Compute the equivalent impedance looking back at terminals ? and ? ((???). 13.Given:bwwwwww73wwww moZ₂wwZGZMMy moor25Z4 = 1 + j5 2Z5 = 2+j6nZ6=52ZqZIDwww f28HEZ12reeeहैं॥=j13³5213=mor· Z₁4 = -j14 aM715 = 15.2Where: Z1 = 2 ΩZ7= 3 + j7 2Z10 = 6 - j10 ΩZ2 = 3 ΩZ8 = 4+j82Z11=j112Z3 = 4 ΩZ9= 5j9Z12=-j120Compute the equivalent impedance looking back at terminals a and b ((Zeq).

Answered: Image /qna-images/answer/45971284-d123-483e-a215-808a4575c29a.jpg

Given: b www www 73 www Z4 un moor Z₂ ww ZG M weeeeww 57 Z4 = 1 + j5 2 Z5=2+j6n Z6=5 Ω Z₁ moon ww Za reee WWW ZID www 28 HE HE Z12 lew 5 13 = = √1352 mor Z14 = M 715 = 152 -j14 e Where: Z1 = 2 Ω Z7= 3 + j7 2 Z10=6-j102 Z2 = 3 Ω Z8 = 4+j8n Z₁1=j110 Z3 = 4 Ω Z9= 5j9 Z12=-j12 Compute the equivalent impedance looking back at terminals a and b ((Zeq).

Given: b www www 73 www Z4 un moor Z₂ ww ZG M weeeeww 57 Z4 = 1 + j5 2 Z5=2+j6n Z6=5 Ω Z₁ moon ww Za reee WWW ZID www 28 HE HE Z12 lew 5 13 = = √1352 mor Z14 = M 715 = 152 -j14 e Where: Z1 = 2 Ω Z7= 3 + j7 2 Z10=6-j102 Z2 = 3 Ω Z8 = 4+j8n Z₁1=j110 Z3 = 4 Ω Z9= 5j9 Z12=-j12 Compute the equivalent impedance looking back at terminals a and b ((Zeq).

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

Chapter13: Transmission Lines: Transient Operation

Section: Chapter Questions

Problem 13.19P

See similar textbooks

Similar questions

Determine: re, Zi, Zo, Av(use exact analysis)
circuit theory please I need the answer through 25 min, I do not have time more than 25 min Question 3) if a=166, b=27, c=3 and d=35, then find Zin for the above circuit. a. Zin = 996,00 + j213,14 ohm b. Zin = 664,00 + j267,14 ohm c. Zin = 863,20 + j186,14 ohm d. Zin = 1328,00 + j321,14 ohm e. Zin = 1660,00 + j294,14 ohm f. Zin = 1660,00 + j429,14 ohm g. Zin = 1460,80 + j280,64 ohm h. Zin = 1195,20 + j483,14 ohm
Answer: 99∠−8.04° (98−j13.8) Show steps please
Answer: y11=75.77mS , y12=y21=-30.3mS , y22=45.47mS
Starting position of an object is represented as5.1±0.2m and finishing position as y=6,9±0.3m.What will be the displacement and errorindisplacement?Select one a. Displacement = 1:8m. Error=0.5mb Displacement = 1.8m, Error= 0.3mc Displacement - 1.8m, Error = 0,36md Displacement = 1.8m, Error = 0.4m
COURSE: Electrical EngineeringSUBJECT: Electrical ApparatusINSTRUCTIONS:• Write the GIVEN with their respective symbols and units.• Do not skip the SOLUTIONS, do it step-by-step with their respective symbols and units• Round up to 3 three decimal places the FINAL ANSWERS and BOX it. PROBLEM: A coil with impedance 8+j6 Ω is connected in series with a capacitive reactance X. The series combination is connected in parallel with a resistor R. Given that the equivalent impedance of the resulting circuit is 5∠0° Ω, find the value of X and R.
The supply voltage of an inductive coil with a resistance of 35 ohms and an inductance of 92.84 millihenry has the equation e = 340 sin (377t). Determine the resulting current's equation. a.) i(t) = 6.867 sin (377t + 45°) b.) i(t) = 6.867 cos (377t - 45°) c.) i(t) = 6.867 cos (377t + 45°) d.) i(t) = 6.867 sin (377t - 45°)
An impulse generator has 8 stages, each having a capacitance of 1.2 μF and 167 KV. If it has to produce a 1/50 (μs) waveform across a load capacitor of 15000 PF, (1) what is its maximum discharge energy, and (2) calculate the regulation? If• t1=3R1C1C2 / (C1+ C 2), and t2=0.7(R1+R2) (C1+ C2) and use exact method
Determine: re, Zi, Zo, Av (use appropriate approximation)
What is the impedance of a transformer coil that has 340' of #18 copper wire and an inductance of 70mH at 60Hz? 1000’ of #18 copper wire is 16 ohms.
termining the impedance and current of the system, the voltage across the resistor, and the voltage across the inductor. Could you please forward the response as a print of a handwritten answer? I got these two answers earlier. Which of them would be correct Impedance Z= (1000 + j 90.836) ohm Current I= 0.219∠-5.19° A Voltage across resistor VR = 219.097∠-5.19° V Voltage across inductor VL = 165.195∠84.81° V Impedance will be Z= 1004.052 ∠ 5.12° Current will be i = 0.221 ∠ -5.19° A. Volatge across resistor will be VR = 211.9∠ -5.19° V Volatge across inductor will be VL = 165.12 ∠ 84.81°V
For the circuit shown, determine:a) Plot pole-zero diagram for H = I / Vsb) Based on H, calculate the forced response for i (t) to an excitation:Vs (t) = 5 e-3tsin (50t) cos (60t) [V]