---
title: "Increase in starting torque with 1000 μF capacitor"
subject: "Electric Machines II"
chapter: 3
key: "semester_5/electric_machines_ii:capacitor-1000uf-torque-increase"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/capacitor-1000uf-torque-increase
site: BCT Engineering
---

# Increase in starting torque with 1000 μF capacitor

Electric Machines II · Chapter 3: Fractional Kilowatt Motors

- Asked in: 2074 Asoj, 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-1000uf-torque-increase
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-3

## Question

A four pole, single phase, 120 V, 50 Hz induction motor gave the following standstill impedances when tested at rated frequency. Main winding: Zm = (1.5+j4) ohms. Auxiliary winding: Za = (3+j6) ohms. If an external capacitor of 1000 μF is inserted in series with the auxiliary winding to obtain higher starting torque, calculate the percentage increase in starting torque.

## Short answer

Inserting a 1000 μF capacitor in series with the auxiliary winding changes its impedance from 6.708 Ω to 4.115 Ω, increasing current from 17.89 A to 29.16 A and widening the phase angle from 6.01° to 26.25°. The starting torque becomes 6.886 times the original, a 589% increase.

## Answer

Starting torque of a two-winding single-phase motor is proportional to the product of the winding currents and the sine of the angle between them:

$$
T_{st} = K I_m I_a \sin\alpha, \qquad I = \frac{V}{|Z|}
$$

**Given:** $V = 120$ V, $f = 50$ Hz, $Z_m = 1.5 + j4\ \Omega$, $Z_a = 3 + j6\ \Omega$, $C = 1000\ \mu\text{F}$.

#### Main winding (unchanged)

$$
Z_m = 1.5 + j4 = 4.272\angle 69.44^\circ\ \Omega,\quad I_m = 28.09\angle -69.44^\circ\ \text{A}
$$

#### Case 1: without capacitor

$$
\begin{aligned}
Z_a &= 3 + j6 = 6.708\angle 63.43^\circ\ \Omega \\
I_a &= \frac{120}{6.708} = 17.89\angle -63.43^\circ\ \text{A} \\
\alpha_1 &= 69.44^\circ - 63.43^\circ = 6.01^\circ \\
T_1 &= K(28.09)(17.89)\sin 6.01^\circ = 52.62\,K
\end{aligned}
$$

#### Case 2: with 1000 μF capacitor

$$
\begin{aligned}
X_C &= \frac{1}{2\pi \times 50 \times 1000\times10^{-6}} = 3.183\ \Omega \\
Z_a' &= 3 + j(6 - 3.183) = 3 + j2.817 = 4.115\angle 43.20^\circ\ \Omega \\
I_a' &= \frac{120}{4.115} = 29.16\angle -43.20^\circ\ \text{A} \\
\alpha_2 &= 69.44^\circ - 43.20^\circ = 26.25^\circ \\
T_2 &= K(28.09)(29.16)\sin 26.25^\circ = 362.3\,K
\end{aligned}
$$

#### Ratio and percentage increase

$$
\begin{aligned}
\frac{T_2}{T_1} &= \frac{29.16 \sin 26.25^\circ}{17.89 \sin 6.01^\circ} = 6.886 \\
\%\ \text{increase} &= (6.886 - 1)\times 100 = 588.6\%
\end{aligned}
$$

| Quantity | Without C | With 1000 μF |
|---|---|---|
| Auxiliary impedance $\lvert Z_a\rvert$ (Ω) | 6.708 | 4.115 |
| Auxiliary current $I_a$ (A) | 17.89 | 29.16 |
| Angle α | 6.01° | 26.25° |
| Relative $T_{st}$ | 1 | 6.886 |

**Answer:** starting torque increases about **6.89 times**, i.e. an increase of about **589%**.

## Check yourself

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

1. Starting torque is proportional to:
   A. I_m I_a sin(alpha)
   B. I_m / I_a
   C. V^2 only
   D. I_m + I_a
   Answer: A. T_st = K I_m I_a sin(alpha), where alpha is the angle between the winding currents. (Key point: Torque proportionality)

2. X_C of 1000 uF at 50 Hz is:
   A. 31.83 ohm
   B. 3.183 ohm
   C. 0.318 ohm
   D. 6 ohm
   Answer: B. X_C = 1/(2 pi x 50 x 1e-3) = 3.183 ohm. (Key point: Capacitive reactance)

3. Adding the capacitor changes the angle between currents from 6.01 degrees to:
   A. 63.43 degrees
   B. 43.20 degrees
   C. 26.25 degrees
   D. 90 degrees
   Answer: C. I_a angle becomes -43.20 degrees, so alpha = 69.44 - 43.20 = 26.25 degrees. (Key point: Effect on angle)

4. The torque increase with the capacitor is about:
   A. 6.9%
   B. 689%
   C. 286%
   D. 589%
   Answer: D. T2/T1 = 6.886, so the increase is (6.886 - 1) x 100 = 588.6%. (Key point: Result)

## Related in Electric Machines II

- [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
- [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
- [Single phase IM principle, torque-slip and starting](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Asingle-phase-principle-torque-slip) — 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
- [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
- [Damper winding to make motor self starting](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Adamper-winding-self-starting) — Electric Machines II, chapter 2

## Related in other subjects

- [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×
- [8-pole motor: speed at max torque (50 N-m, j2)](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aim-8pole-50nm-j2) — Electrical Machines, chapter 5, asked 4×
- [4-pole motor: speed at max torque (50 N-m, j2)](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aim-4pole-50nm-starting-torque-j2) — Electrical Machines, chapter 5, asked 4×
- [Torque of 4-pole induction motor at 1400 rpm](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aim-4pole-2ohm-8ohm-5nm-1400rpm) — Electrical Machines I, chapter 5