an induction motor draws 50 KW at a 0.60 lagging power factor from a 480 v, 60 Hz source. What parallel capacitor will increase the overall power factor to 0.90 lagging?

To solve this problem, we need to follow these steps:

1. Calculate the apparent power (S) and reactive power (Q) before the capacitor is added.

The apparent power S (in kVA) is given by the formula: S = P/PF, where P is the real power (in kW) and PF is the power factor.

So, S = 50 kW / 0.60 = 83.33 kVA.

The reactive power Q (in kVAR) is given by the formula: Q = sqrt(S^2 - P^2), where S is the apparent power (in kVA) and P is the real power (in kW).

So, Q = sqrt((83.33 kVA)^2 - (50 kW)^2) = 66.67 kVAR.

2. Calculate the reactive power (Q_new) after the capacitor is added.

The new apparent power S_new (in kVA) is given by the formula: S_new = P/PF_new, where P is the real power (in kW) and PF_new is the new power factor.

So, S_new = 50 kW / 0.90 = 55.56 kVA.

The new reactive power Q_new (in kVAR) is given by the formula: Q_new = sqrt(S_new^2 - P^2), where S_new is the new apparent power (in kVA) and P is the real power (in kW).

So, Q_new = sqrt((55.56 kVA)^2 - (50 kW)^2) = 25 kVAR.

3. Calculate the reactive power (Q_c) provided by the capacitor.

The reactive power Q_c (in kVAR) provided by the capacitor is given by the formula: Q_c = Q - Q_new, where Q is the original reactive power (in kVAR) and Q_new is the new reactive power (in kVAR).

So, Q_c = 66.67 kVAR - 25 kVAR = 41.67 kVAR.

4. Calculate the capacitance (C) of the capacitor.

The capacitance C (in F) is given by the formula: C = Q_c / (2 * pi * f * V^2), where Q_c is the reactive power provided by the capacitor (in kVAR), f is the frequency (in Hz), and V is the voltage (in V).

So, C = 41.67 kVAR / (2 * pi * 60 Hz * (480 V)^2) = 0.0000023 F or 2.3 µF.

Therefore, a parallel capacitor of 2.3 µF will increase the overall power factor to 0.90 lagging.