Required correction: (Q_c = Q_1 - Q_2 \approx 3.56\ \text{kVAR}) (capacitive).
Fault current: (I_{short} = 110 / 0.0856 \approx 1285\ \text{A}). examples in electrical calculations by admiralty pdf
Then cable cross-section area (A): [ A = \frac{\rho \times L}{R} = \frac{0.0175 \times 45}{0.0194} \approx 40.6\ \text{mm}^2 ] Required correction: (Q_c = Q_1 - Q_2 \approx 3
Using the formula: [ R = \frac{V_{drop}}{I} = \frac{1.65}{85} \approx 0.0194\ \Omega ] examples in electrical calculations by admiralty pdf
For PF=0.90, new apparent power (S_2 = P / 0.90 = 5.2 / 0.90 \approx 5.78\ \text{kVA}) New reactive power (Q_2 = \sqrt{5.78^2 - 5.2^2} \approx 2.52\ \text{kVAR})