---
title: "Starting line current of capacitor start motor"
subject: "Electric Machines II"
chapter: 3
key: "semester_5/electric_machines_ii:capacitor-start-line-current"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/capacitor-start-line-current
site: BCT Engineering
---

# Starting line current of capacitor start motor

Electric Machines II · Chapter 3: Fractional Kilowatt Motors

- Asked in: 2078 Kartik, 6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/capacitor-start-line-current
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-3

## Question

The equivalent impedances of the main and auxiliary windings in a single-phase capacitor start motor are (15 + j22.5) Ω and (50 + j120) Ω respectively, while the capacitance of the capacitor is 12 μF. Determine the line current at starting on a 230 V, 50 Hz supply.

## Short answer

At starting, capacitor reactance Xc = 1/(2π×50×12e-6) = 265.26 Ω. Main current Im = 230/(15+j22.5) = 8.505∠−56.31° A, auxiliary current Ia = 230/(50+j(120−265.26)) = 1.497∠71.01° A. Line current = Im + Ia = 7.69∠−47.4° A, power factor 0.677 lagging.

## Answer

At starting both windings are across the 230 V supply. The capacitor is in series with the auxiliary winding. Line current is the phasor sum of the two winding currents.

**Given:** $Z_m = 15 + j22.5\ \Omega$, $Z_a = 50 + j120\ \Omega$, $C = 12\ \mu\text{F}$, $V = 230\angle 0^\circ$ V, $f = 50$ Hz.

#### Capacitive reactance

$$
X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times 50 \times 12\times10^{-6}} = 265.26\ \Omega
$$

#### Branch impedances

$$
\begin{aligned}
Z_m &= 15 + j22.5 = 27.04\angle 56.31^\circ\ \Omega \\
Z_a' &= 50 + j(120 - 265.26) = 50 - j145.26 \\
&= 153.62\angle -71.01^\circ\ \Omega
\end{aligned}
$$

#### Winding currents

$$
\begin{aligned}
I_m &= \frac{230\angle 0^\circ}{27.04\angle 56.31^\circ} = 8.505\angle -56.31^\circ\ \text{A} \\
&= 4.718 - j7.077\ \text{A} \\
I_a &= \frac{230\angle 0^\circ}{153.62\angle -71.01^\circ} = 1.497\angle 71.01^\circ\ \text{A} \\
&= 0.487 + j1.416\ \text{A}
\end{aligned}
$$

#### Line current

$$
\begin{aligned}
I_L &= I_m + I_a \\
&= (4.718 + 0.487) + j(-7.077 + 1.416) \\
&= 5.205 - j5.661 \\
&= 7.69\angle -47.40^\circ\ \text{A}
\end{aligned}
$$

Phase angle between $I_a$ and $I_m$ $= 71.01^\circ + 56.31^\circ = 127.3^\circ$; starting power factor $= \cos 47.40^\circ = 0.677$ lagging.

```text
        I_a (1.50 A, +71.0 deg)
         ^
         |
  -------+--------------> V
         | \
         |  \  I_L (7.69 A, -47.4 deg)
         |   \
         v I_m (8.51 A, -56.3 deg)
```

**Answer:** starting line current $I_L \approx 7.69\angle -47.4^\circ$ A, i.e. **7.69 A** at 0.677 lagging power factor.

## Check yourself

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

1. At starting, how is the line current of the capacitor-start motor obtained?
   A. Arithmetic sum of |I_m| and |I_a|
   B. Phasor sum I_m + I_a
   C. Difference of |I_m| and |I_a|
   D. Current of the main winding only
   Answer: B. Both windings are across the supply, so the line current is the phasor sum of the two winding currents. (Key point: Line current definition)

2. What is X_C for 12 uF at 50 Hz?
   A. 26.5 ohm
   B. 265.26 ohm
   C. 2.65 ohm
   D. 83.3 ohm
   Answer: B. X_C = 1/(2 pi x 50 x 12e-6) = 265.26 ohm. (Key point: Capacitive reactance)

3. What is the auxiliary branch impedance after adding the capacitor in series with 50 + j120 ohm?
   A. 50 + j385.26 ohm
   B. 50 - j145.26 ohm
   C. 50 + j145.26 ohm
   D. 315.26 ohm resistive
   Answer: B. Net reactance = 120 - 265.26 = -145.26 ohm, so Z_a' = 50 - j145.26 ohm. (Key point: Auxiliary branch impedance)

4. What is the starting line current found?
   A. 8.51 A
   B. 1.50 A
   C. 9.99 A
   D. 7.69 A
   Answer: D. I_m = 8.505 at -56.31 deg plus I_a = 1.497 at +71.01 deg gives 7.69 A at -47.4 deg. (Key point: Result)

## Related in Electric Machines II

- [DFRT, starting methods and capacitor start-run motor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Adfrt-starting-methods-csr) — 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
- [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
- [Starting capacitor and change in starting torque](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Acapacitor-quadrature-torque-change) — Electric Machines II, chapter 3
- [Increase in starting torque with 1000 μF capacitor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Acapacitor-1000uf-torque-increase) — Electric Machines II, chapter 3
- [Capacitor for quadrature currents (4.5+j3.7 Ω)](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Aquadrature-capacitor-250w-j3-7) — Electric Machines II, chapter 3, asked 2×

## Related in other subjects

- [Delta capacitor per phase for 50 kW motor](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Acapacitor-delta-415v-50kw-0-72) — Basic Electrical Engineering, chapter 8
- [Capacitor for single-phase motor pf correction](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Asingle-phase-motor-capacitor-40a-0-7) — Basic Electrical Engineering, chapter 7
- [Single-phase line capacitance, 6 mm, 1.3 m](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Asingle-phase-capacitance-6mm-1-3m) — Power System Analysis I, chapter 4
- [DFRT and making single phase motor self-starting](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Adfrt-self-starting-explain) — Electrical Machines, chapter 7, asked 3×