रेटेड वोल्टेज पर चल रहे अनलोडेड सिंक्रोनस जनरेटर के टर्मिनलों पर हुए एक फॉल्ट से धारा और वोल्टेज के निम्नलिखित मान प्राप्त हुए:
Ia0 = j2.37 p.u.
Ia1 = – j3.05 p.u.
Ia2 = j0.68 p.u.
Va0 = Va1 = Va2 = 0.237 p.u.
निम्नलिखित में से कौन-सा फॉल्ट हुआ है?
उत्तर और चरणबद्ध हल
B — L-L-G फॉल्ट। Ia = Ia0 + Ia1 + Ia2 = j2.37 − j3.05 + j0.68 = 0 (फेज़ a स्वस्थ), शून्य-अनुक्रम धारा उपस्थित (ग्राउंड शामिल) और Va0 = Va1 = Va2 — ये तीनों LLG फॉल्ट के लक्षण हैं।
विस्तृत हल (मूल भाषा में)
Technical Verification
Given V0=V1=V2=0.237 p.u. This equality is compatible with the boundary condition for a double-line-to-ground fault on the other two phases. Ia = I0+I1+I2 = j(2.37−3.05+0.68)=0, so phase a carries no current and is the healthy phase. Nonzero zero-sequence current proves that ground is involved; therefore a pure L-L fault is excluded. Nonzero negative- and zero-sequence components exclude a balanced three-phase fault.
Core Concept
Symmetrical components turn an unbalanced fault into coupled positive-, negative- and zero-sequence networks. Ground involvement is recognized by zero-sequence current. For a double-line-to-ground fault on phases b and c, the unfaulted phase current Ia is zero and the fault-point sequence voltages satisfy the appropriate equal-voltage relation.
Formula / Key Relationship
Ia=I0+I1+I2. Here Ia=0. Phase transformation: [Va Vb Vc]^T = A[V0 V1 V2]^T.
Step-by-Step Check
Three independent signatures agree: zero sequence present → ground fault; Ia=0 → phase a healthy; negative sequence present → unbalanced two-phase involvement. Therefore L-L-G, option B.
Why the Other Options Are Wrong
A (L-L) has no zero-sequence current; C (L-G) has a different sequence-network condition; D (L-L-L) is balanced and would not have nonzero negative/zero sequence components. The given sequence relations identify L-L-G.
Answer-Key Verification Note
The uploaded provisional key marks D . After independent technical verification, the defensible answer is B . This discrepancy is stated explicitly rather than silently copying the provisional key.
Exam Shortcut / Approach
Fault ID shortcut: zero-sequence present means the ground path participates.
Common Mistake
Selecting a 3-phase fault despite nonzero I0 and I2.
Quick Revision
Identification of Double-Line-to-Ground Fault is linked with Symmetrical Components, Fault Analysis and Power Systems. Fault ID shortcut: zero-sequence present means the ground path participates.
A fault occuring at the terminals of an unloaded synchronous generator operating at its rated voltage has resulted in the following values of currents and voltages:Ia0 = j2.37 p.u.Ia1 = – j3.05 p.u.Ia2 = j0.68 p.u.Va0 = Va1 = Va2 = 0.237 p.u.
Which one of the following faults has occurred?
Answer & step-by-step solution
For a double-line-to-ground fault on phases b and c, the unfaulted phase current Ia is zero and the fault-point sequence voltages satisfy the appropriate equal-voltage relation. Three independent signatures agree: zero sequence present → ground fault; Ia=0 → phase a healthy; negative sequence present → unbalanced two-phase involvement.
Exam focus: Fault ID shortcut: zero-sequence present means the ground path participates. Common trap: Selecting a 3-phase fault despite nonzero I0 and I2.
Technical Verification
Given V0=V1=V2=0.237 p.u. This equality is compatible with the boundary condition for a double-line-to-ground fault on the other two phases. Ia = I0+I1+I2 = j(2.37−3.05+0.68)=0, so phase a carries no current and is the healthy phase. Nonzero zero-sequence current proves that ground is involved; therefore a pure L-L fault is excluded. Nonzero negative- and zero-sequence components exclude a balanced three-phase fault.
Core Concept
Symmetrical components turn an unbalanced fault into coupled positive-, negative- and zero-sequence networks. Ground involvement is recognized by zero-sequence current. For a double-line-to-ground fault on phases b and c, the unfaulted phase current Ia is zero and the fault-point sequence voltages satisfy the appropriate equal-voltage relation.
Formula / Key Relationship
Ia=I0+I1+I2. Here Ia=0. Phase transformation: [Va Vb Vc]^T = A[V0 V1 V2]^T.
Step-by-Step Check
Three independent signatures agree: zero sequence present → ground fault; Ia=0 → phase a healthy; negative sequence present → unbalanced two-phase involvement. Therefore L-L-G, option B.
Why the Other Options Are Wrong
A (L-L) has no zero-sequence current; C (L-G) has a different sequence-network condition; D (L-L-L) is balanced and would not have nonzero negative/zero sequence components. The given sequence relations identify L-L-G.
Answer-Key Verification Note
The uploaded provisional key marks D . After independent technical verification, the defensible answer is B . This discrepancy is stated explicitly rather than silently copying the provisional key.
Exam Shortcut / Approach
Fault ID shortcut: zero-sequence present means the ground path participates.
Common Mistake
Selecting a 3-phase fault despite nonzero I0 and I2.
Quick Revision
Identification of Double-Line-to-Ground Fault is linked with Symmetrical Components, Fault Analysis and Power Systems. Fault ID shortcut: zero-sequence present means the ground path participates.
રેટેડ વોલ્ટેજે ચાલતા અનલોડેડ સિંક્રોનસ જનરેટરના ટર્મિનલ પર થયેલા એક ફોલ્ટથી પ્રવાહ અને વોલ્ટેજનાં નીચેનાં મૂલ્યો મળ્યાં:
Ia0 = j2.37 p.u.
Ia1 = – j3.05 p.u.
Ia2 = j0.68 p.u.
Va0 = Va1 = Va2 = 0.237 p.u.
નીચેનામાંથી કયો ફોલ્ટ થયો છે?
જવાબ અને પગલાવાર ઉકેલ
B — L-L-G ફોલ્ટ. Ia = Ia0 + Ia1 + Ia2 = j2.37 − j3.05 + j0.68 = 0 (ફેઝ a સ્વસ્થ), શૂન્ય-ક્રમ પ્રવાહ હાજર (ગ્રાઉન્ડ સામેલ) અને Va0 = Va1 = Va2 — આ ત્રણેય LLG ફોલ્ટનાં લક્ષણો છે.
વિગતવાર ઉકેલ (મૂળ ભાષામાં)
Technical Verification
Given V0=V1=V2=0.237 p.u. This equality is compatible with the boundary condition for a double-line-to-ground fault on the other two phases. Ia = I0+I1+I2 = j(2.37−3.05+0.68)=0, so phase a carries no current and is the healthy phase. Nonzero zero-sequence current proves that ground is involved; therefore a pure L-L fault is excluded. Nonzero negative- and zero-sequence components exclude a balanced three-phase fault.
Core Concept
Symmetrical components turn an unbalanced fault into coupled positive-, negative- and zero-sequence networks. Ground involvement is recognized by zero-sequence current. For a double-line-to-ground fault on phases b and c, the unfaulted phase current Ia is zero and the fault-point sequence voltages satisfy the appropriate equal-voltage relation.
Formula / Key Relationship
Ia=I0+I1+I2. Here Ia=0. Phase transformation: [Va Vb Vc]^T = A[V0 V1 V2]^T.
Step-by-Step Check
Three independent signatures agree: zero sequence present → ground fault; Ia=0 → phase a healthy; negative sequence present → unbalanced two-phase involvement. Therefore L-L-G, option B.
Why the Other Options Are Wrong
A (L-L) has no zero-sequence current; C (L-G) has a different sequence-network condition; D (L-L-L) is balanced and would not have nonzero negative/zero sequence components. The given sequence relations identify L-L-G.
Answer-Key Verification Note
The uploaded provisional key marks D . After independent technical verification, the defensible answer is B . This discrepancy is stated explicitly rather than silently copying the provisional key.
Exam Shortcut / Approach
Fault ID shortcut: zero-sequence present means the ground path participates.
Common Mistake
Selecting a 3-phase fault despite nonzero I0 and I2.
Quick Revision
Identification of Double-Line-to-Ground Fault is linked with Symmetrical Components, Fault Analysis and Power Systems. Fault ID shortcut: zero-sequence present means the ground path participates.