Examples In Electrical Calculations By Admiralty Pdf ✦ [ PREMIUM ]

Maximum allowable drop per core: 1.65 V (two cores in series).

Fault current: (I_{short} = 110 / 0.0856 \approx 1285\ \text{A}). examples in electrical calculations by admiralty pdf

Cable data: 16 mm² copper, length 30 m round trip. Resistance: [ R_{cable} = \rho \times \frac{L}{A} = 0.0175 \times \frac{60}{16} \approx 0.0656\ \Omega ] Maximum allowable drop per core: 1

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}) the prospective fault current matters.

Required correction: (Q_c = Q_1 - Q_2 \approx 3.56\ \text{kVAR}) (capacitive).

Load current: (I = P/V = 3000/110 \approx 27.3\ \text{A}). The fuse was rated 40 A — fine for overload. But for short-circuit, the prospective fault current matters.