A single−phase motor is connected to 400V, 50Hz supply. The motor draws a current of 31.7A at a power factor 0.7 lag. The additional reactive power (in VAR) to be supplied by the capacitor bank will be ___________
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The motor apparent power is S=VI=400×31.7=12.68 kVA. At pf=0.7, real power is 8.876 kW and reactive power is Q=S√(1−0.7²)≈9.06 kVAR. For full correction to unity power factor, the capacitor must supply approximately 9055 VAR, matching the final key.
Formula / Key Relation
S = VI
= 400 × 31.7
= 12.68 kVA
P = S cosφ
= 12.68 × 0.7
= 8.876 kW
sinφ = √(1−0.7²) ≈ 0.71414
Q = S sinφ
≈ 12.68 × 0.71414
≈ 9.055 kVAR
Qc ≈ 9055 VAR
Detailed Explanation
Concept & ReasoningPower factor 0.7 lag means the motor absorbs inductive reactive power. A shunt capacitor supplies equal leading VARs if the target is unity power factor. Compute either Q=P tanφ or Q=S sinφ. The problem does not state a different target power factor, so the final key implies complete compensation.How to Solve It in the ExamFor unity correction, capacitor kVAR = original inductive kVAR.Important Exam PointState the target power factor; here the final answer assumes unity.Editorial / Answer-Key NoteThe official GPSC final answer key revised the provisional outcome for this question from A to D.Related Revision PathThis question sits…
Concept & Reasoning
Power factor 0.7 lag means the motor absorbs inductive reactive power. A shunt capacitor supplies equal leading VARs if the target is unity power factor. Compute either Q=P tanφ or Q=S sinφ. The problem does not state a different target power factor, so the final key implies complete compensation.
How to Solve It in the Exam
For unity correction, capacitor kVAR = original inductive kVAR.
Important Exam Point
State the target power factor; here the final answer assumes unity.
Editorial / Answer-Key Note
The official GPSC final answer key revised the provisional outcome for this question from A to D.
Related Revision Path
This question sits in the revision path Power System → Power Factor Correction → Capacitor Banks .
Quick Trick
For unity correction, capacitor kVAR = original inductive kVAR.
Common Mistake
Calculating capacitor VAR as P(1−pf) instead of Ptanφ.
Exam Tip
State the target power factor; here the final answer assumes unity.
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