---
title: "Capacitor for quadrature currents (4.5+j3.7 Ω)"
subject: "Electric Machines II"
chapter: 3
key: "semester_5/electric_machines_ii:quadrature-capacitor-250w-j3-7"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/quadrature-capacitor-250w-j3-7
site: BCT Engineering
---

# Capacitor for quadrature currents (4.5+j3.7 Ω)

Electric Machines II · Chapter 3: Fractional Kilowatt Motors

- Asked in: 2079 Bhadra, 7 marks; 2076 Chaitra, 6 marks (asked 2 times)
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/quadrature-capacitor-250w-j3-7
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-3

## Question

A 250 W, 230 V, 50 Hz single phase capacitor start induction motor has the following constants for the main and starting windings.
Main Winding: Zm = (4.5 + j3.7) Ω
Starting Winding: Zs = (9.5 + j3.5) Ω
Determine the value of capacitor to be connected in series with starting winding that will make the main and starting windings current in quadrature at starting. [2076 Chaitra paper prints the rating as 250 KW.]

## Short answer

To make currents quadrature, starting branch impedance angle must be -50.6°, requiring capacitive reactance 15.05 Ω, giving capacitor value 211 µF.

## Answer

For the two winding currents to be in quadrature, the starting-winding current must **lead** the main-winding current by $90^\circ$. The capacitor makes the starting-winding branch capacitive.

**Given:** $Z_m = 4.5 + j3.7\ \Omega$, $Z_s = 9.5 + j3.5\ \Omega$, $f = 50$ Hz. (The rating, 250 W or "250 kW", does not affect the result.)

#### Angle of main winding current

$$
\theta_m = \tan^{-1}\frac{3.7}{4.5} = 39.43^\circ \quad (I_m \text{ lags } V \text{ by } 39.43^\circ)
$$

#### Required angle of starting branch

$I_s$ must lead $I_m$ by $90^\circ$, so it must **lead** $V$ by $90^\circ - 39.43^\circ = 50.57^\circ$. The impedance angle of the starting branch must be $-50.57^\circ$:

$$
\begin{aligned}
\tan(-50.57^\circ) &= \frac{X_s - X_C}{R_s} = \frac{3.5 - X_C}{9.5} \\
3.5 - X_C &= 9.5 \times (-1.2162) = -11.554 \\
X_C &= 3.5 + 11.554 = 15.054\ \Omega
\end{aligned}
$$

#### Capacitance

$$
C = \frac{1}{2\pi fX_C} = \frac{1}{2\pi \times 50 \times 15.054} = 211.4\ \mu\text{F}
$$

```text
           Is (leads V by 50.6 deg)
          /
         /  90 deg between Is and Im
  ------+--------------------> V
         \
          \ Im (lags V by 39.4 deg)
```

Check: angle between currents $= 50.57^\circ + 39.43^\circ = 90^\circ$.

**Answer:** $X_C = 15.05\ \Omega$, so a capacitor of about **211 µF** must be connected in series with the starting winding.

## Check yourself

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

1. For the main and starting winding currents to be in quadrature, the starting-winding current must:
   A. Lag the main current by 90 degrees
   B. Lead the main-winding current by 90 degrees
   C. Be in phase with the main current
   D. Have the same magnitude as the main current
   Answer: B. The capacitor makes the starting branch capacitive, so Is leads Im by 90 degrees. (Key point: Condition for quadrature)

2. For Zm = 4.5 + j3.7 ohm, Im lags V by:
   A. 50.57 degrees
   B. 39.43 degrees
   C. 45 degrees
   D. 82.2 degrees
   Answer: B. tan^-1(3.7/4.5) = 39.43 degrees. (Key point: Angle of main winding)

3. Is must lead V by 50.57 degrees. Which gives Xc?
   A. tan(-50.57) = (3.5 - Xc)/9.5, giving Xc = 15.05 ohm
   B. tan(50.57) = (3.5 - Xc)/9.5, giving Xc = -8 ohm
   C. tan(-39.43) = (3.5 - Xc)/9.5, giving Xc = 9.2 ohm
   D. tan(-50.57) = (3.5 + Xc)/9.5, giving Xc = -15.05 ohm
   Answer: A. 3.5 - Xc = 9.5 x (-1.2162) = -11.55, so Xc = 15.05 ohm. (Key point: Required starting branch angle and Xc)

4. With Xc = 15.054 ohm at 50 Hz, the capacitor is about:
   A. 21.1 microfarad
   B. 211 microfarad
   C. 2.11 millifarad
   D. 67 microfarad
   Answer: B. C = 1/(2 pi x 50 x 15.054) = 211.4 microfarad. (Key point: Capacitance value)

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