When calculating voltage drop please make sure that you are referring to the correct parameters. Don't get confused between voltage drop and voltage difference. Consider a busbar at the source at Voltage Vs and the voltage at the load as Vl. The voltage drop across a line impedance is deltaV and is equal to a vector quantity of the product of current and line impedance. The source voltage is then equal to the load voltage plus the voltage drop across the line added vectorially. The voltage difference between the source and the load is equal to the difference of the moduli |Vs| - |Vl|. This is not necessarily equal to the voltage drop across the line.**The most practical approximate formula is**: dV = k. [R.cos (fi) + X.sen (fi)]. I. L

where R and X (ohms / km) of the cable and L is the length (km); I in (A).

K = sqrt (3) for 3 phases or K = 2 for single-phase with 2 conductors.

Usually this approach is enough, but the exact value of the voltage drop can be obtained if you calculate the voltage in the two points with a load flow study and make the difference of the modules of the two voltage values |V1| at the entrance and |V2| on output.**Note that the phasor equation** V1-V2 = Z. I which is the direct OHM law for AC, where Z = R + jX does not provide the value that usually interests of dV, since the difference that matters is usually the difference of modules

dV = | V1 | - | V2 | and it is not | V1-V2 | = | Z.I | that this equation provides.

A voltmeter connected between V1 and V2 would read | V1-V2 | = | Z.I | but the voltage drop that usually matters is the difference: measurement of the voltmeter at the entrance | V1 | and measurement of the output | V2 |, ie dV = | V1 | - | V2 |.**About concepts you might think as follows**:

- Voltage drop can be understood as the voltage difference in one point of the circuit (e.g., a busbar) but for two different load situations.

The fact is that in the situation without load the output voltage equals the input (source or V thevenin), thereby calculating the voltage drop on the output from the condition without load and the situation under load corresponds to calculate the module of the difference between | the input voltage | and | output voltage with load | where Z=R+jX is the impedance between the input (source) and the output (busbar).

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