AC source of 110 Vrms, 60 Hz supplies an ideal single-phase phase-controlled half-wave rectifier that charges a battery. Assume the battery is modelled using a voltage source of 100 V with a series resistance of 2Ω Choose the correct option. Statement-1: Minimum value of control firing angle is α = 40° (approx.) Statement-2: For firing angle, α = 60°, the battery current start falling from its peak value after 30° which is measured from the angle at which current starts rising.
Answer & step-by-step solution
For a controlled half-wave battery charger, current can flow only after the source instantaneous voltage exceeds the battery emf and the SCR is fired. The conduction interval is bounded by the firing instant and the point where source voltage again falls to the battery voltage; evaluating the average current with the given source and load data gives the value in option A.
Concept and Reasoning
For a controlled half-wave battery charger, current can flow only after the source instantaneous voltage exceeds the battery emf and the SCR is fired. The conduction interval is bounded by the firing instant and the point where source voltage again falls to the battery voltage; evaluating the average current with the given source and load data gives the value in option A.
Calculation / Key Relation
During conduction: i(θ)=[Vm sinθ−E]/R
Vm=√2 Vrms
Exam Takeaway
For a battery load, conduction requires source voltage > battery emf as well as SCR triggering.
Common Mistake
Do not use the resistive-load half-wave average formula without accounting for battery emf.
Quick Trick
For a battery load, conduction requires source voltage > battery emf as well as SCR triggering.
110 Vrms, 60 Hz का AC स्रोत एक आदर्श सिंगल-फेज़ फेज़-नियंत्रित हाफ-वेव रेक्टिफायर को आपूर्ति करता है जो एक बैटरी चार्ज करता है। मान लीजिए बैटरी 2Ω श्रेणी प्रतिरोध के साथ 100 V वोल्टेज स्रोत के रूप में मॉडल की गई है। सही विकल्प चुनिए।
कथन-1: नियंत्रण फायरिंग कोण का न्यूनतम मान α = 40° (लगभग) है।
कथन-2: फायरिंग कोण α = 60° के लिए, बैटरी धारा अपने शिखर मान से उस कोण से 30° बाद गिरना शुरू करती है जिस पर धारा बढ़ना शुरू होती है।
उत्तर और चरणबद्ध हल
A — दोनों कथन सत्य। धारा तभी बहती है जब स्रोत का तात्क्षणिक वोल्टेज बैटरी emf से अधिक हो: sin α_min = 100/(√2×110) ≈ 0.643, α_min ≈ 40°। α = 60° पर धारा (Vm sinωt − E)/R का शिखर ωt = 90° पर आता है, यानी चालन शुरू होने के 30° बाद।
विस्तृत हल (मूल भाषा में)
Concept and Reasoning
For a controlled half-wave battery charger, current can flow only after the source instantaneous voltage exceeds the battery emf and the SCR is fired. The conduction interval is bounded by the firing instant and the point where source voltage again falls to the battery voltage; evaluating the average current with the given source and load data gives the value in option A.
Calculation / Key Relation
During conduction: i(θ)=[Vm sinθ−E]/R
Vm=√2 Vrms
Exam Takeaway
For a battery load, conduction requires source voltage > battery emf as well as SCR triggering.
Common Mistake
Do not use the resistive-load half-wave average formula without accounting for battery emf.
Quick Trick
For a battery load, conduction requires source voltage > battery emf as well as SCR triggering.
110 Vrms, 60 Hz નો AC સ્રોત એક આદર્શ સિંગલ-ફેઝ ફેઝ-નિયંત્રિત હાફ-વેવ રેક્ટિફાયરને પુરવઠો આપે છે જે એક બેટરી ચાર્જ કરે છે. ધારો કે બેટરી 2Ω શ્રેણી અવરોધ સાથે 100 V વોલ્ટેજ સ્રોત તરીકે મોડેલ થયેલી છે. સાચો વિકલ્પ પસંદ કરો.
વિધાન-1: નિયંત્રણ ફાયરિંગ કોણનું લઘુતમ મૂલ્ય α = 40° (આશરે) છે.
વિધાન-2: ફાયરિંગ કોણ α = 60° માટે, બેટરી પ્રવાહ તેના શિખર મૂલ્યથી, પ્રવાહ વધવાનું શરૂ થાય તે કોણથી 30° પછી ઘટવાનું શરૂ કરે છે.
જવાબ અને પગલાવાર ઉકેલ
A — બંને વિધાનો સાચાં. પ્રવાહ ત્યારે જ વહે છે જ્યારે સ્રોતનું તાત્ક્ષણિક વોલ્ટેજ બેટરી emf કરતાં વધુ હોય: sin α_min = 100/(√2×110) ≈ 0.643, α_min ≈ 40°. α = 60° પર પ્રવાહ (Vm sinωt − E)/R નું શિખર ωt = 90° પર આવે છે, એટલે વહન શરૂ થયાના 30° પછી.
વિગતવાર ઉકેલ (મૂળ ભાષામાં)
Concept and Reasoning
For a controlled half-wave battery charger, current can flow only after the source instantaneous voltage exceeds the battery emf and the SCR is fired. The conduction interval is bounded by the firing instant and the point where source voltage again falls to the battery voltage; evaluating the average current with the given source and load data gives the value in option A.
Calculation / Key Relation
During conduction: i(θ)=[Vm sinθ−E]/R
Vm=√2 Vrms
Exam Takeaway
For a battery load, conduction requires source voltage > battery emf as well as SCR triggering.
Common Mistake
Do not use the resistive-load half-wave average formula without accounting for battery emf.
Quick Trick
For a battery load, conduction requires source voltage > battery emf as well as SCR triggering.
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