Delmar's Standard Textbook of Electricity (MindTap Course List)
6th Edition
ISBN: 9781285852706
Author: Stephen L. Herman
Publisher: Cengage Learning
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Textbook Question
Chapter 18, Problem 15PP
An R-L series circuit contains two resistors and two inductors. The resistors are 68
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-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
Chapter 18 Solutions
Delmar's Standard Textbook of Electricity (MindTap Course List)
Ch. 18 - 1. What is the relationship of voltage and current...Ch. 18 - What is the relationship of voltage and current...Ch. 18 - 3. What is power factor?
Ch. 18 - 4. A circuit contains a 20- resistor and an...Ch. 18 - 5. An R-L series circuit has a power factor of...Ch. 18 - 6. An R-L series circuit has an apparent power of...Ch. 18 - 7. The resistor in an R-L series circuit has a...Ch. 18 - 8. An R-L series circuit has a reactive power of...Ch. 18 - 9. An R-L series circuit contains a resistor and...Ch. 18 - 10. What is the voltage drop across the resistor...
Ch. 18 - 11. A phase angle meter connected in an R-L series...Ch. 18 - 12. An R-L series circuit has a power factor of...Ch. 18 - An AC electric motor is connected to a 240-V,...Ch. 18 - You are a journeyman electrician working in an...Ch. 18 - Prob. 1PPCh. 18 - Assume that the voltage drop across the resistor,...Ch. 18 - Prob. 3PPCh. 18 - Prob. 4PPCh. 18 - In an R-L series circuit,...Ch. 18 - In an R-L series circuit,...Ch. 18 - In an R-L series circuit,...Ch. 18 - In an R-L series circuit, the apparent power is...Ch. 18 - 9. An R-L series circuit is connected to a 60-Hz,...Ch. 18 - In an R-L series circuit, Z=88,R=32.FindXL.Ch. 18 - In an R-L series circuit, apparent power = 450 VA,...Ch. 18 - In an R-L series circuit, =22, true power = 94...Ch. 18 - An R-L series circuit contains two resistors and...Ch. 18 - An R-L series circuit contains two resistors and...Ch. 18 - An R-L series circuit contains two resistors and...Ch. 18 - An R-L series circuit contains two resistors and...
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