Electrical Engineering (EE) Exam  >  Electrical Engineering (EE) Tests  >  Test: Line to Line & Double Line to Ground Fault - Electrical Engineering (EE) MCQ

Test: Line to Line & Double Line to Ground Fault - Electrical Engineering (EE) MCQ


Test Description

10 Questions MCQ Test - Test: Line to Line & Double Line to Ground Fault

Test: Line to Line & Double Line to Ground Fault for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Test: Line to Line & Double Line to Ground Fault questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Line to Line & Double Line to Ground Fault MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Line to Line & Double Line to Ground Fault below.
Solutions of Test: Line to Line & Double Line to Ground Fault questions in English are available as part of our course for Electrical Engineering (EE) & Test: Line to Line & Double Line to Ground Fault solutions in Hindi for Electrical Engineering (EE) course. Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free. Attempt Test: Line to Line & Double Line to Ground Fault | 10 questions in 30 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study for Electrical Engineering (EE) Exam | Download free PDF with solutions
Test: Line to Line & Double Line to Ground Fault - Question 1

In which type of fault, zero sequence currents do not exist?

Detailed Solution for Test: Line to Line & Double Line to Ground Fault - Question 1

Line to Line fault:

When two conductors of a 3-phase system are short-circuited line to line fault occurs.
IR = 0
IF = IY = -IB

From equation (i), we found that zero sequence currents do not exist in the LL fault.

Test: Line to Line & Double Line to Ground Fault - Question 2

At the terminal of a 3ϕ, 6.6 kV, 10 MVA alternator, a load R = 200 Ω is connected between two phases, and other phase is kept open. The sequence impedance of the alternator is Z1 = Z2 = j 5Ω and Z0 = j2Ω.

What is the current through the load resistance?

Detailed Solution for Test: Line to Line & Double Line to Ground Fault - Question 2

Concept:
In a double line fault, the fault current is given by

Where Z1 is the positive sequence impedance
Z2 is the negative sequence impedance

Calculation:
V = 6.6 kV
Z1 = Z2 = j5Ω
Zf = 200 Ω
Z0 = j2Ω
This is the case of the line to line fault.

Ia = 0 (∵ open circuited)
From the formula of a double line fault or L-L fault,

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: Line to Line & Double Line to Ground Fault - Question 3

In a three phase system for a line to line fault the positive, negative and zero sequence voltage Va1, Va2, Va0 respectively, for phase voltage Va are related as

Detailed Solution for Test: Line to Line & Double Line to Ground Fault - Question 3

For line to line fault, there is no ground.
Va0 = 0 and Va1 = Va2

Test: Line to Line & Double Line to Ground Fault - Question 4

The positive, negative and zero sequence impedances of a three-phase generator are Z1, Z2 and Z0 respectively. For a line-to-line fault with fault impedance Zf, the fault current is If1 = kIf, where If is the fault current with zero fault impedance. The relation between Zf and k is

Detailed Solution for Test: Line to Line & Double Line to Ground Fault - Question 4

In line to line fault,
fault current with fault impedance zf is

fault current with zero fault impedance is,


⇒ z1 + z2 = k (z1 + z2 + zf)
⇒ z1 + z2 - k z1 - k z2 = k zf 

*Answer can only contain numeric values
Test: Line to Line & Double Line to Ground Fault - Question 5

An 11 kV, 25 MVA, 3-phase Y-connected alternator was subjected to three different types of fault at its terminals. The fault currents were:
5610 A for a three phase fault
6760 A for line to line fault
8630 A for single line to ground fault.
If the alternator neutral is solidly grounded, the zero-sequence reactance of the alternator (in pu) is _________


Detailed Solution for Test: Line to Line & Double Line to Ground Fault - Question 5

Voltage (E) = 11 kV
Phase voltage (Eph
Three phase fault current = 5610 A

⇒ X1 = 1.132 Ω
L-L fault current = 6760 A

⇒ X2 = 0.495 Ω
L-G fault current = 8630 A

Test: Line to Line & Double Line to Ground Fault - Question 6

Determine the fault current in the system following a double line to ground short circuit fault at the terminal of a star connected synchronous generator operating initially on an open circuit voltage of 1.0 pu. The positive, negative, and zero sequences reactance of the generator are, respectively j0.35, j0.25, and j0.20, and the star point is isolated from the ground.

Detailed Solution for Test: Line to Line & Double Line to Ground Fault - Question 6

Concept:
Here star point is isolated so the flow of fault current is not possible in neutral or ground ( also zero sequence current is zero because zero-sequence current flow through the ground and here the ground is not available) so current is absent in neutral but current available in phase so this fault is considered as line to line  (L-L) fault.

Positive sequence current (i1) for L-L fault is given by;
i1 = voc/(x1 + x2)
Where;
voc = open circuit voltage, x1 = Positive sequence reactance, x2 = Negative sequence reactance
Fault current (if) for L-L fault is given by:
if = √3 × i1

Calculation:
Given that:

voc = 1.0 pu, x1 = j0.35, x2 = j0.25 pu, xo = 0.20pu, xn = ∞ pu  (since Neutral is isolated from ground)
Here, voc = Open circuit voltage, x1 = Positive sequence reactance, x2 = Negative sequence reactance, xo = Zero sequence reactance, xn = Neutral reactance
Here L-L (line to line) fault happened so we will first calculate positive sequence current (i1) then calculate fault current (if) by help of positive sequence current.
Calculation of Positive sequence current (i1);
 i1 = 1/(j0.35+j0.25)
i1 = -j1.6666 pu
So fault current (if) ;
if = √3 × i1
If = -j2.887 pu

Test: Line to Line & Double Line to Ground Fault - Question 7

The reactance data for a power system is shown in the figure in per unit on a common base is as follows.

G1: X1 = X2 = j0.1, X0 = j0.05
G2: X1 = X2 = j0.1, X0 = j0.05
T1: X1 = X2 = j0.25, X0 = j0.25
T2: X1 = X2 = j0.25, X0 = j0.25
Line: X1 = X2 = j0.3, X0 = j0.5

If a double line to ground fault occurs at bus 1, the fault current is _______

Detailed Solution for Test: Line to Line & Double Line to Ground Fault - Question 7

For the given power system network, the positive sequence impedance network is given below.

The impedance at a point of fault is,
X1eq = (j0.35)∥(j0.65) 
= j0.2275 pu
Negative sequence reactance is the same as positive sequence reactance
X2eq = X1eq = j0.2275pu 
The zero sequence reactance network is,

Zero sequence impedance
Z0eq = j0.25 ∥ j0.75 = j0.1875 pu
For a double line to fault at bus 1 is,

*Answer can only contain numeric values
Test: Line to Line & Double Line to Ground Fault - Question 8

A 30 MV A, 3-phase, 50Hz,13.8kV, star-connected synchronous generator has positive, negative and zero sequence reactances, 15%,15% and 5% respectively. A reactance (Xn) is connected between the neutral of the generator and ground. A double line to ground fault takes place involving phases  ′b′and′c′, with a fault impedance of j0.1p.u. The value of  Xn (in p.u.) that will limit the positive sequence generator current to 4270 A is _________.


Detailed Solution for Test: Line to Line & Double Line to Ground Fault - Question 8

We have the base current 

The positive sequence current, Ip = 4270 A

Fault impedance = 0.1 pu
Xn = 3zf + 0.05 + 3xn = 0.35 + 3xn 

From above equation, we get  Xn = 1.07 p.u

Test: Line to Line & Double Line to Ground Fault - Question 9

A three-phase star-connected alternator rated at 30 MVA, 13.8 kV has positive, negative, and zero sequence reactance values as 0.25 pu, 0.35 pu, and 0.10 pu respectively. The neutral of the alternator is solidly grounded. Find the current in phase B when a double line to ground fault occurs on its terminals B and C.

Detailed Solution for Test: Line to Line & Double Line to Ground Fault - Question 9

The equivalent circuit for LLG fault can be drawn as:

Z1 = j0.25 pu, E = 1 pu
Z2 = j0.35 pu
Z0 = j0.10 pu

α = 1∠120°
α2 = 1∠-120°
⇒ Ib = 4.804 pu   

*Answer can only contain numeric values
Test: Line to Line & Double Line to Ground Fault - Question 10

A 30 MVA, 13.2 KV synchronous generator has solid grounded neural. It positive, negative and zero sequence impedance are 0.30, 0.40 and 0.05 PU respectively. What value of reactance (in ohm) must be placed in the neutral of the generator to restrict the fault current to ground to rated line current for a double line to ground fault.


Detailed Solution for Test: Line to Line & Double Line to Ground Fault - Question 10

First find rated current of generator

Taking rated voltage and MVA as base

Z1 = j0.30 PU, Z2 = J 0.40 PU Z0 = J 0
Double line to ground fault

∴ 0.12 + 0.7 (0.05 + 3Xn) = 1.2
Xn = 0.5 PU = 0.5 × 5.88 = 2.94 Ω 

Information about Test: Line to Line & Double Line to Ground Fault Page
In this test you can find the Exam questions for Test: Line to Line & Double Line to Ground Fault solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Line to Line & Double Line to Ground Fault, EduRev gives you an ample number of Online tests for practice

Top Courses for Electrical Engineering (EE)

Download as PDF

Top Courses for Electrical Engineering (EE)