--- title: "Split phase motor added resistance and capacitance" subject: "Electric Machines II" chapter: 3 key: "semester_5/electric_machines_ii:split-phase-resistance-capacitance" source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/split-phase-resistance-capacitance site: BCT Engineering --- # Split phase motor added resistance and capacitance Electric Machines II · Chapter 3: Fractional Kilowatt Motors - Asked in: 2078 Bhadra, 6 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/split-phase-resistance-capacitance - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-3 ## Question A 50 Hz split phase induction motor has a resistance 5 Ω and an inductive reactance of 20 Ω in both main and auxiliary windings. Determine a) the value of resistance b) capacitance to be added in series with auxiliary windings to send the same current in each winding with a phase difference of 90°. ## Short answer The auxiliary impedance must have magnitude 20.616 Ω and angle –14.04° for equal current and 90° shift. Added series resistance R = 15 Ω and capacitive reactance Xc = 25 Ω, corresponding to C = 127.3 µF. ## Answer To send the **same current** in both windings with a **90° phase shift**, the auxiliary branch impedance must have the **same magnitude** as the main winding impedance but an angle that is 90° less. A resistance alone cannot shift the current by 90°, so both a resistance and a capacitor are added in series with the auxiliary winding. **Given:** $f = 50$ Hz, $Z_m = Z_a = 5 + j20\ \Omega$ (before adding anything). #### Main winding $$ \begin{aligned} Z_m &= 5 + j20 = 20.616\angle 75.96^\circ\ \Omega \end{aligned} $$ So $I_m$ lags $V$ by $75.96^\circ$. #### Required auxiliary impedance For $|I_a| = |I_m|$ and $I_a$ leading $I_m$ by $90^\circ$: $$ \begin{aligned} \theta_a &= 75.96^\circ - 90^\circ = -14.04^\circ \\ Z_a' &= 20.616\angle -14.04^\circ \\ &= 20.616(\cos 14.04^\circ - j\sin 14.04^\circ) \\ &= 20 - j5\ \Omega \end{aligned} $$ Let $R$ and $X_C$ be added in series: $Z_a' = (5 + R) + j(20 - X_C)$. #### a) Resistance to be added $$ \begin{aligned} 5 + R &= 20 \\ R &= 15\ \Omega \end{aligned} $$ #### b) Capacitance to be added $$ \begin{aligned} 20 - X_C &= -5 \;\Rightarrow\; X_C = 25\ \Omega \\ C &= \frac{1}{2\pi f X_C} = \frac{1}{2\pi \times 50 \times 25} \\ &= 127.3 \times 10^{-6}\ \text{F} \end{aligned} $$ #### Check $|Z_a'| = \sqrt{20^2 + 5^2} = 20.616\ \Omega = |Z_m|$, so the currents are equal; angle difference $= 75.96^\circ - (-14.04^\circ) = 90^\circ$. ```text I_a (leads V by 14.04 deg) ^ | / | / | / -------+-----------------> V \ \ 75.96 deg \ v I_m (I_a and I_m 90 deg apart) ``` **Answer:** add $R = 15\ \Omega$ and $C \approx 127.3\ \mu\text{F}$ ($X_C = 25\ \Omega$) in series with the auxiliary winding. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. To get equal currents with 90 degrees phase difference, the auxiliary impedance must: A. Have the same magnitude as Zm and an angle 90 degrees less B. Have half the magnitude of Zm C. Be purely resistive D. Have the same angle as Zm Answer: A. Equal magnitude gives equal currents, and the angle lowered by 90 degrees gives the shift. (Key point: Condition for equal currents at 90 degrees) 2. Zm = 5 + j20 = 20.616 at 75.96 degrees. The required Za' is: A. 20 - j5 ohm B. 5 - j20 ohm C. 15 - j5 ohm D. 20 + j5 ohm Answer: A. 20.616 at -14.04 degrees = 20 - j5. (Key point: Required auxiliary impedance) 3. The resistance to be added is: A. 15 ohm B. 20 ohm C. 5 ohm D. 25 ohm Answer: A. 5 + R = 20, so R = 15 ohm. (Key point: Resistance) 4. 20 - Xc = -5 gives Xc = 25 ohm, so C is about: A. 127.3 microfarad B. 12.7 microfarad C. 637 microfarad D. 1273 microfarad Answer: A. C = 1/(2 pi 50 x 25) = 127.3 microfarad. 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