Every Power & Power Factor question asked in GPSC DEE Electrical, Supervisor Instructor — solve, check instantly, then read the formula and exam tip.
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View answer & solution
B — Watt. Active power is the real, useful power consumed (P = VI cosφ) and its SI unit is the watt (W).
View answer & solutionA — The product of RMS voltage and RMS current. S = V_rms × I_rms, measured in VA.
View answer & solutionA — 80% of the apparent power. Power factor = P/S, so P = 0.8 S.
View answer & solutionFor a fixed real-power load and supply voltage, poor power factor requires higher current. Raising power factor with shunt capacitors or an over-excited synchronous machine reduces kVA demand, line current and I²R losses and generally improves voltage regulation. Key relation: Single phase: I=P/(V·PF). Three phase: I=P/(√3 V_L·PF).
Exam focus: For the same kW, current is inversely proportional to power factor. Common trap: Capacitors supply leading reactive power; they do not supply lagging VARs.
View answer & solutionIts magnitude |S| is apparent power in VA, while P is measured in W and Q in var. S=P/PF=1000/0.8=1250 VA. Q=√(S²−P²)=750 var. Key relation: S=P+jQ; |S|=√(P²+Q²); PF=P/|S|=cosφ.
Exam focus: Draw the power triangle before choosing a power-factor relation. Common trap: Do not confuse P/Q with power factor.
View answer & solutionC — Volt-ampere reactive (VAR). Reactive power Q = VI sinφ is measured in VAR.
