Electric Motor Control
Electric Motor Control
10th Edition
ISBN: 9781305177611
Author: Herman
Publisher: Cengage
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Chapter 21, Problem 12SQ

Convert the control circuit only, Figure 21–11, from the wiring diagram to an elementary diagram. Include the limit switches (RLS, FLS) as operating in the control circuit.

Chapter 21, Problem 12SQ, Convert the control circuit only, Figure 2111, from the wiring diagram to an elementary diagram.

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I need help checking if its correct -E1 + VR1 + VR4 – E2 + VR3 = 0 -------> Loop 1 (a) R1(I1) + R4(I1 – I2) + R3(I1) = E1 + E2 ------> Loop 1 (b) R1(I1) + R4(I1) - R4(I2) + R3(I1) = E1 + E2 ------> Loop 1 (c) (R1 + R3 + R4) (I1)  - R4(I2)    = E1 + E2 ------> Loop 1 (d) Now that we have loop 1 equation will procced on finding the equation of I2 current loop. However, a reminder that because we are going in a clockwise direction, it goes against the direction of the current. As such we will get an equation for the matrix that will be:   E2 – VR4 – VR2 + E3 = 0 ------> Loop 2 (a) -R4(I2 – I1) -R2(I2) = -E2 – E3  ------> Loop 2 (b) -R4(I2) + R4(I1) - R2(I2) = -E2 – E3  -----> Loop 2 (c)                                     R4(I1) – (R4 + R2)(I2) = -E2 – E3  -----> Loop 2 (d) These two equations will be implemented to the matrix formula I = inv(A) * b       R11                        R12   (R1 + R3 + R4)     -R4     -R4     R4 + R2
10.2 For each of the following groups of sources, determineif the three sources constitute a balanced source, and if it is,determine if it has a positive or negative phase sequence.(a) va(t) = 169.7cos(377t +15◦) Vvb(t) = 169.7cos(377t −105◦) Vvc(t) = 169.7sin(377t −135◦) V(b) va(t) = 311cos(wt −12◦) Vvb(t) = 311cos(wt +108◦) Vvc(t) = 311cos(wt +228◦) V(c) V1 = 140 −140◦ VV2 = 114 −20◦ VV3 = 124 100◦ V
Apply single-phase equivalency to determine the linecurrents in the Y-D network shown in Fig. P10.13. The loadimpedances are Zab = Zbc = Zca = (25+ j5) W
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