A three-phase full converter feeds a highly inductive load. The input is 415 V (Line-to-Line), 50 Hz. For α = 60°, verify.
1. The average output voltage is 280 V.
2. The fundamental displacement factor is 0.5.
3. Each thyristor conducts for 6.67 ms.
Which of the statements are correct?
Answer & step-by-step solution
In three-phase bridges each thyristor conducts for 120 electrical degrees under continuous current. Vdc≈1.35×415×cos60°≈280.1 V. Displacement factor=cos60°=0.5. Key relation: 1φ: Vdc=(2Vm/π)cosα≈0.9Vrms cosα; 3φ: Vdc≈1.35V_LL cosα.
Exam focus: Large load inductance → nearly constant DC current. Common trap: Distinguish displacement factor cosα from total power factor including distortion.
Statement-wise Verification
- Statement 1 - Correct.
The average output voltage is 280 V.
Firing angle controls average DC voltage - Statement 2 - Correct.
The fundamental displacement factor is 0.5.
Displacement factor=cos60°=0.5. - Statement 3 - Correct.
Each thyristor conducts for 6.67 ms.
Each thyristor conducts 120°; at 50 Hz that is (120/360)×20 ms≈6.67 ms.
Core Concept
A fully controlled bridge feeding a highly inductive load operates with nearly continuous DC current. Firing angle controls average DC voltage; the AC-side current is non-sinusoidal and its fundamental is displaced approximately by α. In three-phase bridges each thyristor conducts for 120 electrical degrees under continuous current.
Formula / Key Relationship
1φ: Vdc=(2Vm/π)cosα≈0.9Vrms cosα; 3φ: Vdc≈1.35V_LL cosα.
Step-by-Step Check
Vdc≈1.35×415×cos60°≈280.1 V. Displacement factor=cos60°=0.5. Each thyristor conducts 120°; at 50 Hz that is (120/360)×20 ms≈6.67 ms.
Why the Other Options Are Wrong
Option A is incorrect because it omits correct statement(s) 1. Option B is incorrect because it omits correct statement(s) 3. Option C is incorrect because it omits correct statement(s) 2.
Exam Shortcut / Approach
Large load inductance → nearly constant DC current.
Common Mistake
Distinguish displacement factor cosα from total power factor including distortion.
Quick Revision
Firing Angle and Thyristor Conduction is linked with Three-Phase Full Converter, Controlled Rectifiers and Power Electronics. Large load inductance → nearly constant DC current.
एक तीन-फेज़ फुल कन्वर्टर अत्यधिक प्रेरणिक लोड को पोषित करता है। इनपुट 415 V (लाइन-टू-लाइन), 50 Hz है। α = 60° के लिए सत्यापित कीजिए।
1. औसत आउटपुट वोल्टेज 280 V है।
2. मूल विस्थापन गुणांक 0.5 है।
3. प्रत्येक थाइरिस्टर 6.67 ms तक चालन करता है।
कौन-से कथन सही हैं?
उत्तर और चरणबद्ध हल
D — 1, 2 और 3। Vdc ≈ 1.35×415×cos60° ≈ 280 V; विस्थापन गुणांक = cos60° = 0.5; प्रत्येक थाइरिस्टर 120° चालन करता है = 20 ms/3 ≈ 6.67 ms।
विस्तृत हल (मूल भाषा में)
Statement-wise Verification
- Statement 1 - Correct.
The average output voltage is 280 V.
Firing angle controls average DC voltage - Statement 2 - Correct.
The fundamental displacement factor is 0.5.
Displacement factor=cos60°=0.5. - Statement 3 - Correct.
Each thyristor conducts for 6.67 ms.
Each thyristor conducts 120°; at 50 Hz that is (120/360)×20 ms≈6.67 ms.
Core Concept
A fully controlled bridge feeding a highly inductive load operates with nearly continuous DC current. Firing angle controls average DC voltage; the AC-side current is non-sinusoidal and its fundamental is displaced approximately by α. In three-phase bridges each thyristor conducts for 120 electrical degrees under continuous current.
Formula / Key Relationship
1φ: Vdc=(2Vm/π)cosα≈0.9Vrms cosα; 3φ: Vdc≈1.35V_LL cosα.
Step-by-Step Check
Vdc≈1.35×415×cos60°≈280.1 V. Displacement factor=cos60°=0.5. Each thyristor conducts 120°; at 50 Hz that is (120/360)×20 ms≈6.67 ms.
Why the Other Options Are Wrong
Option A is incorrect because it omits correct statement(s) 1. Option B is incorrect because it omits correct statement(s) 3. Option C is incorrect because it omits correct statement(s) 2.
Exam Shortcut / Approach
Large load inductance → nearly constant DC current.
Common Mistake
Distinguish displacement factor cosα from total power factor including distortion.
Quick Revision
Firing Angle and Thyristor Conduction is linked with Three-Phase Full Converter, Controlled Rectifiers and Power Electronics. Large load inductance → nearly constant DC current.
એક ત્રણ-ફેઝ ફુલ કન્વર્ટર અત્યંત પ્રેરક લોડને પોષે છે. ઇનપુટ 415 V (લાઇન-ટુ-લાઇન), 50 Hz છે. α = 60° માટે ચકાસો.
1. સરેરાશ આઉટપુટ વોલ્ટેજ 280 V છે.
2. મૂળભૂત વિસ્થાપન ગુણાંક 0.5 છે.
3. દરેક થાઇરિસ્ટર 6.67 ms સુધી વહન કરે છે.
કયાં વિધાનો સાચાં છે?
જવાબ અને પગલાવાર ઉકેલ
D — 1, 2 અને 3. Vdc ≈ 1.35×415×cos60° ≈ 280 V; વિસ્થાપન ગુણાંક = cos60° = 0.5; દરેક થાઇરિસ્ટર 120° વહન કરે છે = 20 ms/3 ≈ 6.67 ms.
વિગતવાર ઉકેલ (મૂળ ભાષામાં)
Statement-wise Verification
- Statement 1 - Correct.
The average output voltage is 280 V.
Firing angle controls average DC voltage - Statement 2 - Correct.
The fundamental displacement factor is 0.5.
Displacement factor=cos60°=0.5. - Statement 3 - Correct.
Each thyristor conducts for 6.67 ms.
Each thyristor conducts 120°; at 50 Hz that is (120/360)×20 ms≈6.67 ms.
Core Concept
A fully controlled bridge feeding a highly inductive load operates with nearly continuous DC current. Firing angle controls average DC voltage; the AC-side current is non-sinusoidal and its fundamental is displaced approximately by α. In three-phase bridges each thyristor conducts for 120 electrical degrees under continuous current.
Formula / Key Relationship
1φ: Vdc=(2Vm/π)cosα≈0.9Vrms cosα; 3φ: Vdc≈1.35V_LL cosα.
Step-by-Step Check
Vdc≈1.35×415×cos60°≈280.1 V. Displacement factor=cos60°=0.5. Each thyristor conducts 120°; at 50 Hz that is (120/360)×20 ms≈6.67 ms.
Why the Other Options Are Wrong
Option A is incorrect because it omits correct statement(s) 1. Option B is incorrect because it omits correct statement(s) 3. Option C is incorrect because it omits correct statement(s) 2.
Exam Shortcut / Approach
Large load inductance → nearly constant DC current.
Common Mistake
Distinguish displacement factor cosα from total power factor including distortion.
Quick Revision
Firing Angle and Thyristor Conduction is linked with Three-Phase Full Converter, Controlled Rectifiers and Power Electronics. Large load inductance → nearly constant DC current.
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