A 30 kV, 50 Hz, 50 MVA generator has the positive, negative, and zero sequence reactance’s of 0.25 pu, 0.15 pu, and 0.05 pu, respectively. The neutral of the generator is grounded with a reactance so that the fault current for a bolted LG fault and that of a bolted three-phase fault at the generator terminal are equal. The value of grounding reactance is:

This question was previously asked in

UPPCL AE EE Previous Paper 3 (Held On: 4 November 2019 Shift 1)

Option 2 : 1.8 Ω

ST 1: Logical reasoning

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20 Questions
20 Marks
20 Mins

**Concept:**

Fault current for LG fault \(= \frac{{3\;Ea}}{{{X_1} + {X_2} + {X_0} + 3{X_n}}}\)

Fault current for three phase fault = \(\frac{{Ea}}{{{X_1}}} \)

where,

X_{1 }= Positive sequence reactance

X_{2} = Negative sequence reactance

X_{0 }= Zero sequence reactance

X_{n} = Grounding reactance

**Calculation:**

Given that,

X1 = 0.25 pu

X2 = 0.15 pu

X0 = 0.05 pu

Fault current for LG fault is

\({\left( {{I_f}} \right)_{LG}} = \frac{{3\;Ea}}{{0.25 + 0.15 + 0.05 + 3{X_n}}} = \frac{{3\;Ea}}{{0.45 + 3{X_n}}}\)

Fault current for three phase fault,

\(= \frac{{Ea}}{{{X_1}}} = \frac{{Ea}}{{0.25}}\)

Given that both the fault currents are equal.

\(\Rightarrow \frac{{3\;Ea}}{{0.45 + 3{X_n}}} = \frac{{Ea}}{{0.25}}\)

⇒ X_{n} = 0.1 pu

Base kV = 30 kV

Base MVA = 50 MVA

\({X_n}\;in\;ohms = {X_n}\;in\;pu \times \frac{{k{V^2}}}{{MVA}}\)