circ\ \Omega \\
Z_s &= \sqrt{7.5^2 + 4^2}\angle\tan^{-1}\tfrac{4}{7.5} = 8.5\angle 28.07^\circ\ \Omega
\end{aligned}
$$

#### a) Main winding current

$$
I_m = \frac{230\angle 0^\circ}{8.5\angle 61.93^\circ} = 27.06\angle -61.93^\circ\ \text{A} = 12.73 - j23.88\ \text{A}
$$

#### b) Starting winding current

$$
I_s = \frac{230\angle 0^\circ}{8.5\angle 28.07^\circ} = 27.06\angle -28.07^\circ\ \text{A} = 23.88 - j12.73\ \text{A}
$$

#### c) Phase angle between $I_s$ and $I_m$

$$
\alpha = 61.93^\circ - 28.07^\circ = 33.86^\circ \quad (I_s \text{ leads } I_m)
$$

#### d) Line current

$$
\begin{aligned}
I_L &= I_m + I_s = (12.73 + 23.88) - j(23.88 + 12.73) \\
&= 36.61 - j36.61 = 51.77\angle -45^\circ\ \text{A}
\end{aligned}
$$

#### e) Power factor

$$
\text{pf} = \cos 45^\circ = 0.707\ \text{lagging}
$$

**Answer:** $I_m = 27.06$ A, $I_s = 27.06$ A, $\alpha = 33.86^\circ$, $I_L = 51.77$ A, pf $= 0.707$ lagging.

## Check yourself

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

1. How are the winding currents found at starting?
   A. Each is V/Z because both windings are across the supply
   B. Using the slip
   C. Using a series capacitor
   D. Using the turns ratio only
   Answer: A. Both windings connect directly across 230 V, so each current is V divided by its own impedance. (Key point: Method)

2. What is the main winding current for Zm = 4 + j7.5 ohm at 230 V?
   A. 27.06 A at -61.93°
   B. 27.06 A at -28.07°
   C. 51.77 A at -45°
   D. 13.5 A at -61.93°
   Answer: A. |Zm| = 8.5 ohm, so Im = 230/8.5 = 27.06 A at an angle of -tan^-1(7.5/4) = -61.93°. (Key point: Winding currents)

3. What is the angle between Is and Im?
   A. 90°
   B. 33.86°, Is leading Im
   C. 33.86°, Im leading Is
   D. 0°
   Answer: B. Im lags by 61.93° and Is by 28.07°, so Is leads Im by 33.86°. (Key point: Phase angle between currents)

4. What are the line current and the power factor?
   A. 54.12 A, 0.5
   B. 51.77 A, 0.707 lagging
   C. 51.77 A, 0.707 leading
   D. 27.06 A, 0.866 lagging
   Answer: B. IL = Im + Is = 36.61 - j36.61 = 51.77 A at -45°, so pf = cos 45° = 0.707 lagging. (Key point: Line current and pf)

## Related in Electric Machines II

- [Main winding current and pf at 5% slip](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Asingle-phase-im-main-current) — Electric Machines II, chapter 3
- [Starting line current of capacitor start motor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Acapacitor-start-line-current) — Electric Machines II, chapter 3
- [Split phase motor added resistance and capacitance](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Asplit-phase-resistance-capacitance) — Electric Machines II, chapter 3
- [Not self starting and permanent split capacitor motor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Anot-self-starting-psc-motor) — Electric Machines II, chapter 3
- [Input current, pf and torque of single phase IM](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Asingle-phase-im-input-torque) — Electric Machines II, chapter 3
- [Synchronous motor starting methods and damper winding](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Amotor-starting-methods-damper) — Electric Machines II, chapter 2

## Related in other subjects

- [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
- [Split phase induction motor](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Asplit-phase-im) — Electrical Machine, chapter 7
- [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
- [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