GATE 2025 EE – Question 63
Using shunt capacitors, the power factor of a 3-phase, 4 kV induction motor (drawing 390 kVA at 0.77 pf lag) is to be corrected to 0.85 pf lag. The line current of the capacitor bank, in A, is ___________ (round off to one decimal place).
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Correct answer: 9.05 to 9.15
Explanation
$P=390\times0.77=300.3$ kW and $Q_1=390\times0.638=248.8$ kVAr. At 0.85 pf, $Q_2=P\tan(\cos^{-1}0.85)=300.3\times0.6197=186.1$ kVAr. The capacitors supply $Q_c=248.8-186.1=62.7$ kVAr, so the line current is $I=\dfrac{Q_c}{\sqrt3V_L}=\dfrac{62700}{\sqrt3\times4000}=9.1$ A.