GATE 2025 EE Q.63: solution and answer

GATE 2025 · EEQ.63 · 2 marks · NAT

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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Answer and solution

ANSWER

The line current of the capacitor bank is about 9.1 A.

Official answer: 8.5 to 10.0Source: Official answer key, GATE 2025 EE (IIT Roorkee)

Real power of the motor is unchanged by the capacitors: $P = 390 \times 0.77 = 300.3 kW$ .
Initial reactive power: $Q_{1} = 390 sin (cos^{- 1} 0.77) = 390 \times 0.6381 = 248.85 kVAr$ (lagging).
Reactive power needed at 0.85 lag: $Q_{2} = P tan (cos^{- 1} 0.85) = 300.3 \times 0.6197 = 186.1 kVAr$ .
Three-phase capacitor bank rating: $Q_{c} = Q_{1} - Q_{2} = 248.85 - 186.1 \approx 62.7 kVAr$ .
Line current of the bank: $I_{c} = \frac{Q _{c}}{3 V _{L}} = \frac{62.7 \times 1 0 ^{3}}{3 \times 4000} \approx 9.05 A \approx 9.1 A$ , which lies in the key range 8.5 to 10.0.

KEY IDEA

Capacitors change only reactive power, so the bank supplies the difference between the motor's reactive power before and after correction. Then use the three-phase relation $Q_{c} = 3 V_{L} I_{c}$ to get the line current.

WATCH OUT FOR

Using 390 kVA instead of the real power $P$ in the $tan$ formula for the final reactive power. Another is using the phase voltage in $Q_{c} = 3 V_{L} I_{c}$ , or treating the bank as single-phase and dividing by 4 kV alone.

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GATE 2025 EE Q.63: solution and answer

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