એક સિંગલ-ફેઝ ફુલ કન્વર્ટર 230 V, 50 Hz સાથે જોડાયેલ છે. લોડ R = 10 Ω સાથે ખૂબ મોટું પ્રેરકત્વ છે. α = 30° માટે ચકાસો.
1. સરેરાશ આઉટપુટ વોલ્ટેજ આશરે 179.3 V છે.
2. ઇનપુટ પ્રવાહનું RMS મૂલ્ય 17.93 A છે.
3. મૂળભૂત પાવર ફેક્ટર 0.866 છે.
કયાં વિધાનો સાચાં છે?
જવાબ અને પગલાવાર ઉકેલ
C — 1, 2 અને 3. Vdc = (2√2×230/π)cos30° ≈ 179.3 V; મોટા પ્રેરકત્વથી Id ≈ 17.93 A અને સ્ક્વેર-વેવ ઇનપુટ પ્રવાહનું RMS = Id = 17.93 A; મૂળભૂત વિસ્થાપન ગુણાંક = cos30° = 0.866.
વિગતવાર ઉકેલ (મૂળ ભાષામાં)
Statement-wise Verification
- Statement 1 - Correct.
Average output voltage is approximately 179.3 V.
Vdc=(2√2×230/π)cos30°≈179.3 V. - Statement 2 - Correct.
The RMS value of the input current is 17.93 A.
With R=10 Ω and highly inductive continuous current, Id≈17.93 A. - Statement 3 - Correct.
The fundamental power factor is 0.866.
The fundamental displacement factor is cos30°≈0.866.
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=(2√2×230/π)cos30°≈179.3 V. With R=10 Ω and highly inductive continuous current, Id≈17.93 A. The fundamental displacement factor is cos30°≈0.866.
Why the Other Options Are Wrong
Option A is incorrect because it omits correct statement(s) 3. Option B is incorrect because it omits correct statement(s) 1. Option D 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
Average Output and Input Current is linked with Single-Phase Full Converter, Controlled Rectifiers and Power Electronics. Large load inductance → nearly constant DC current.
A single-phase full converter is connected to 230 V, 50 Hz. The load is R = 10 Ω with a very large inductance. For α = 30°, verify.
1. Average output voltage is approximately 179.3 V.
2. The RMS value of the input current is 17.93 A.
3. The fundamental power factor is 0.866.
Which of the statements are correct?
Answer & step-by-step solution
Firing angle controls average DC voltage; the AC-side current is non-sinusoidal and its fundamental is displaced approximately by α. Vdc=(2√2×230/π)cos30°≈179.3 V. With R=10 Ω and highly inductive continuous current, Id≈17.93 A. 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.
Average output voltage is approximately 179.3 V.
Vdc=(2√2×230/π)cos30°≈179.3 V. - Statement 2 - Correct.
The RMS value of the input current is 17.93 A.
With R=10 Ω and highly inductive continuous current, Id≈17.93 A. - Statement 3 - Correct.
The fundamental power factor is 0.866.
The fundamental displacement factor is cos30°≈0.866.
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=(2√2×230/π)cos30°≈179.3 V. With R=10 Ω and highly inductive continuous current, Id≈17.93 A. The fundamental displacement factor is cos30°≈0.866.
Why the Other Options Are Wrong
Option A is incorrect because it omits correct statement(s) 3. Option B is incorrect because it omits correct statement(s) 1. Option D 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
Average Output and Input Current is linked with Single-Phase Full Converter, Controlled Rectifiers and Power Electronics. Large load inductance → nearly constant DC current.
एक सिंगल-फेज़ फुल कन्वर्टर 230 V, 50 Hz से जुड़ा है। लोड R = 10 Ω के साथ बहुत बड़ा प्रेरकत्व है। α = 30° के लिए सत्यापित कीजिए।
1. औसत आउटपुट वोल्टेज लगभग 179.3 V है।
2. इनपुट धारा का RMS मान 17.93 A है।
3. मूल पावर फैक्टर 0.866 है।
कौन-से कथन सही हैं?
उत्तर और चरणबद्ध हल
C — 1, 2 और 3। Vdc = (2√2×230/π)cos30° ≈ 179.3 V; बड़े प्रेरकत्व से Id ≈ 17.93 A और स्क्वेयर-वेव इनपुट धारा का RMS = Id = 17.93 A; मूल विस्थापन गुणांक = cos30° = 0.866।
विस्तृत हल (मूल भाषा में)
Statement-wise Verification
- Statement 1 - Correct.
Average output voltage is approximately 179.3 V.
Vdc=(2√2×230/π)cos30°≈179.3 V. - Statement 2 - Correct.
The RMS value of the input current is 17.93 A.
With R=10 Ω and highly inductive continuous current, Id≈17.93 A. - Statement 3 - Correct.
The fundamental power factor is 0.866.
The fundamental displacement factor is cos30°≈0.866.
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=(2√2×230/π)cos30°≈179.3 V. With R=10 Ω and highly inductive continuous current, Id≈17.93 A. The fundamental displacement factor is cos30°≈0.866.
Why the Other Options Are Wrong
Option A is incorrect because it omits correct statement(s) 3. Option B is incorrect because it omits correct statement(s) 1. Option D 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
Average Output and Input Current is linked with Single-Phase Full Converter, Controlled Rectifiers and Power Electronics. Large load inductance → nearly constant DC current.
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