--- title: "Input current, pf and torque of single phase IM" subject: "Electric Machines II" chapter: 3 key: "semester_5/electric_machines_ii:single-phase-im-input-torque" source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/single-phase-im-input-torque site: BCT Engineering --- # Input current, pf and torque of single phase IM Electric Machines II · Chapter 3: Fractional Kilowatt Motors - Asked in: 2071 Shrawan, 6 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/single-phase-im-input-torque - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-3 ## Question A single phase induction motor has R1 = 2 Ω, X1 = X2' = 3.1 Ω, R2' = 1.98 Ω and Xmag = 40.17 Ω. If the motor is supplied from 240 V single phase ac supply, determine input current, power factor and torque developed by motor. ## Short answer Assuming 4-pole, 50 Hz, s=0.05: Z_F=9.29+j9.94 ohm, Z_B=0.44+j1.45 ohm, total Z=11.72+j14.49 ohm, I1=240/|Z|=12.88 A, pf=cos(51.02 deg)=0.629 lag. Torque T = (I1^2(R_F-R_B))/omega_s = 1467.6/157.08 = 9.34 N.m. ## Answer The slip, poles and frequency are not given. **Assume** a 4-pole, 50 Hz motor running at slip $s = 0.05$, and neglect core loss. Use the double revolving field equivalent circuit. **Given:** $V = 240$ V, $R_1 = 2\ \Omega$, $X_1 = X_2' = 3.1\ \Omega$, $R_2' = 1.98\ \Omega$, $X_{mag} = 40.17\ \Omega$. #### Half values $$ 0.5X_m = 20.085\ \Omega,\quad 0.5X_2' = 1.55\ \Omega,\quad \frac{0.5R_2'}{s} = \frac{0.99}{0.05} = 19.8\ \Omega,\quad \frac{0.5R_2'}{2-s} = \frac{0.99}{1.95} = 0.5077\ \Omega $$ #### Forward and backward impedances $$ \begin{aligned} Z_F &= \frac{(j20.085)(19.8 + j1.55)}{19.8 + j21.635} = 9.287 + j9.938\ \Omega \\ Z_B &= \frac{(j20.085)(0.5077 + j1.55)}{0.5077 + j21.635} = 0.437 + j1.449\ \Omega \end{aligned} $$ #### Input current and power factor $$ \begin{aligned} Z &= (2 + j3.1) + Z_F + Z_B = 11.724 + j14.487 \\ &= 18.64\angle 51.02^\circ\ \Omega \\ I_1 &= \frac{240\angle 0^\circ}{18.64\angle 51.02^\circ} = 12.88\angle -51.02^\circ\ \text{A} \\ \text{pf} &= \cos 51.02^\circ = 0.629\ \text{lagging} \end{aligned} $$ Input power $= 240 \times 12.88 \times 0.629 = 1944$ W. #### Torque developed $$ \begin{aligned} P_{gF} &= I_1^2 R_F = 12.88^2 \times 9.287 = 1540.1\ \text{W} \\ P_{gB} &= I_1^2 R_B = 12.88^2 \times 0.437 = 72.5\ \text{W} \\ P_g &= P_{gF} - P_{gB} = 1467.6\ \text{W} \\ \omega_s &= \frac{2\pi N_s}{60} = \frac{2\pi \times 1500}{60} = 157.08\ \text{rad/s} \\ T &= \frac{P_g}{\omega_s} = \frac{1467.6}{157.08} = 9.34\ \text{N·m} \end{aligned} $$ (Mechanical power developed $= (1-s)P_g = 1394$ W.) **Answer (for s = 0.05, 4-pole, 50 Hz):** input current $\approx 12.88$ A, power factor $\approx 0.629$ lagging, torque developed $\approx 9.34$ N·m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What had to be assumed since slip, poles and frequency were not given? A. 4-pole, 50 Hz, s = 0.05 B. 2-pole, 60 Hz, s = 1 C. 6-pole, 50 Hz, s = 0.2 D. No assumption needed Answer: A. The solution assumes a 4-pole, 50 Hz motor at s = 0.05 and neglects core loss. (Key point: Assumptions) 2. The backward rotor resistance term 0.5R2'/(2-s) = 0.99/1.95 is about: A. 19.8 ohm B. 0.5077 ohm C. 1.98 ohm D. 0.05 ohm Answer: B. 0.99/1.95 = 0.5077 ohm. (Key point: Backward branch) 3. With Z = 18.64 at 51.02 degrees ohm and 240 V, the current and pf are: A. 12.88 A, 0.629 lagging B. 12.88 A, 0.777 lagging C. 6.4 A, 0.629 lagging D. 12.88 A, 0.629 leading Answer: A. I = 240/18.64 = 12.88 A, pf = cos 51.02 = 0.629 lagging. (Key point: Current and pf) 4. Torque developed is computed as: A. (P_gF + P_gB)/omega_s B. (P_gF - P_gB)/omega_s with omega_s = 157.08 rad/s C. P_gF x omega_s D. (1-s)P_g/omega_s only Answer: B. T = (1540.1 - 72.5)/157.08 = 9.34 N·m. 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