primary) oi J50 onm. the no-load primary voitage is 250 KV (me-to-ine), the connected balance load of 1.5+ j0.25 ohms, the secondary voltage (line to neutral) would be: O A. 12 kV O B. 22.2 kV O C. None of the other choices are correct O D. 13.2 kV O E. 23 kV

Q: The transient stability analysis should be performed based on which fault?

A: Transient stability analysis is used in the transfer function under various disturbances.

Q: A transformer is connected to a primary impedance of 2000 and a secondary impedance of 4500. The…

A: From given primary impedance and secondary impedance calculate turns ratio. Now , using turns ratio…

Q: H-ws A Soleoid loco emindinmeter&75.o cm long is made Svom copper wire of diameter o lo em with very…

A: Given data, Length (L) = 75 cm diameter (d) = 0.100 cm Field (B) = 8.00 mT

Q: QUESTION TWO Power transformers are used in the power system to change the voltage levels to a level…

A:

Q: Kindly send solution. An inductance of 1 henry and a resistance of 2 ohms are connected in series…

A: The inductance,L=1 H The value of the resistor, R=2 Ω The battery voltage, VB=6e-0.0001 V

Q: Three 100kVA 34.5kV primary 12/6.9kV secondary. Connect the transformers delta-wye. What is the…

A: Given: A transformer with Three 100kVA 34.5kV primary 12/6.9kV secondary. Primary is delta…

Q: In a system of 132 kV, the circuit phase to ground capacitance is 0.2 uF and the circuit inductance…

A:

Q: 11 kV at a power factor of 0.867lagging. Each machine supplies half the total power. The synchronous…

A:

Q: sformer 1200 KVA, 11000/2200 V, 50 Hz e reactance/phase of the primary winding ar ding values for…

A:

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A: The external characteristics of the dc generator are drawn between the load current IL and the load…

Q: 5. Obtain the symmetrical components of a set of unbalanced currents I-1.7 25°, I,=0.761800 and…

A: In this question we will find symmetrical components of current...

Q: An alternahng with a value f 16 A. uns tantaneous value eand find fugueney . 50HZ U has an ms Wsuite…

A: (5) f=50 HzIrms=10 A

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A:

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A:

Q: Referring o figure shown below, determine the total impedance at point a and b. 42-90°0 2430°0

A:

Q: 10. Refer to the graphic appendix B. The total impedance of the circuit is most near: 100 + j5714…

A: The solution is given below

Q: There phase power supply load R is 1699 VA, S is 15584 VA and T is 12818 VA. Find (a) the…

A:

Q: Referring o figure shown below, determine the a. total impedance b. branch current IT, I1, and I2…

A:

Q: High-voltage transmission with lower power losses is possible with A. alternating current B. cross…

A: Given: High voltage transmission

Q: The one line diagram of the system and data: Gl and G2:reactance=0.2 p.u ,11kv, 100MVA T1 and T2:…

A: Fault analysis- Transmission line are represented using series reactance. So, series resistance of…

Q: The terminal voltage for cases where the cosQ value is reverse, forward and 1, respectively,…

A: It is asked to draw the phasor diagrams for the terminal voltage for case where power factor  is…

Q: Given the following Resistances and Inductors in Series and Voltage Source: R1 = 1,664 ohms R2 =…

A: inductive reactance = 2π×frequency×inductance. if two impedances Z1 and Z2 are connected in series…

Q: The pour inpul to a p00 voltoHz phase motor Junning on full lend ot On ffniemy d 9 in 92% Fuso…

A: Given, Input voltage, V= 2200 V Frequency, f = 50 Hz Wattmeter readings, W1=300 kWW2=120 kW

Q: The total impedance of the circuit iz Lw Żtotal = R+ Žequiv = R – j- %3D w2 LC – 1

A: Z⏞total=R+Z⏞equivZ⏞total=R-jLωω2LC-1

Q: Use the node voltage method to find the fault at node 2. Voltages before fault: V₁ = 8027.1∠0. 685°…

A: before fault, the voltage at node 1 and node 2,

Q: what will be the total and unit for impedance if the given is 8mA and 9mV?

A: Electrical impedance is the amount of opposition that a circuit presents to current or voltage…

Q: three-phase AY transformer bank shown in E TOOFYA to a balanced delta conr foll

A: Three phase transformers are used in transmission and distribution system. There are two ways to…

Q: on A Load absorbs 10 KVA at a leading power factor of 0.6 The complex power is: Select one: O a…

A: The is a problem of circuits.  Given:- Load power |SL|= 10 kVA  Power factor, cos(phi)= 0.6 leading…

Q: What are these load profiles? Explain their differences. Why is it important to know these load…

A: Load Profile: These load profiles are the graphs between the electrical peak load versus time. It…

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A:

Q: A 10A current is lagging the voltage by 47° and a 12A current is lagging the voltage by 58°. What is…

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A:

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Q: EX The 3 bidliretiomat cantrollery fronected- IL A Ty N) VA Van VEN Ts UBN to toad resistarnce toA…

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Q: 9. What is the total load in kVA on this transformer? Leeeeeee 0004000 120 load 1 600 120 -240- load…

A: In this question we need to find the total load in KVA on this transformer.

Q: Two 300-VA parallel-connected transformers can supply a maximum load of a. 150 VA b. 300 VA C. 450…

A: Two 300 VA parallel connected transformer can supply a maximum load of, equal to the sum of their…

Q: False True the slip-rings replaced by separated segments In order to generate a bidirectional…

A: "The slip rings replaced by seperated segment in order to generate a bidirectional voltage" This…

Q: + Vourlt) ३10 H InF O A. 1591.5 Hz O B. 333 Hz OC. 5032.9 Hz O D.316 Hz

A: The frequency of a RLC circuit is given by  f=12πLC=12π10*1*10-9=1591.5 Hz

Fault Analysis

In an electrical engineering system, a fault is an abnormal electric current flowing through the system, and fault analysis is the study of these faults. For instance, a short-circuit is a fault where the live wire touches the ground or neutral wire. If a circuit is interrupted by a failure of a current-carrying wire then an open circuit fault occurs. The open-circuit faults are also called series faults. One or more phases or ground may be involved in a three-phase fault or may occur only between two phases. Current flows into the earth in case of a ground fault or earth fault. For most situations, the prospective short circuit current of a fault that is predictable can be calculated. To limit the loss of service because of a failure, the protective devices can detect fault conditions and operate circuit breakers and also other devices. All phases equally can be affected in a poly-phase system which is known as symmetric electrical fault. The asymmetric fault becomes more complicated to analyze if only some phases are affected. By using methods such as symmetrical components, the power system analysis of these types of faults is often simplified. The main objective of power system protection is to design, detect, and interrupt power system faults.