Voltage drop formula

Voltage drop calculation is one of the necessary step to tell whether the electrical system is designed properly.

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Three voltage drop formulas are presented in this article:

  • Voltage drop formula in circuit with resistive elements
  • Voltage drop formula in circuit with resistive and reactive components
  • Voltage drop formula using per-unit system

Voltage drop formula in circuit with resistive elements

Voltage drop occurs according to the Ohm’s law in AC and DC electrical system. Voltage drop is caused only by a resistance in AC systems where reactance is negligible (usually at low voltage) and in DC systems. As a current flows through a resistance, a voltage drop occurs:

\begin{equation} U_d = I \cdot R \end{equation}

where:
\(V_d\) - voltage drop, V
\(I\) - current, A
\(R\) - resistance, \(\Omega\)

Voltage drop formula in circuit with resistive and reactive components

Voltage drop in AC electrical system with reactive elements should be calculated taking into account impedance, which includes a real component (resistance) and and imaginary component (reactance):

\begin{equation} U_d = I \cdot Z \end{equation}

where:
\(Z\) - impedance, \(\Omega\)

The following example shows voltage drop calculation for three phase medium voltage cable, which can be presented as a resistance and inductance element shown below:

scheme1

The voltage drop formula used: \begin{equation} U_d = \sqrt{3} \cdot l \cdot I \cdot (R' \cdot cos(\varphi) + 2 \cdot \pi \cdot f \cdot L' \cdot sin(\varphi)) \end{equation} \begin{equation} U_d = \sqrt{3} \cdot 2 \cdot 254 \cdot (0,165 \cdot 0,95 + 2 \cdot \pi \cdot 50 \cdot 0,00038 \cdot 0,31) = 170,48 V \end{equation}

where:
\(l\) - cable length, km
\(R'\) - unitary resistance, \(\Omega/km\)
\(L'\) - unitary inductance, H/km
\(f\) - frequency, Hz
\(cos(\varphi\)) - power factor.

Voltage drop formula using per-unit system

This type of calculation is valuable, when there is a need to specify a voltage drop for a transformer. The voltage drop accross transformer occurs due due to impedance of transformer (where reactance is substantially higher than resistance) and due to turns ratio. In order to better understand this calculation formula it is worth to analyze the below example.

scheme1

The base value of apparent power, which is chosen for the calculation purposes is 10 MVA (any value can be chosen) and it applies to the whole analyzed system. The next arbitrary base value is Bus 2 voltage, which equals to 6000 V (the same value as the nominal voltage of generator). The transformer turns ratio should be used in order to get base voltage of Bus 2: \begin{equation} U_{baseBus2} = U_{baseBus1} \cdot \dfrac{15000}{6600} = 6000 \cdot \dfrac{15000}{6600} = 13 636,36 V \end{equation}

The transformer reactance is equal to 9,0 %. In order to convert it for the proper base, we need to make following calculation: \begin{equation} X_{transformerpu} = 0,09 \cdot (\dfrac{6000}{6600})^2 \cdot \dfrac{10}{12} = 0,62 p.u. \end{equation} The base impedance: \begin{equation} Z_{base} = \dfrac{(U_{base})^2}{S_{base}} = \dfrac{(13,636)^2}{10} = 18,59 \Omega \end{equation} The reactance of the line in per-unit: \begin{equation} X_{linepu} = \dfrac{0,005}{Z_{base}} = \dfrac{(13,636)^2}{10} = 2,69e-4 p.u. \end{equation} The Bus 1 voltage in p.u.: \begin{equation} U_{Bus1pu} = \dfrac{6000}{6600} = 0,91 p.u. \end{equation} The base current can be calculated: \begin{equation} I_{base} = \dfrac{S_{base}}{\sqrt{3} \cdot U_{base}} = \dfrac{10}{\sqrt{3} \cdot 6,6} = 874,77 A \end{equation} The per unit current is calculated: \begin{equation} I_{pu} = \dfrac{300}{I_{base}} = 0,343 p.u. \end{equation} The next step is to calculate a voltage drop: \begin{equation} U_{Bus3pu} = U_{Bus1pu} - I(X_{transformerpu}+X_{linepu}) = 0,91 - 0,343(0,62+2,69e-4) = 0,697 p.u. \end{equation} In order to get value expressed in volts: \begin{equation} U_{Bus3} = 0,697 p.u. \cdot 13 636,36 V = 9 504,54 V \end{equation}