एक पारेषण लाइन Z = (2 + j8) Ω का 0.8 लैगिंग लोड पर वोल्टेज नियमन 10% है। यदि लोड 0.707 लीडिंग हो, तो V.R. ________ होगा (मान लें दोनों स्थितियों में धारा समान है)
उत्तर और चरणबद्ध हल
A — −6.63%। लघु लाइन में V.R. ∝ I(R cosφ ± X sinφ); लैगिंग: 2×0.8 + 8×0.6 = 6.4 → 10%। लीडिंग: 2×0.707 − 8×0.707 = −4.24। अतः V.R. = 10 × (−4.24/6.4) ≈ −6.63%।
विस्तृत हल (मूल भाषा में)
Concept & Reasoning
With R=2 Ω and X=8 Ω, the lagging 0.8 power factor has sinφ=0.6. The proportional drop factor is 2(0.8)+8(0.6)=6.4. For 0.707 leading, cosφ≈sinφ≈0.707 and the factor is 2(0.707)−8(0.707)=−4.242. Holding current and receiving voltage constant, regulation scales by −4.242/6.4, so 10% becomes about −6.63%.
How to Solve It in the Exam
Leading pf flips the sign of the X sinφ term.
Important Exam Point
Use the given 10% case as a ratio; no absolute line voltage is required.
Quick Trick
Leading pf flips the sign of the X sinφ term.
Common Mistake
Forgetting the negative reactive-drop term at leading power factor.
A transmission line Z = (2 + j8) Ω has 10% of the voltage regulation with the lagging load of 0.8. If the load is 0.707 leading, then the V.R. is ________ (Assume the current is same in both cases)
Answer & step-by-step solution
For a short-line approximation, voltage regulation is proportional to I(R cosφ + X sinφ) for lagging current, while the reactive term changes sign for leading current. Using the given 10% lagging case to eliminate the common I/V factor gives a negative regulation for 0.707 leading, about −6.63%.
Concept & Reasoning
With R=2 Ω and X=8 Ω, the lagging 0.8 power factor has sinφ=0.6. The proportional drop factor is 2(0.8)+8(0.6)=6.4. For 0.707 leading, cosφ≈sinφ≈0.707 and the factor is 2(0.707)−8(0.707)=−4.242. Holding current and receiving voltage constant, regulation scales by −4.242/6.4, so 10% becomes about −6.63%.
How to Solve It in the Exam
Leading pf flips the sign of the X sinφ term.
Important Exam Point
Use the given 10% case as a ratio; no absolute line voltage is required.
Quick Trick
Leading pf flips the sign of the X sinφ term.
Common Mistake
Forgetting the negative reactive-drop term at leading power factor.
એક પારેષણ લાઇન Z = (2 + j8) Ω નું 0.8 લેગિંગ લોડે વોલ્ટેજ નિયમન 10% છે. જો લોડ 0.707 લીડિંગ હોય, તો V.R. ________ થશે (ધારો કે બંને સ્થિતિમાં પ્રવાહ સમાન છે)
જવાબ અને પગલાવાર ઉકેલ
A — −6.63%. ટૂંકી લાઇનમાં V.R. ∝ I(R cosφ ± X sinφ); લેગિંગ: 2×0.8 + 8×0.6 = 6.4 → 10%. લીડિંગ: 2×0.707 − 8×0.707 = −4.24. તેથી V.R. = 10 × (−4.24/6.4) ≈ −6.63%.
વિગતવાર ઉકેલ (મૂળ ભાષામાં)
Concept & Reasoning
With R=2 Ω and X=8 Ω, the lagging 0.8 power factor has sinφ=0.6. The proportional drop factor is 2(0.8)+8(0.6)=6.4. For 0.707 leading, cosφ≈sinφ≈0.707 and the factor is 2(0.707)−8(0.707)=−4.242. Holding current and receiving voltage constant, regulation scales by −4.242/6.4, so 10% becomes about −6.63%.
How to Solve It in the Exam
Leading pf flips the sign of the X sinφ term.
Important Exam Point
Use the given 10% case as a ratio; no absolute line voltage is required.
Quick Trick
Leading pf flips the sign of the X sinφ term.
Common Mistake
Forgetting the negative reactive-drop term at leading power factor.
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