..andNowZ₁ = x₁ + 38 = 1·2+38 = 5522₂ = x₂+4·1=09+4·1=552}Z₁ = (5) 2; Z₂ = (5) 2; Z= (50+ j40) 2210⁰ = 5774 210° volts =E₂Referring to Fig. 12.87, we have,1₁ Z₁ + (1₁+1₂)Z = E₁and1₂ *Z₁ + (1₁ + 1₂) ZE₂Subtracting eq. (ii) from eq. (i), we have,1₁ - 1₂..1₁-1₂ =Adding eqs. (i) and (), we have,=1₁ + 1₂-10,000√√312000=E₁-E₂Z₁-1242 + j1003j5(200+j250) A20⁰ = (6928+j0) volts1001₁ + 1₂On solvin eqs. (iii) and (iv), we have,=E₁ + E₂2 Z+Z₁12614+j103(5686+j1003) - (6928+ j0)j5= (200+j250) A(5686+j1003) + (6928+ j0)2 (50+ j40) + j512654 24.5°785131-2240-4°96-4 2-35-9° A = (78-1-j56-6) A(78-1-j56-6) A(5686+j1003) volts169 A ; I, = 164 A597...(1).... (iii)...(iv)Hello expert, I want you to explain howhe got the law inside the circle. Can youexplain step by step and a clear line, please?

Answered: Image /qna-images/answer/2b2d3e73-441e-4c68-881d-a82dc29fed32.jpg

Referring to Fig. 12.87, we have, 1₁ Z₁ + (1₁ +1₂)Z = E₁ 1₂ *Z₁ + (1₁ +1₂)Z = E₂ and Subtracting eq. (ii) from eq. (i), we have, 1₁-1₂ :. 1₁-1₂ Adding eqs. (i) and (), we have, = 1₁ + 1₂ E₁-E₂ Z₁ -1242 + j1003 j5 (200+j250) A = = (5686+j1003) - (6928+ j 0) j5 = (200+j250) A E₁ + E₂ 2 Z+Z₁ 12614+j103 (5686+j1003) + (6928+ j0) 2 (50+ j40) + j5 12654 24.5⁰ 85 131-2240-4⁰ = 96-42-35-9° A = (78-1-i56-6) A ...(1) ... (iii)

Referring to Fig. 12.87, we have, 1₁ Z₁ + (1₁ +1₂)Z = E₁ 1₂ *Z₁ + (1₁ +1₂)Z = E₂ and Subtracting eq. (ii) from eq. (i), we have, 1₁-1₂ :. 1₁-1₂ Adding eqs. (i) and (), we have, = 1₁ + 1₂ E₁-E₂ Z₁ -1242 + j1003 j5 (200+j250) A = = (5686+j1003) - (6928+ j 0) j5 = (200+j250) A E₁ + E₂ 2 Z+Z₁ 12614+j103 (5686+j1003) + (6928+ j0) 2 (50+ j40) + j5 12654 24.5⁰ 85 131-2240-4⁰ = 96-42-35-9° A = (78-1-i56-6) A ...(1) ... (iii)

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter2: Fundamentals

Section: Chapter Questions

Problem 2.17MCQ: Consider the load convention that is used for the RLC elements shown in Figure 2.2 of the text. A....

See similar textbooks

Similar questions

A 100km, 60Hz transmission line has a resistance of 2.07Ω/km, an inductance of 3.108mH/km, and a capacitance of 1.4774mF/f. This line supplies a 100MW three-phase star load with a 0.9 power factor lagging at 215kV. Using the π-circuit, we ask: a-) Calculate the voltage at the emitter terminal;b-) Calculate the losses in the transmission line;c-) Determine whether the line is absorbing or supplying reactive power. PLEASE, TYPE, HAND WRITING GETS UNDERSTANDING.
Figure a shows the ignition circuit of an automobile system. It consists of four Resistors R, as well as the energy storing elements of Inductor and Capacitor . Meanwhile Figure b shows the equivalent circuit of the same system of Figure a. Let Z1 and Z2 be the Impedances seen from the respective terminals. The variables and are the mesh currents flowing in the loops 1 and 2, respectively. Let the values of Resistors are 2, while the value of the Capacitor and Inductor are 1. i) The total impedance of Z1=ii) The total impedance of Z2=iii) Current I2(S) in function of Vs(S) = iv} Find the transfer function of Vz2(S)/Vs(S) =
Use DFT method to compute the circular convolution of x(n)={4,4} and h(n)={2, 6}.
Solve the impedances of the circuit using mesh analysis. Complete the table given. Note: If there are no coefficients for a particular parameter, input zero. Round off only the final answer to five decimal places if possible. @mesh 1: Z11I1 + Z12I2 +Z13I3 + Z14I4 = Source1 @mesh 2: Z21I1 + Z22I2 + Z23I3 + Z24I4 = Source2 @mesh 3: Z31I1 + Z32I2 + Z33I3 + Z34I4 = Source3 @mesh 4: Z41I1 + Z42I2 + Z43I3 + Z44I4 = Source4
the following electrical circuit questions 1. Check the Reciprocity theorem for the network, with source and output response positions being at input and output terminals. 2.Draw an oriented graph for the network and obtain the corresponding incidence matrix. Then, find all cut sets and loops of the network. 3.For the network, assign arbitrary branch voltages and branch currents subject only to KVL and KCL constraints. Then, verify the conclusion of Tellegen’s theorem.
In the circuit shown below, IS(t) = 10√(2) cos (104t+30°) A, RA=1Ω. A.) Determine the Thevenin equivalent circuit seen by RA. B.) Set up the nodal equations for determining the node voltages. Express each equation in simplest form with variables,V1, V2, V3.
A HV transmission line having receiving end load is 300MVA, 50 Hz line from poin A to point B. Resistance 0.13 ohm/KM , for 160 KM Inductance XL= 0.4 ohm per kilometer Parallel admittance j318 × 10-6 Power Factor 0.8770, 0.8477 lagging, unity and 0.8770, 0.8477 leading Assume desired voltage at the receiving end of the transmission line is not less than 380kV., if the line is supplying rated voltage and apparent power at various power factors given Estimate : a. Sending end voltages b. Voltage regulation c. Efficiency of the transmission line d. Give a technical explanations on the results obtained
After solving for lb, kindly find Z in the circuit shown below, where vg = 25/15° V and la= 5/15°A. TY
B.1. Attenuation is measured inA. parts per million (ppm).B. watts (W).C. decibels (dB).D. amperes (A). B,2. While you're wiring portions of a structured cabling system, you add a conductor termination thatautomatically removes the insulation from the conductor as it's seated into the connector. Which of thefollowing connection techniques did you use?A. Fiber optic terminationB. CleavingC. Plug-compatible connectionD. Insulation displacement connectionB.C. Which of the following is a specialty electrical wiring system that's used in patient care areas of healthcare facilities to provide added protection against the possibility of electrical shock?A. An isolated power systemB. An intrinsically safe circuit systemC. A redundant grounding systemD. An automatic transfer switching system
TOPIC: Transmission Lines, Power Systems, and Powerplants:INSTRUCTIONS:- Answer in this format: Given, Illustration, Required Conversion, Solution, Final Answer.- Step-by-step solution, do not skip even simple calculations to avoid confusion.- If answered in written form, make sure it is readable.PROBLEM:A three-phase, three-wire, 500-V, 60-Hz source supplies a three-phase induction motor, a wye-connected capacitor bank that draws 2 KVAr per phase, and a balanced three-phase heater that draws a total of 10 kW. The induction motor is operating at its rated 75 Hp and has an efficiency and power factor of 90.5 and 89.5 percent, respectively. Determine the system KVA. 78.12 75.98 24.81 83.17
please explain steps-by-steps solution tq
A coil of wire having a value of (5 + j8) ohms is connected in series with a capacitive reactance Xc, and this series combination is then connected in parallel with a resistor R. If the equivalent impedance of the circuit is 4 /0o ohms, calculate the values of Xc and R.