
Electric Motor Control
10th Edition
ISBN: 9781305177611
Author: Herman
Publisher: Cengage
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Chapter 48, Problem 11SQ
To determine
Explain when the power supply is lost for 10 minutes, what happens to the motor while running in the forward direction in Figure 48-8.
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Solve the following nonlinear system using Newton's method
1
f1(x1, x2, x3)=3x₁ = cos(x2x3)
-
-
2
f2(x1, x2, x3) = x² - 81(x2 +0.1)² + sin x3 + 1.06
f3(x1, x2, x3) = ex1x2 +20x3 +
Using x (0)
X1 X2 X3
10π-3
3
= 0.1, 0.1, 0.1 as initial conditio
Use Newton-Raphson method to solve the system
x²
-
2x-y+0.5= 0
x² + 4y² 4 = 0
-
with the starting value (xo,yo) = (2,0.25) and two iteration number.
Reversing 3⍉ Motors using manual starters with wiring diagram of forward contacts and reverse contacts.
Chapter 48 Solutions
Electric Motor Control
Ch. 48 - In Figure 482, what is the purpose of normally...Ch. 48 - Can a time-delay relay be used satisfactorily in a...Ch. 48 - Prob. 3SQCh. 48 - Prob. 4SQCh. 48 - Prob. 5SQCh. 48 - What is the purpose of the lockout relay or safety...Ch. 48 - What happens if the reverse push button is closed...Ch. 48 - Prob. 8SQCh. 48 - Prob. 9SQCh. 48 - In Figure 487, is it necessary to push the stop...
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- Determine (a) the input impedance and (b) the reflectedimpedance, both at terminals (a,b) in the circuit of Fig. P11.14.arrow_forward11.4 Determine Vout in the circuit shown in Fig. P11.4.arrow_forwardFor the circuit in Fig. P11.1, determine (a) iL(t) and (b) theaverage power dissipated in RL.arrow_forward
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- Use Newton-Raphson method to solve the system x² - 2x-y+0.5= 0 x² + 4y² 4 = 0 - with the starting value (xo,yo) = (2,0.25) and two iteration number.arrow_forwardSolve the following systems using Gauss Seidal and Jacobi iteration methods for n=8 and initial values X0=(000). - 2x16x2 x3 = -38 - -3x1 x2+7x3 = −34 -8x1 + x2 - 2x3 = -20arrow_forwardSolve the following nonlinear system using Newton's method 1 f1(x1, x2, x3)=3x₁ = cos(x2x3) - - 2 f2(x1, x2, x3) = x² - 81(x2 +0.1)² + sin x3 + 1.06 f3(x1, x2, x3) = ex1x2 +20x3 + Using x (0) X1 X2 X3 10π-3 3 = 0.1, 0.1, 0.1 as initial conditioarrow_forward
- A single phase a.c. distributor AB has: The distance from A to B is 500 m. The distance from A to C is 800 m. The impedance of each section is (6+j 8) /km. A B C The voltage at the far end is maintained at 250 volt. Find: sending voltage, sending current, supply power factor and 80 A 60 A total voltage drop. 0.8 lag. P.f 0.6 lead. p.farrow_forwardA 3-phase, 4-wire distributor supplies a balanced voltage of 400/230 V to a load consisting of 8 A at p.f. 0-7 lagging for R-phase, 10 A at p.f. 0-8 leading for Y phase and 12 A at unity p.f. for B phase. The resistance of each line conductor is 0.4 2. The reactance of neutral is 0.2 2. Calculate the neutral current, the supply voltage for R phase and draw the phasor diagram. The phase sequence is RYB.arrow_forwardThe three line leads of a 400/230 V, 3-phase, 4-wire supply are designated as R, Y and B respectively. The fourth wire or neutral wire is designated as N. The phase sequence is RYB. Compute the currents in the four wire when the following loads are connected to this supply: From R to N: 25 kW, unity power factor. From Y to N: 20 kVA, 0-7 lag. From B to N: 30 kVA, 0-6 lead.arrow_forward
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