neal power output of the 185 W, 110 V, 60 Hz, single-phase induction motor whose constants referred to stator are: Stator resistance 1-86 2 ; Stator reactance 3-56 2 Rotor resistance 3.56 2 ; Standstill rotor reactance 2-56 %3D Magnetising reactance of stator winding 53-52 [202 WI

neal power output of the 185 W, 110 V, 60 Hz, single-phase induction motor whose constants referred to stator are: Stator resistance 1-86 2 ; Stator reactance 3-56 2 Rotor resistance 3.56 2 ; Standstill rotor reactance 2-56 %3D Magnetising reactance of stator winding 53-52 [202 WI

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

See similar textbooks

Similar questions

Equivalent circuit parameters of 3 HP (HP) 220V, 50Hz, 6 poles, split phase wound induction motor are given below. The loss of the motor is = 4W. Since the slip is s=5%; Calculate the stator current, output torque, motor speed and efficiency. R1 = 2 , R2` = 3 , X1=0 , X2` =0 , Xm=∞
A 200V, 3, star connected induction motor takes 30A at a line voltage of 35V with rotor locked. With this line voltage, power input to motor is 450W and core loss is 60W.The dc resistance between a pair of stator terminals is 0.15Ω. If the ratio of ac to dc resistance is 1.6, find the equivalent leakage reactance/phase of the motor and the stator and rotor resistance per phase.
The equivalent circuit parameters per phase of a 3-phase, star-connected, 220V, 7.5 kW, 60Hz, 6-pole induction motor are as follows. Total friction-ventilation losses are 403 W. for 2% slip; R1=0.294 R2=0.144 X1=0.503 X2=0.209 Xm=13.25 Find the impedance, speed, shaft torque, output power, stator current, power factor and efficiency values.
A three-phase Y-connected 230-V (line-to-line) 7.5-kW 60-Hz six-pole induction motor has the following parameter values in Ω/phase referred to the stator: R1 = 0.294, R2 = 0.144 X1 = 0.503, X2 = 0.209, Xm = 13.25 The total friction, windage, and core losses may be assumed to be constant at 403 W, independent of load. For a slip of 4 percent, determine: (a) stator current, power factor, (b) output torque and power,(c) efficiency when the motor is operated at rated voltage and frequency, (d) the load component 12 of the stator current, the induced torque Tind, and the converted power Pconv for a slip s= 0.04,(e) the maximum torque and the corresponding speed; and (f) the starting torque Tst and the corresponding stator load current 12 start
QI: The parameters of a 4 kW, 220V. 2-phase induction motor, in ohms per phase, are: r1=0.53, x1 =2.4, XM= 70, r2 = 0.96, x2 = 3.0. The motor is operated from unbalanced 2- phase source whose voltages are 230V and 210V, the smaller voltage is lagging the larger by 800. For a slip of 5% and rotational losses of 30 W, find the positive- and negative- sequence components of: (a) the applied voltages. (b) Find the values of phase currents (c) the input current. (d) Find the developed mechanical power. (e) Find the efficiency.
The stator resistance of a 1-phase induction motor is 2.5 ohms and its leakage reactance is 2.0 ohms. Onno-load, the motor takes 4 A at 96 V and at 0.25 lagging power factor. The no-load friction andwindage loss is negligible. Under the blocked-rotor condition, the input power is 130 W at 6 A and42 V. Obtain the equivalent circuit parameters.
A 440V, 60Hz, 10 pole delta connected induction motor is rated at 100KW. The equivalent parameter for the motor are: Rs = 0.15 Ω R2` = 0.15 Ω Xs = 0.3 Ω X2` = 0.2 Ω Xm = 6 Ω R c = 120 Ω At full load condition, the friction and windage losses are 600W. The slip of the motor is 0.04. Calculate: The stator line current Is ? The input power ? Stator copper losses ? The air gap power PG ? The converted power Pconv? The torque induced by the motor ? The load torque ? The starting torque? The efficiency of the motor ?
A 220 V, single-phase induction motor gave the following test results:Blocked-rotor test: 120 V, 9.6 A, 460 WNo-load test: 220 V, 4.6 A, 125 WThe stator winding resistance is 1.5 Ω, and during the blocked-rotor test, the starting winding is open. Determine the equivalent circuit parameters. Also, find the core, friction and windage losses.
The stator resistance of a 1-phase induction motor is 2.5 ohm and its leakage reactance is 2.0 ohm. Onno-load, the motor takes 4 A at 96 V and at 0.25 lagging power factor. The no-load friction andwindage loss is negligible. Under the blocked-rotor condition, the input power is 130 W at 6 A and42 V. obtain the equivalent circuit parameters.
Please i need correct answer. Find the main dimensions, number of stator turns per phase & number of stator slots for a 25KW 415 V, 50 Hz, 2880 rpm, 3-phase delta connected induction motor. Assume that: specific magnetic loading 0.48wb/m², specific electric loading = 21000 ac/m, Efficiency = 88%, power factor = 0.87, stator winding Factor = 0.9451 Take ratio (L /τ) = 0.7.
A 480V, 60Hz, six-pole, Y connected cage rotor induction motor has the followingparameters in ohms per phase referred to the stator: R1=0.5ΩX1=1.1Ω R2=0.45ΩX2=0.6Ω Xm=30Ω The total rotational losses (friction, windage, and core losses) are 1100W and are assumedto be constant. For a rotor slip of 2.2% at the rated voltage and rated frequency, calculate:d) The air-gap power (Pag) and the developed mechanical power (Pconv)e) Developed electromechanical torque, and the output mechanical torquef) Efficiency
230 V (phase - phase), The equivalent circuit parameters of a 60 Hz, 6 unipolar star connected 3-phase asynchronous motor reduced to the stator side in ohms R1=12,5 X1=21,3, R2=28,6, X2=13,6, Rc=4200, Xm=1800 . If the friction and wind losses are 3% of the converted power, for 1125 rpm rotor mechanical speed, a) Stator copper loss b) Input power c) Air gap strength d) Rotor copper loss e) Converted power f) shaft moment g) Yield (efficiency) calculate.