--- title: "Induction motor plugging: braking current and torque" subject: "Utilization of Electrical Energy" chapter: 3 key: "semester_7/utilization_of_electrical_energy:induction-motor-plugging-braking-current" source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/induction-motor-plugging-braking-current site: BCT Engineering --- # Induction motor plugging: braking current and torque Utilization of Electrical Energy · Chapter 3: Control of Electric Drive - Asked in: 2081 Baisakh, 4+4 marks - Video: [plugging braking of dc motor | braking in induction motor | in hindi | rheostatic braking | L-13](https://www.youtube.com/watch?v=L5xKZteSjJE) — Mind of Engineer - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/induction-motor-plugging-braking-current - Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-3 ## Question A 400 V, star connected, 3-phase, 6 pole, 50 Hz induction motor with parameters referred to the stator are given below: r1 = r2' = 1 Ω and x1 = x2' = 2 Ω. This motor is to be braked by plugging from initial full load speed of 950 rpm. Stator to rotor turn ratio is 2:3. Find: i) Calculate the initial braking current and torque as the ratio of their full load values. ii) What resistance must be inserted in the rotor circuit to reduce the maximum braking current to 1.5 times full load current? What will be the initial braking torque now? ## Short answer Initial braking current is 5.0 times full-load current (54.0 A) and braking torque is 0.641 T_fl. To limit maximum braking current to 1.5 times full-load current, insert 53.4 Ω per phase in the rotor circuit (23.7 Ω referred to stator); initial braking torque becomes 1.43 T_fl. ## Answer Data: $V_{ph} = 400/\sqrt3 = 230.94$ V, $r_1 = r_2' = 1\ \Omega$, $x_1 = x_2' = 2\ \Omega$, 6 poles, 50 Hz. Magnetising branch neglected (approximate circuit). $$ \begin{aligned} N_s &= \frac{120 f}{P} = \frac{120\times50}{6} = 1000\ \text{rpm} \\ s_{fl} &= \frac{1000-950}{1000} = 0.05 \\ s_b &= \frac{N_s + N}{N_s} = \frac{1000+950}{1000} = 1.95 \ \ (\text{plugging}) \end{aligned} $$ #### i) Initial braking current and torque $$ I = \frac{V_{ph}}{\sqrt{(r_1 + r_2'/s)^2 + (x_1+x_2')^2}} $$ $$ \begin{aligned} I_{fl} &= \frac{230.94}{\sqrt{(1+20)^2 + 4^2}} = \frac{230.94}{21.378} = 10.80\ \text{A} \\ I_b &= \frac{230.94}{\sqrt{(1+0.5128)^2 + 4^2}} = \frac{230.94}{4.276} = 54.00\ \text{A} \\ \frac{I_b}{I_{fl}} &= 5.00 \end{aligned} $$ Torque $T = \dfrac{3 I^2 r_2'}{\omega_s s}$, so $$ \begin{aligned} \frac{T_b}{T_{fl}} &= \left(\frac{I_b}{I_{fl}}\right)^2 \frac{s_{fl}}{s_b} = 5.00^2 \times \frac{0.05}{1.95} = 0.641 \end{aligned} $$ **Answer (i): $I_b = 5.0\, I_{fl}$ (54.0 A), $T_b = 0.641\, T_{fl}$.** #### ii) Rotor resistance for $I_b = 1.5\, I_{fl}$ The braking current is largest at the start ($s = 1.95$). Required impedance: $$ \begin{aligned} Z &= \frac{230.94}{1.5\times10.80} = 14.25\ \Omega \\ 1 + \frac{R_2'}{1.95} &= \sqrt{14.25^2 - 4^2} = 13.68 \\ R_2' &= 24.72\ \Omega \ \ (\text{total, referred to stator}) \end{aligned} $$ External resistance referred to stator $= 24.72 - 1 = 23.72\ \Omega$. Referring to the rotor (stator : rotor turns = 2 : 3, so $R_{rotor} = R' \times (3/2)^2$): $$ R_{ext} = 23.72 \times 2.25 = 53.38\ \Omega \ \text{per phase} $$ New initial braking torque: $$ \begin{aligned} \frac{T_b'}{T_{fl}} &= \frac{(1.5 I_{fl})^2 (24.72/1.95)}{I_{fl}^2 (1/0.05)} = \frac{2.25 \times 12.68}{20} = 1.43 \end{aligned} $$ **Answer (ii): insert 53.4 Ω per phase in the rotor (23.7 Ω referred to stator); initial braking torque = 1.43 × full-load torque.** Adding rotor resistance reduces the current but raises the torque, because the rotor power factor improves at high slip. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The motor runs at 950 rpm (Ns = 1000 rpm) and is plugged. What is the slip at the start of braking? A. 0.05 B. 1.05 C. 0.95 D. 1.95 Answer: D. Plugging reverses the field direction, so s = (Ns + N)/Ns = 1.95. (Key point: Slip during plugging) 2. What is Ib/Ifl for the initial braking current? A. 1.0 B. 5.0 C. 0.64 D. 20 Answer: B. Ifl = 10.80 A at s = 0.05 and Ib = 54.0 A at s = 1.95, a ratio of 5.0. (Key point: Current ratio) 3. How is the torque ratio found from the current ratio? A. Tb/Tfl = (Ib/Ifl) x (sfl/sb) B. Tb/Tfl = (Ib/Ifl)^2 x (sfl/sb) C. Tb/Tfl = Ib/Ifl D. Tb/Tfl = (Ib/Ifl)^2 x (sb/sfl) Answer: B. T = 3I^2 r2'/(ws s), so the ratio is (Ib/Ifl)^2 x sfl/sb = 25 x 0.05/1.95 = 0.641. (Key point: Torque ratio) 4. With stator : rotor turns 2:3, how is the 23.72 Ω (referred to stator) converted to the rotor? A. Multiply by 3/2 B. Multiply by (3/2)^2 = 2.25 C. Multiply by (2/3)^2 D. Leave unchanged Answer: B. Resistance scales with the square of the turns ratio, so R_rotor = R' x 2.25 = 53.4 Ω. 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