DELMAR'S STANDARD TEXT OF ELECTRICITY
DELMAR'S STANDARD TEXT OF ELECTRICITY
6th Edition
ISBN: 9780357323380
Author: Herman
Publisher: CENGAGE C
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Chapter 8, Problem 5PP

Find the unknown values in the circuit if the total current is 0.65 A and the resistors have the following values:

R 1 = 22 k Ω  R 2 = 15 Ω  R 3 = 22 Ω  R 4 = 33 Ω E T _ _ _ _ _  E 1 _ _ _ _ _  E 2 _ _ _ _ _  E 3 _ _ _ _ _  E 4 _ _ _ _ _ R T _ _ _ _ _  I 1 _ _ _ _ _  I 2 _ _ _ _ _  I 3 _ _ _ _ _  I 4 _ _ _ _ _

Chapter 8, Problem 5PP, 5. Find the unknown values in the circuit if the total current is 0.65 A and the resistors have the , example  1

Chapter 8, Problem 5PP, 5. Find the unknown values in the circuit if the total current is 0.65 A and the resistors have the , example  2

FIGURE 8-25 A parallel-series circuit.

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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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