---
title: "Starting capacitor and change in starting torque"
subject: "Electric Machines II"
chapter: 3
key: "semester_5/electric_machines_ii:capacitor-quadrature-torque-change"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/capacitor-quadrature-torque-change
site: BCT Engineering
---

# Starting capacitor and change in starting torque

Electric Machines II · Chapter 3: Fractional Kilowatt Motors

- Asked in: 2081 Bhadra, 8 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/capacitor-quadrature-torque-change
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-3

## Question

The main winding and starting winding of a 50 Hz capacitor start single phase induction motor have impedances as follow:
Main winding: (3+j3) ohm
Starting winding: (7.5+j3) ohm
Calculate the value of capacitor to be connected in series with the starting winding to produce a phase difference of 90 degree between main winding current and starting winding current at starting. Also calculate the percentage change in starting torque.

## Short answer

To achieve 90° phase difference, capacitive reactance Xc = 10.5 Ω gives C = 303.2 µF. Starting torque increases by 93.3% compared to operation without the capacitor because the winding currents become orthogonal.

## Answer

The capacitor must make the starting-winding current **lead** the main-winding current by $90^\circ$. Starting torque is $T_{st} = K I_m I_s \sin\alpha$, where $\alpha$ is the angle between the two currents. The change in torque is found by comparing with the motor **without the capacitor** (plain resistance split-phase start), with the same supply voltage $V$.

#### Capacitor value

$$
\theta_m = \tan^{-1}\frac{3}{3} = 45^\circ \ (I_m \text{ lags } V)
$$

For $\alpha = 90^\circ$, $I_s$ must lead $V$ by $90^\circ - 45^\circ = 45^\circ$, so the starting branch angle is $-45^\circ$:

$$
\begin{aligned}
\frac{3 - X_C}{7.5} &= \tan(-45^\circ) = -1 \\
X_C &= 3 + 7.5 = 10.5\ \Omega \\
C &= \frac{1}{2\pi \times 50 \times 10.5} = 303.2\ \mu\text{F}
\end{aligned}
$$

#### Starting torque without capacitor

$$
\begin{aligned}
\theta_s &= \tan^{-1}\frac{3}{7.5} = 21.80^\circ,\quad |Z_s| = 8.0777\ \Omega \\
\alpha_1 &= 45^\circ - 21.80^\circ = 23.20^\circ \\
I_{s1} &= \frac{V}{8.0777} = 0.1238V \\
T_1 &\propto I_m \times 0.1238V \times \sin23.20^\circ = 0.04877\,V I_m
\end{aligned}
$$

#### Starting torque with capacitor

$$
\begin{aligned}
|Z_s'| &= |7.5 - j7.5| = 10.6066\ \Omega,\quad I_{s2} = 0.0943V \\
T_2 &\propto I_m \times 0.0943V \times \sin90^\circ = 0.09428\,V I_m
\end{aligned}
$$

$I_m$ is the same in both cases, so

$$
\frac{T_2}{T_1} = \frac{0.09428}{0.04877} = 1.9333
$$

Percentage change $= (1.9333 - 1) \times 100 = 93.33\ \%$ (increase).

```text
        Is (with C, leads V by 45 deg)
       /
      / 90 deg
 ----+-----------------> V
     |\  Is (no C, lags 21.8 deg)
     | \
     |  Im (lags 45 deg)
```

**Answer:** $C = 303.2$ µF (series with starting winding); starting torque increases by about **93.3 %** compared with the same motor started without the capacitor.

## Check yourself

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

1. Main winding 3 + j3 lags V by 45 degrees. For 90 degrees between currents, the starting branch angle must be:
   A. -45 degrees
   B. +45 degrees
   C. -90 degrees
   D. -21.8 degrees
   Answer: A. Is must lead V by 45 degrees, so the impedance angle is -45 degrees. (Key point: Required angle)

2. (3 - Xc)/7.5 = -1 gives:
   A. Xc = 10.5 ohm, C about 303 microfarad
   B. Xc = 4.5 ohm, C about 707 microfarad
   C. Xc = 3 ohm, C about 1061 microfarad
   D. Xc = 10.5 ohm, C about 30 microfarad
   Answer: A. Xc = 3 + 7.5 = 10.5 ohm; C = 1/(2 pi 50 x 10.5) = 303.2 microfarad. (Key point: Capacitor value)

3. The change in starting torque is found by comparing T = K Im Is sin(alpha):
   A. With and without the capacitor, same V and the same Im
   B. With different supply voltages
   C. Ignoring the angle alpha
   D. Only using Is
   Answer: A. Im is unchanged, so only Is sin(alpha) changes. (Key point: Torque comparison basis)

4. The starting torque with the capacitor changes by about:
   A. 93.3 % increase
   B. 19.3 % increase
   C. 50 % decrease
   D. 193 % decrease
   Answer: A. T2/T1 = 0.09428/0.04877 = 1.9333, an increase of 93.3 %. (Key point: Result)

## Related in Electric Machines II

- [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
- [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×
- [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
- [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
- [Capacitor for quadrature currents (4.5+j3.5 Ω)](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Aquadrature-capacitor-250w-j3-5) — Electric Machines II, chapter 3

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