Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN: 9781305632134
Author: J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher: Cengage Learning
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Chapter 2, Problem 2.39P
To determine
(a)
The nodal equation in matrix form.
To determine
(b)
The nodal equation in matrix form and suggest method to solve for voltages.
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Chapter 2 Solutions
Power System Analysis and Design (MindTap Course List)
Ch. 2 - The rms value of v(t)=Vmaxcos(t+) is given by a....Ch. 2 - If the rms phasor of a voltage is given by V=12060...Ch. 2 - If a phasor representation of a current is given...Ch. 2 - Prob. 2.4MCQCh. 2 - Prob. 2.5MCQCh. 2 - Prob. 2.6MCQCh. 2 - Prob. 2.7MCQCh. 2 - Prob. 2.8MCQCh. 2 - Prob. 2.9MCQCh. 2 - The average value of a double-frequency sinusoid,...
Ch. 2 - The power factor for an inductive circuit (R-L...Ch. 2 - The power factor for a capacitive circuit (R-C...Ch. 2 - Prob. 2.13MCQCh. 2 - The instantaneous power absorbed by the load in a...Ch. 2 - Prob. 2.15MCQCh. 2 - With generator conyention, where the current...Ch. 2 - Consider the load convention that is used for the...Ch. 2 - Prob. 2.18MCQCh. 2 - The admittance of the impedance j12 is given by...Ch. 2 - Consider Figure 2.9 of the text, Let the nodal...Ch. 2 - The three-phase source line-to-neutral voltages...Ch. 2 - In a balanced three-phase Y-connected system with...Ch. 2 - In a balanced system, the phasor sum of the...Ch. 2 - Consider a three-phase Y-connected source feeding...Ch. 2 - For a balanced- load supplied by a balanced...Ch. 2 - A balanced -load can be converted to an...Ch. 2 - When working with balanced three-phase circuits,...Ch. 2 - The total instantaneous power delivered by a...Ch. 2 - The total instantaneous power absorbed by a...Ch. 2 - Under balanced operating conditions, consider the...Ch. 2 - One advantage of balanced three-phase systems over...Ch. 2 - While the instantaneous electric power delivered...Ch. 2 - Given the complex numbers A1=630 and A2=4+j5, (a)...Ch. 2 - Convert the following instantaneous currents to...Ch. 2 - The instantaneous voltage across a circuit element...Ch. 2 - For the single-phase circuit shown in Figure...Ch. 2 - A 60Hz, single-phase source with V=27730 volts is...Ch. 2 - (a) Transform v(t)=75cos(377t15) to phasor form....Ch. 2 - Let a 100V sinusoidal source be connected to a...Ch. 2 - Consider the circuit shown in Figure 2.23 in time...Ch. 2 - For the circuit shown in Figure 2.24, compute the...Ch. 2 - For the circuit element of Problem 2.3, calculate...Ch. 2 - Prob. 2.11PCh. 2 - The voltage v(t)=359.3cos(t)volts is applied to a...Ch. 2 - Prob. 2.13PCh. 2 - A single-phase source is applied to a...Ch. 2 - Let a voltage source v(t)=4cos(t+60) be connected...Ch. 2 - A single-phase, 120V(rms),60Hz source supplies...Ch. 2 - Consider a load impedance of Z=jwL connected to a...Ch. 2 - Let a series RLC network be connected to a source...Ch. 2 - Consider a single-phase load with an applied...Ch. 2 - A circuit consists of two impedances, Z1=2030 and...Ch. 2 - An industrial plant consisting primarily of...Ch. 2 - The real power delivered by a source to two...Ch. 2 - A single-phase source has a terminal voltage...Ch. 2 - A source supplies power to the following three...Ch. 2 - Consider the series RLC circuit of Problem 2.7 and...Ch. 2 - A small manufacturing plant is located 2 km down a...Ch. 2 - An industrial load consisting of a bank of...Ch. 2 - Three loads are connected in parallel across a...Ch. 2 - Prob. 2.29PCh. 2 - Figure 2.26 shows three loads connected in...Ch. 2 - Consider two interconnected voltage sources...Ch. 2 - Prob. 2.35PCh. 2 - Prob. 2.36PCh. 2 - Prob. 2.37PCh. 2 - Prob. 2.38PCh. 2 - Prob. 2.39PCh. 2 - A balanced three-phase 240-V source supplies a...Ch. 2 - Prob. 2.41PCh. 2 - A balanced -connected impedance load with (12+j9)...Ch. 2 - A three-phase line, which has an impedance of...Ch. 2 - Two balanced three-phase loads that are connected...Ch. 2 - Two balanced Y-connected loads, one drawing 10 kW...Ch. 2 - Three identical impedances Z=3030 are connected in...Ch. 2 - Two three-phase generators supply a three-phase...Ch. 2 - Prob. 2.48PCh. 2 - Figure 2.33 gives the general -Y transformation....Ch. 2 - Consider the balanced three-phase system shown in...Ch. 2 - A three-phase line with an impedance of...Ch. 2 - A balanced three-phase load is connected to a...Ch. 2 - What is a microgrid?Ch. 2 - What are the benefits of microgrids?Ch. 2 - Prob. CCSQCh. 2 - Prob. DCSQ
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- A linear electrical load draws 11 A at a 0.72 lagging power factor./1 153. When a capacitor is connected, the line current dropped to 122 A and the power factor improved to 0.98 lagging. Supply frequency is 50 Hz. a. Let the current drawn from the source before and after introduction of the capacitor be 11 and 12 respectively. Take the source voltage as the reference and express 11 and 12 as vector quantities in polar form. b. Obtain the capacitor current, IC = 12 - 11, graphically as well as using complex number manipulation. Compare the results. c. Express the waveforms of the source current before (11(t)) and after (12(t)) introduction of the capacitor in the form Im sin(2лft + 0). Hand sketch them on the same graph. Clearly label your plots. d. Analytically solve i2(t) – i1(t) using the theories of trigonometry to obtain the capacitor current in the form, ¡C(t) = ICm sin(2πft + OC). Compare the result with the result in Part b.arrow_forwardTransmission line data:Data:• Active power of the load (P): 50 kW• Power factor of the load (PF): 0.8 (lagging)• Line-to-line voltage at the load (V_LC): 13.8 kV• Line resistance (R): 2 Ω• Line inductance (L): 0.8 H• Line capacitance (C): 0.0003 F• Required series compensation: 60% of the line impedance.• Line length: 250 kmDetermine:1. Characteristic impedance and propagation constant.2. The generalized long line constants A, B, C, D.3. Total voltage, current and power at the generating end.4. Voltage regulation.5. Parameters A, B, C, D of the compensation circuit.6. New generalized constants of the power system afterseries compensation.7. Conclusion of the results obtained.arrow_forward3.18 In a single-phase half-wave ac-dc converter, the average value of the load current is 1.78 A. If the converter is operated from a 240 V, 50 Hz supply and if the average value of the output voltage is 27% of the maximum possible value, calculate the following, assume the load to be resistive. (a) Load resistance (b) Firing angle (c) Average output voltage (d) The rms load voltage (e) The rms load current (f) DC power (g) AC power (h) Rectifier efficiency (i) Form factor (j) Ripple factorarrow_forward
- Disassemmble In another way f(t) = [sin(200πt) - cos(300πt)]2arrow_forward☑ 口 ☐ : Homework help starts here! Home | bartleby → https://www.bartleby.com Answered: Decompose using relationships 3 S(+) = 50 sin 3 (500πiz) | bartleby ASK AN EXPERT √ MATH SOLVER Decompose using relationships 3 S(+) = 50 sin 3 (500... Steps "Disassemble in another way." f(+)= 50 Sin³ (500+) Step 1 To decompose f(t)=50 Sin³ (500x+), use trigonometric indentity for Sin³ (e) Sin³ (0)- 3 Sin (e)-Sin (30) 4 Step2: Apply indentity to f(t). f(+)= 50 Sin³ (500+) = 50. 3 Sin (500+) - Sin (3*500π+) = 50. = = 12.5 4 3 Sin (500+)-Sin (1500+) 4 [3 Sin (580x+)-Sin (1500x+)] 37.5 Sin (500+)-12.5 Sin (1500π+) final decomposed form is, PM 05:2 19/12/2024 દ ))^ C W E × م اكتب هنا للبحثarrow_forward6) The principal of DIT hire you to design PBAX with 120 phones. Assuming the number of call is 3/hour/line, the average call duration is 4 minutes, and 55% of all call are made external via a T-1 trunk (24 channels) to the PSTN. Determine carried traffic and channel usage.arrow_forward
- Need a solufor number 2arrow_forwardDecompose using relationships 3 S(+) = 50 sin ³ (500πiz)arrow_forwardA linear electrical load draws I₁ A at a 0.72 lagging power factor. 11 = 153 When a capacitor is connected, the line current dropped to 122 A and the power factor improved to 0.98 lagging. Supply frequency is 50 Hz. a. Let the current drawn from the source before and after introduction of the capacitor be 1₁ and 12 respectively. Take the source voltage as the reference and express 11 and 12 as vector quantities in polar form. b. Obtain the capacitor current, Ic = I2 − I₁, graphically as well as using complex number manipulation. Compare the results. c. Express the waveforms of the source current before (i₁(t)) and after (i2(t)) introduction of the capacitor in the form Im sin(2лft + 0). Hand sketch them on the same graph. Clearly label your plots. d. Analytically solve i₂ (t) - i₁ (t) using the theories of trigonometry to obtain the capacitor current in the form, ic (t) = 1cm sin(2´ft + 0c). Compare the result with the result in Part b.arrow_forward
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