Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN: 9780133923605
Author: Robert L. Boylestad
Publisher: PEARSON
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Transcribed Image Text:An induction motor has a fullI load slip of 0.04. It's starting current 5 times its full load
current. What is the ratio of starting torque to the full load torque?
Select one:
а. 1
b. 2
c. 0.2
d. 0.5
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- 1. A fundamental DC motor is shown below. VTerminal = 90V. The running current, I = 5A and the rotor resistance, R = 4 ohms. Answer the following: CEMF or E at the running condition = , Power delivered by the source at the running condition = Power converted to mechanical power at the running condition Power lost in heat by the rotor resistance = Show your work for each calculation. Suggest that you clearly mark the schematic. frutarrow_forwardProblem 7 A three-phase, 60 Hz, six-pole, Y-connected, 480-V induction motor has the following parameters: R₁ = 0.202, R2 = 0.102, Xeq = 50 The load of the motor is a drilling machine. At 1150 rpm, the load torque is 150Nm. The motor is driven by a constant v/f technique. When the frequency of the supply voltage is reduced to 50 Hz, calculate the following: a. Motor speed b. Maximum torque at 60 Hz and 50 Hz c. Motor current at 50 Hz Hint: For a drilling-machine load (an inverse-speed-characteristics load) T₁/T₂ = n₂/n₁ = (1-S₂)/(1-S₁)arrow_forwardA three-phase induction motor used for a commercial application in Oman runs at 580 rpm at no load and at 552 rpm at full load. a) How many poles do this motor have? b) What is the percentage slip at full load? c) The frequency of the rotor currents when slip is 5 % d) The frequency of the rotor currents at standstill. a) b) c) d)arrow_forward
- Please question 4arrow_forwardA large 3-phase, 4160 V, 60 Hz, squirrel-cage induction motor draws a current of 400 A and the total active power of 2500 kW when operating at full-load. The corresponding speed is accurately measured and is found to be 1178 r/min. The stator is connected in wye and the resistance between two stator terminals is 0.15 0. The total iron losses are 26.5 kW and the windage and friction losses are 15 kW. Calculate the following: The power factor at full-load The active power supplied to the rotor The total I²R losses in the rotor The load mechanical power [kW], torque [kN.m], and efficiency [%] a. b. C. d.arrow_forwardA 240 V series motor develops a torque of 120 Nm at a speed of 900 rpm. The armature winding resistance is 0.20 and the series field winding resistance is 0.1 Q. For a constant load torque find the new speed of the motor when a diverter resistance of 0.1 Q is placed across the series field winding.arrow_forward
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