--- title: "Main winding current and pf at 5% slip" subject: "Electric Machines II" chapter: 3 key: "semester_5/electric_machines_ii:single-phase-im-main-current" source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/single-phase-im-main-current site: BCT Engineering --- # Main winding current and pf at 5% slip Electric Machines II · Chapter 3: Fractional Kilowatt Motors - Asked in: 2074 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/single-phase-im-main-current - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-3 ## Question A 230 V, 50 Hz, 4 pole, class A, single phase induction motor has the following parameters: r1m = 2.51 Ω, r2' = 7.81 Ω, Xm = 150.88 Ω, x1m = 4.62 Ω, x2' = 4.62 Ω. Determine the main winding current and power factor when the motor is running at a slip of 0.05. ## Short answer Using double-field theory, forward and backward impedances are computed from given parameters. Total input impedance Z = 61.55∠48.24° Ω. Main winding current I = 230/61.55 = 3.74 A, power factor = cos 48.24° ≈ 0.666 lagging. ## Answer Using double field revolving theory, the rotor and magnetising branches are split into **forward** and **backward** halves. Core loss is neglected (no core-loss resistance is given). **Given:** $V = 230$ V, $r_1 = 2.51\ \Omega$, $x_1 = 4.62\ \Omega$, $r_2' = 7.81\ \Omega$, $x_2' = 4.62\ \Omega$, $X_m = 150.88\ \Omega$, $s = 0.05$. ```text I -> r1 jx1 o--/\/\/--mmm--+--------+---------+ | | | V 0.5jXm 0.5r2'/s | Forward | +j0.5x2' | Z_F +--------+---------+ | | | 0.5jXm 0.5r2'/(2-s) | Backward | +j0.5x2' | Z_B o--------------+--------+---------+ ``` #### Forward impedance $$ \begin{aligned} 0.5 X_m &= 75.44\ \Omega,\quad \frac{0.5r_2'}{s} = \frac{3.905}{0.05} = 78.1\ \Omega,\quad 0.5x_2' = 2.31\ \Omega \\ Z_F &= \frac{(j75.44)(78.1 + j2.31)}{78.1 + j77.75} \\ &= 36.60 + j39.01\ \Omega \end{aligned} $$ #### Backward impedance $$ \begin{aligned} \frac{0.5 r_2'}{2-s} &= \frac{3.905}{1.95} = 2.003\ \Omega \\ Z_B &= \frac{(j75.44)(2.003 + j2.31)}{2.003 + j77.75} \\ &= 1.884 + j2.290\ \Omega \end{aligned} $$ #### Total input impedance $$ \begin{aligned} Z &= (r_1 + jx_1) + Z_F + Z_B \\ &= (2.51 + 36.60 + 1.884) + j(4.62 + 39.01 + 2.290) \\ &= 40.99 + j45.92 \\ &= 61.55\angle 48.24^\circ\ \Omega \end{aligned} $$ #### Main winding current and power factor $$ \begin{aligned} I_m &= \frac{V}{Z} = \frac{230\angle 0^\circ}{61.55\angle 48.24^\circ} = 3.737\angle -48.24^\circ\ \text{A} \\ \text{pf} &= \cos 48.24^\circ = 0.666\ \text{lagging} \end{aligned} $$ (For reference: air-gap powers are $P_{gF} = I^2 R_F = 511.0$ W and $P_{gB} = I^2 R_B = 26.3$ W.) **Answer:** main winding current $I_m \approx 3.74$ A, power factor $\approx 0.666$ lagging. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the DFRT equivalent circuit, how are the magnetising and rotor branches treated? A. Split into forward and backward halves B. Doubled C. Ignored D. Placed in parallel with the supply Answer: A. Each is split into half-value forward and backward branches. (Key point: Equivalent circuit method) 2. With r2' = 7.81 ohm and s = 0.05, the forward rotor resistance term 0.5r2'/s is: A. 3.905 ohm B. 7.81 ohm C. 78.1 ohm D. 156.2 ohm Answer: C. 3.905/0.05 = 78.1 ohm. (Key point: Forward branch resistance) 3. The backward rotor resistance term 0.5r2'/(2-s) is about: A. 78.1 ohm B. 3.905 ohm C. 2.003 ohm D. 0.05 ohm Answer: C. 3.905/1.95 = 2.003 ohm. (Key point: Backward branch resistance) 4. The input impedance is 61.55 at 48.24 degrees ohm with V = 230 V. The main current and pf are: A. 3.74 A, 0.666 lagging B. 3.74 A, 0.666 leading C. 6.1 A, 0.75 lagging D. 3.74 A, 0.9 lagging Answer: A. I = 230/61.55 = 3.74 A, pf = cos 48.24 = 0.666 lagging. (Key point: Total impedance and result) ## Related in Electric Machines II - [Split-phase motor winding currents at starting](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Asplit-phase-starting-currents) — Electric Machines II, chapter 3 - [Input current, pf and torque of single phase IM](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Asingle-phase-im-input-torque) — Electric Machines II, chapter 3 - [Speed, input current and output of 2/3 HP motor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Atwo-thirds-hp-motor-performance) — Electric Machines II, chapter 3 - [Single phase IM principle, torque-slip and starting](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Asingle-phase-principle-torque-slip) — Electric Machines II, chapter 3 - [Current, pf, power and efficiency of 3.3 kV motor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Amotor-3-3kv-efficiency) — Electric Machines II, chapter 2 - [Split phase motor added resistance and capacitance](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Asplit-phase-resistance-capacitance) — Electric Machines II, chapter 3 ## Related in other subjects - [Starting current and rotor resistance for 150 kW slip ring motor](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aim-150kw-3000v-starting-current-30a) — Electrical Machines I, chapter 5 - [6-pole induction motor: slip, losses, efficiency](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aim-4pole-975rpm-slip-losses) — Electrical Machines, chapter 5 - [250 HP 4-pole slip ring motor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Aslip-ring-250hp-4pole) — Electric Machine Design, chapter 4, asked 5× - [200 kW 4-pole slip ring motor (delta)](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Aslip-ring-200kw-delta) — Electric Machine Design, chapter 4, asked 4×