--- title: "Capacitor for quadrature currents (4.5+j3.5 Ω)" subject: "Electric Machines II" chapter: 3 key: "semester_5/electric_machines_ii:quadrature-capacitor-250w-j3-5" source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/quadrature-capacitor-250w-j3-5 site: BCT Engineering --- # Capacitor for quadrature currents (4.5+j3.5 Ω) Electric Machines II · Chapter 3: Fractional Kilowatt Motors - Asked in: 2075 Chaitra, 6 marks - 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-j3-5 - 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 its main and starting windings: Zm = (4.5+j3.5) Ω and Zs = (9.5+j3.5) Ω. Determine the value of the starting capacitor that will place the main and starting winding currents in quadrature at starting. ## Short answer Main winding angle = tan⁻¹(3.5/4.5) = 37.87°. For quadrature, starting-branch angle must be −52.13°, yielding Xc = 15.71 Ω. Capacitance C = 1/(2π×50×15.71) ≈ 202.6 μF. ## Answer For the currents to be in quadrature, the auxiliary (starting) winding current must lead the main winding current by $90^\circ$. So the angle of the starting branch impedance must be $90^\circ$ less than that of the main winding. **Given:** $Z_m = 4.5 + j3.5\ \Omega$, $Z_s = 9.5 + j3.5\ \Omega$, $V = 230$ V, $f = 50$ Hz. #### Main winding angle $$ \theta_m = \tan^{-1}\frac{3.5}{4.5} = 37.87^\circ \quad (I_m \text{ lags } V \text{ by } 37.87^\circ) $$ #### Required angle of starting branch $$ \theta_s = 37.87^\circ - 90^\circ = -52.13^\circ $$ With capacitor: $Z_s' = 9.5 + j(3.5 - X_C)$, so $$ \begin{aligned} \frac{3.5 - X_C}{9.5} &= \tan(-52.13^\circ) = -\frac{4.5}{3.5} = -1.2857 \\ 3.5 - X_C &= -12.214 \\ X_C &= 15.714\ \Omega \end{aligned} $$ #### Capacitance $$ \begin{aligned} C &= \frac{1}{2\pi f X_C} = \frac{1}{2\pi \times 50 \times 15.714} \\ &= 202.6 \times 10^{-6}\ \text{F} \end{aligned} $$ ```text o----+-------------+ | | Z_m=4.5+j3.5 Z_s=9.5+j3.5 | | 230 V | [C] ~203 uF | | | CS (centrifugal switch) o----+-------------+ ``` **Check:** $Z_s' = 9.5 - j12.214 = 15.47\angle -52.13^\circ$; angle between currents $= 37.87^\circ + 52.13^\circ = 90^\circ$. **Answer:** starting capacitor $C \approx 202.6\ \mu\text{F}$ ($X_C = 15.71\ \Omega$). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For main and starting currents to be in quadrature, the starting branch current must: A. Be in phase with I_m B. Lead I_m by 90 degrees C. Lag I_m by 90 degrees D. Equal I_m in magnitude Answer: B. The capacitor makes the auxiliary current lead the main current by 90 degrees. (Key point: Quadrature condition) 2. The main winding angle is 37.87 degrees lagging. What impedance angle must the starting branch have? A. +127.87 degrees B. +52.13 degrees C. -52.13 degrees D. -90 degrees Answer: C. theta_s = 37.87 - 90 = -52.13 degrees. (Key point: Required angle) 3. What equation gives X_C? A. (3.5 - X_C)/9.5 = tan(-52.13 deg) B. (3.5 + X_C)/9.5 = tan(52.13 deg) C. X_C = 3.5 ohm D. (X_C - 3.5)/4.5 = 1 Answer: A. Net reactance (3.5 - X_C) over resistance 9.5 equals tan(-52.13 deg) = -1.2857, so X_C = 15.71 ohm. (Key point: Required X_C) 4. The capacitance for X_C = 15.714 ohm at 50 Hz is about: A. 20.3 uF B. 202.6 uF C. 2026 uF D. 15.7 uF Answer: B. C = 1/(2 pi x 50 x 15.714) = 202.6 uF. 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