Electric Motor Control
Electric Motor Control
10th Edition
ISBN: 9781133702818
Author: Herman
Publisher: CENGAGE L
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Chapter 31, Problem 9SQ

What is meant by the “compelling action” produced by an accelerating relay?

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In the circuit shown, let Vs-9, R₁-8, R2-2, and R3-4. Use Nodal analysis to determine the current lo. In particular find: V2= 10= A The relative tolerance for this problem is 5 %. R₁ V₁ + ww R₂ Vs V₂ 21 x R3
1. Choose all nodes that must be included, if any, to construct the supernode for Nodal analysis. OV1, V3 OV1, V2 ○ V2, V3 OV1, V2, V4 OV1, V2, V3 O V2, V3, V4 2. Write KCL equation (Nodal equation) at super-node. Write your expression in terms of node voltages V1, V2, V3 and V4 and of the form (G11 V1+G12 V2+G13 V3+G14 V4 = 11), then enter the corresponding values: At super-node KCL: 1/Q G11 1/0 G12 1/Ω G13 1/Q G14 A 3. Use the above equation, the circuit and and super-node inner expression to calculate V3 and then lo : V3= V 10 = R3 Vst + A V₁ + VS2 V₂ V3 w W R₁ R₂ R4 ww R5 V4 V$3
Enter the matrix values (numerical) to solve for voltages at nodes v1, and v2, for the circuit shown, using Nodal equations. In the matrix, row 1, and row 2, correspond to node v1, and node v2 current expressions, respectively. Let Is1=14, Is2=7, R₁=5, R₂-8, R3=2, and R4-5. [G11 G12] [Vi₁ The matrix values are shown here: = G21 G22 [V2] [41] [12] {Hint: As discussed in class and to avoid sign errors, assume nodal currents are locally defined at each node (leaving) and use node labeling as indicated in the circuit. } The relative tolerance for this problem is 5%. VI R2 ww Isl 12 NODE v1 G11 G12 RI 1/Q 1/0 A 4= NODE v2 G21- 1/Q G22 1/0 12 W A === www R3 R4 www Use Cramer's rule (matrix), substitution, or any other method to calculate the voltages: v1 = V v2= V Is2
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