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

The different type of faults in power systems are:

• Single line to ground fault (LG)
• Line to line fault (LL)
• Double line to ground fault (LLG)
• Three-phase faults (LLL or LLLG)

Frequency of occurrence:
Among the given faults, LG or line to ground fault is most common
and occurs frequently.
The order of frequency of occurrence is given below.
LG > LL > LLG > LLL
Severity of faults:
Among the given faults, LLLG or 3 phase faults are the most severe. LG
or line to ground fault is least severe.
Line to line fault is more severe than the line to a ground fault while the
double line to ground fault is one level severe than LL.
The order of frequency of occurrence is given below.
LLL > LLG > LL > LG

Single line-to-ground fault:

On a transmission line when one conductor drops to the ground or comes in contact with the neutral conductor, then there is an insulation breakdown between one of the phases and earth, due to which single line to ground fault takes place. Such types of failures may occur in a power system due to many reasons like high-speed wind, falling off a tree, lightning, etc.

Unsymmetrical faults: 
These are faults that lead to unequal currents with unequal phase shifts in a three-phase system. The unsymmetrical fault occurs in a system due to the presence of an open circuit or short circuit of transmission or distribution line. These are classified as

• Single line to ground fault (LG)
• Double line fault (LL)
• Double line to ground fault (LLG)

The decreasing order of fault currents at terminals of a synchronous generator is given below
I_f(LG) > I_f(LLG) > I_f(3ϕ) > I_f(LL)
The fault current is maximum for line to ground fault.

Given that,
Z₁ = j 0.1 pu
Z₂ = j 0.05 pu
Z₀ = j 0.01 pu
Z_n = j 0.01 pu
For LG fault, fault current is given by

I_f in Amperes = (I_f in pu) × I_base

Given that,
X₁ = 0.25 pu
X₂ = 0.15 pu
X₀ = 0.05 pu
Fault current for LG fault


Fault current for three phase fault,

Given that both the fault currents are equal.

⇒ X_n = 0.1 pu
Base kV = 30 kV
Base MVA = 50 MVA

Concept:
In a single line to ground (LG) fault, the fault current is

Where,
E_a = terminal voltage
Z₁ = positive sequence impedance
Z₂ = negative sequence impedance
Z₀ = zero sequence impedance
Z_n = impedance in neutral grounding
The fault current can also be written as I_f = 3I_a1  

Calculation:
Single line to ground fault, Fault current in I_f = 3I_a1

Fault current I_f = 15396 A.

The following circuit shown is a system with a line to ground fault.

Here, V_a = 0, I_b = 0, I_c = 0

By substituting I_b and I_c values,

Hence in LG faults, all of the three components l_a0, l_a1, and l_a2 are equal.
V_a = 0 = V_a1 + V_a2 + V_a0
Now substituting the values of V_a0, V_a1 and V_a2 from the sequence network equation,
E_a – I_a1Z₁ – I_a2Z₂ – I_a0Z₀ = 0
⇒ E_a – I_a1Z₁ – I_a1Z₂ – I_a1Z₀ = 0

From the above equation, it is clear that to simulate an L-G fault all the three sequence networks are required and since the currents are equal in magnitude and phase angle, the three sequence networks must be connected in series. The voltage across each sequence network corresponds to the same sequence component of Va.
The interconnection of the sequence network is shown in the figure.

Zero sequence current in LG fault
For LG fault, all the sequence currents are equal.

Where
IR₀ is the zero-sequence current
i_f is the fault current

Concept:
In a line to ground fault, the current through the neutral during fault is given by
If = 3I_a0
Where I_a0 is the zero sequence current of generator

Calculation:
Given that, the zero sequence current of generator for line to ground fault = j2.4 pu
The current through the neutral during the fault = 3 × j2.4 = j7.2 pu

For L − G fault,

Positive require a network is


Negative recurrence network is


Zero sequence network is