--- title: "Capacitor for quadrature currents (7+j5 Ω)" subject: "Electric Machines II" chapter: 3 key: "semester_5/electric_machines_ii:capacitor-quadrature-250w-7-j5" source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/capacitor-quadrature-250w-7-j5 site: BCT Engineering --- # Capacitor for quadrature currents (7+j5 Ω) Electric Machines II · Chapter 3: Fractional Kilowatt Motors - Asked in: 2083 Baisakh (new course), 3 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-250w-7-j5 - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-3 ## Question A 250 W, 230 V, 50 Hz capacitor start motor has following impedances at standstill. Zm = 7+j5 Ω and Za = 11.5+j5 Ω. Find the value of the capacitor to be connected in series with the auxiliary winding to give a quadrant phase displacement between the currents in two windings. Draw the circuit and phasor diagram for motor. ## Short answer Set Xc so that auxiliary current leads main current by 90°. Compute main winding angle θm=35.54°, required auxiliary branch angle -54.46°, Xc=21.1 Ω, giving C=150.9 µF. ## Answer For quadrature, $I_a$ must lead $I_m$ by $90^\circ$, so the auxiliary branch angle must be $90^\circ$ less than that of the main winding. **Given:** $Z_m = 7 + j5\ \Omega$, $Z_a = 11.5 + j5\ \Omega$, $V = 230$ V, $f = 50$ Hz. $$ \begin{aligned} \theta_m &= \tan^{-1}\frac{5}{7} = 35.54^\circ \\ \theta_a' &= 35.54^\circ - 90^\circ = -54.46^\circ \\ 5 - X_C &= 11.5\tan(-54.46^\circ) = -11.5 \times \frac{7}{5} = -16.1 \\ X_C &= 21.1\ \Omega \\ C &= \frac{1}{2\pi \times 50 \times 21.1} = 150.9\ \mu\text{F} \end{aligned} $$ Currents: $I_m = \dfrac{230}{8.602\angle 35.54^\circ} = 26.74\angle -35.54^\circ$ A, $I_a = \dfrac{230}{19.78\angle -54.46^\circ} = 11.62\angle 54.46^\circ$ A. ```text Circuit: Phasors: o---+--------+ I_a (54.5 deg) | | ^ Z_m Z_a | 90 deg | | -------+-------> V 230V| [C] \ | | v I_m (-35.5 deg) | CS o---+--------+ ``` **Answer:** $C \approx 150.9\ \mu\text{F}$ ($X_C = 21.1\ \Omega$) in series with the auxiliary winding. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For quadrature phase displacement, what must be true of the auxiliary-branch angle compared with the main winding angle? A. It must equal it B. It must be 90° less (Ia leads Im by 90°) C. It must be 90° more D. It must be 180° less Answer: B. Ia must lead Im by 90°, so the auxiliary branch angle is the main angle minus 90°, i.e. 35.54° - 90° = -54.46°. (Key point: Quadrature condition) 2. Which relation gives the series capacitive reactance? A. 5 - XC = 11.5 tan(-54.46°) B. 5 + XC = 11.5 tan(35.54°) C. 11.5 - XC = 5 tan(-54.46°) D. XC = 7 + 11.5 Answer: A. The auxiliary reactance is 5 - XC and its resistance is 11.5, so tan of the branch angle (-54.46°) = (5 - XC)/11.5. (Key point: Equation for XC) 3. What are the capacitive reactance and the capacitor value? A. XC = 16.1 ohm, C = 197.7 µF B. XC = 21.1 ohm, C = 150.9 µF C. XC = 21.1 ohm, C = 15.09 µF D. XC = 5 ohm, C = 636.6 µF Answer: B. 5 - XC = -16.1 gives XC = 21.1 ohm, and C = 1/(2π x 50 x 21.1) = 150.9 µF. 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