Understanding Voltage
Drop
Voltage drop occurs when the voltage at the end of a cable is lower than at the beginning due to the
resistance in the wire. It is an essential factor in electrical installations, ensuring the efficient
transfer of power and minimizing energy losses.
Voltage Drop Formula
The voltage drop (\( V_{\text{drop}} \)) is calculated using the formula:
\[
V_{\text{drop}} = I \times R \times L \times k
\]
Where:
- \( V_{\text{drop}} \): Voltage drop (in volts)
- \( I \): Current (in amperes)
- \( R \): Wire resistance per meter (\(\Omega/\text{m}\))
- \( L \): Total wire length (round trip, in meters)
- \( k \): Current type factor:
- \( k = 1 \) for DC systems
- \( k = 2 \) for AC single-phase systems
- \( k = 1.732 \) for AC three-phase systems
Theoretical Insight
- Material Resistivity: Conductors like copper and aluminum have specific resistivity
values that affect the voltage drop.
- Wire Cross-Section: A thicker wire (larger cross-sectional area) reduces resistance
and minimizes voltage drop.
- Wire Length: Longer wires result in higher resistance, leading to greater voltage
drop.
Example Calculations
Example 1: DC System
Input:
- Material: Copper
- Current (\( I \)): 15 A
- Wire diameter (\( d \)): 2.5 mm
- Wire length (\( L \)): 40 m (round trip: \( 2 \times 40 = 80 \, \text{m} \))
- Voltage: 120 V
Calculation:
Resistivity of copper:
\[
\rho = 1.68 \times 10^{-8} \, \Omega \cdot \text{m}
\]
Cross-sectional area:
\[
A = \frac{\pi d^2}{4} = \frac{\pi (2.5 \times 10^{-3})^2}{4} = 4.91 \times 10^{-6} \, \text{m}^2
\]
Resistance per meter:
\[
R = \frac{\rho}{A} = \frac{1.68 \times 10^{-8}}{4.91 \times 10^{-6}} = 3.42 \times 10^{-3} \,
\Omega/\text{m}
\]
Voltage drop:
\[
V_{\text{drop}} = I \times R \times L \times k = 15 \times 3.42 \times 10^{-3} \times 80 \times 1 = 4.10
\, \text{V}
\]
Output:
- Voltage Drop: \( 4.10 \, \text{V} \)
- Percentage Voltage Drop: \( \frac{4.10}{120} \times 100 = 3.42\% \)
Example 2: AC Single-Phase System
Input:
- Material: Aluminum
- Current (\( I \)): 20 A
- Wire diameter (\( d \)): 3 mm
- Wire length (\( L \)): 50 m (round trip: \( 2 \times 50 = 100 \, \text{m} \))
- Voltage: 240 V
Calculation:
Resistivity of aluminum:
\[
\rho = 2.82 \times 10^{-8} \, \Omega \cdot \text{m}
\]
Cross-sectional area:
\[
A = \frac{\pi d^2}{4} = \frac{\pi (3 \times 10^{-3})^2}{4} = 7.07 \times 10^{-6} \, \text{m}^2
\]
Resistance per meter:
\[
R = \frac{\rho}{A} = \frac{2.82 \times 10^{-8}}{7.07 \times 10^{-6}} = 3.99 \times 10^{-3} \,
\Omega/\text{m}
\]
Voltage drop:
\[
V_{\text{drop}} = I \times R \times L \times k = 20 \times 3.99 \times 10^{-3} \times 100 \times 2 =
15.96 \, \text{V}
\]
Output:
- Voltage Drop: \( 15.96 \, \text{V} \)
- Percentage Voltage Drop: \( \frac{15.96}{240} \times 100 = 6.65\% \)
Example 3: AC Three-Phase System
Input:
- Material: Copper
- Current (\( I \)): 50 A
- Wire diameter (\( d \)): 5 mm
- Wire length (\( L \)): 75 m (round trip: \( 2 \times 75 = 150 \, \text{m} \))
- Voltage: 415 V
Calculation:
Resistivity of copper:
\[
\rho = 1.68 \times 10^{-8} \, \Omega \cdot \text{m}
\]
Cross-sectional area:
\[
A = \frac{\pi d^2}{4} = \frac{\pi (5 \times 10^{-3})^2}{4} = 19.63 \times 10^{-6} \, \text{m}^2
\]
Resistance per meter:
\[
R = \frac{\rho}{A} = \frac{1.68 \times 10^{-8}}{19.63 \times 10^{-6}} = 0.86 \times 10^{-3} \,
\Omega/\text{m}
\]
Voltage drop:
\[
V_{\text{drop}} = I \times R \times L \times k = 50 \times 0.86 \times 10^{-3} \times 150 \times 1.732 =
11.18 \, \text{V}
\]
Output:
- Voltage Drop: \( 11.18 \, \text{V} \)
- Percentage Voltage Drop: \( \frac{11.18}{415} \times 100 = 2.69\% \)
