Voltage drop sin phi

With small cables at 50Hz, the resistance dominates over the inductive reactance, so the voltage drop due to current flowing along the cable is mostly attributable to the resistance, and the inductive element can be ignored in calculations. As cables get larger, the resistance decreases in inverse proportion to CSA but the inductance does not, so the inductive reactance increasingly plays a part in determining the voltage drop.

Therefore, for larger cables (e.g. 25mm² and up) there are separate figures tabulated for the resistive (r) and reactive (x) components of the voltage drop and the result of the overall impedance (z). The example shows the voltage drop being proportional to the length L, the load current Ib, and the sum of the resistive (r) and reactive (x) components - note the suffixes - multiplied by the cosine and sine of the load phase angle respectively which accounts for the 90° between their contributions to the voltage drop.

Why does the load phase angle play a part? The voltage drop is the effective change in supply voltage that the load sees, but the drop in the cable isn't necessarily in phase with the supply voltage. It occurs as a result of the load current, which is displaced from the supply voltage by the load phase angle Φ. Therefore the magnitude of the drop cannot just be algebraically subtracted from the magnitude of the supply voltage if Φ is significant i.e. if the pf is much below unity.

To summarise:
A load of power factor cos(Φ) draws a current displaced in phase from the supply voltage by Φ.
That current creates a voltage drop in the cable which can be resolved into resistive and reactive components.
For cables over around 16mm², the reactive component is a large enough fraction of the total to deserve consideration.
Tabulated drop figures (r) and (x) are multiplied by cos(Φ) and sin(Φ) respectively and added algebraically to give the total drop.