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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)

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Answer & Explanation
Correct AnswerA. − 6.63%

Quick Explanation

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%.

Formula / Key Relation

Lagging factor = Rcosφ + Xsinφ
= 2(0.8)+8(0.6)=6.4
Leading factor = Rcosφ − Xsinφ
= 2(0.707)−8(0.707)=−4.242
VRlead = 10% × (−4.242/6.4)
≈ −6.63%

Detailed Explanation

Concept & ReasoningWith 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 ExamLeading pf flips the sign of the X sinφ term.Important Exam PointUse the given 10% case as a ratio; no absolute line voltage is required.Related Revision PathThis question sits in the revision path Power System → Transmission Lines → Voltage Regulation .

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