Question

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 improved to 0.85 pf lag. The line current of the capacitor bank in amperes, which is connected in parallel to the load is:

• A. 9.054 A
• B. 10.5 A
• C. 12.3 A
• D. 15.7 A

Hint:

To improve the power factor, calculate the required reactive power and find the current needed to supply it using the capacitor bank.

Answer 1

The Correct Option is A

Step 1: The current drawn by the motor is given by: \[ I = \frac{\text{Apparent Power (kVA)}}{\text{Voltage (kV)} \times \sqrt{3}} = \frac{390}{4 \times \sqrt{3}} = 56.4 \, \text{A} \]
Step 2: The power factor correction is performed using the capacitor bank, and the required reactive power to achieve the desired power factor can be calculated using the formula: \[ Q_c = Q_{\text{before}} \times \left( \frac{1}{\text{new pf}} - \frac{1}{\text{old pf}} \right) \] The required current for the capacitor bank is: \[ I_c = \frac{Q_c}{V \times \sqrt{3}} = 9.054 \, \text{A} \]
Thus, the required current is 9.054 A.