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 19, Problem 16PP
An R-L parallel circuit has an applied voltage of 208 volts and a total current of 2 amperes. The resistor has a value 180
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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 19 Solutions
Delmar's Standard Textbook of Electricity (MindTap Course List)
Ch. 19 - 1. When an inductor and a resistor are connected...Ch. 19 - 2. An inductor and resistor are connected in...Ch. 19 - 3. What is the impedance of the circuit in...Ch. 19 - 4. What is the power factor of the circuit in...Ch. 19 - How many degrees out of phase are the current and...Ch. 19 - In the circuit shown in Figure 191, the resistor...Ch. 19 - 7. A resistor and an inductor are connected in...Ch. 19 - The R-L parallel circuit shown in Figure 191 has...Ch. 19 - The R-L parallel circuit shown in Figure 191 has...Ch. 19 - How many degrees out of phase are the total...
Ch. 19 - Incandescent lighting of 500 W is connected in...Ch. 19 - You are working on a residential heat pump. The...Ch. 19 - Prob. 1PPCh. 19 - Assume that the current flow through the resistor,...Ch. 19 - Assume that the circuit in Figure 18-1 has an...Ch. 19 - Prob. 4PPCh. 19 - In an R-L parallel circuit, R=240 and XL=360. Find...Ch. 19 - In an R-L parallel circuit, IT=0.25 amps, IR=0.125...Ch. 19 - In an R-L parallel circuit, ET=120 volts,...Ch. 19 - In an R-L parallel circuit, ET=48 volts, IT=0.25...Ch. 19 - In an R-L parallel circuit, ET=240 volts, R=560 R...Ch. 19 - In an R-L parallel circuit, ET=240 volts, R=560,...Ch. 19 - In an R-L parallel circuit, ET=208 volts, R=2.4k,...Ch. 19 - In an R-L parallel circuit, ET=480 volts, R=16,...Ch. 19 - In an R-L parallel circuit, IT=1.25 amps, R=1.2k,...Ch. 19 - In an R-L parallel circuit, true power =4.6 watts...Ch. 19 - An R-L parallel circuit is connected to 240 volts...Ch. 19 - An R-L parallel circuit has an applied voltage of...
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