Calculate current, voltage, area, resistivity or conductor length from the other four values using I = V×A÷(ρ×L) in any units.
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Current Through a Conductor Formula
The calculator is based on Ohm’s law and the resistance of a uniform conductor. The conductor resistance is:
R = \rho*L/A
Substituting this into I = V/R gives the main current formula:
I = (V*A)/(\rho*L)
The calculator can also rearrange the same relationship to solve for any one missing value:
V = (I*\rho*L)/A
A = (I*\rho*L)/V
\rho = (V*A)/(I*L)
L = (V*A)/(I*\rho)
- I = current through the conductor, in amperes A
- V = voltage across the conductor, in volts V
- A = cross-sectional area of the conductor, in square meters m²
- ρ = resistivity of the conductor material, in ohm-meters Ω·m
- L = conductor length, in meters m
- R = electrical resistance, in ohms Ω
To calculate current, enter voltage, area, resistivity, and length. A higher voltage or larger cross-sectional area increases current. A longer conductor or a material with higher resistivity lowers current.
To calculate voltage, the formula finds the voltage needed to drive a known current through a conductor with the given material, length, and area.
To calculate cross-sectional area, the formula finds the area needed to carry a known current at a given voltage for the selected conductor length and resistivity.
To calculate resistivity, the formula solves for the material property using measured voltage, current, length, and area.
To calculate length, the formula finds the conductor length that matches the entered voltage, current, area, and resistivity.
Common Resistivity Values for Conductors
Resistivity depends on material and temperature. The values below are typical approximate values near 20°C.
| Material | Typical resistivity, Ω·m | Conductivity note |
|---|---|---|
| Silver | 1.59 × 10-8 | Very high conductivity |
| Copper | 1.68 × 10-8 | Common wire conductor |
| Aluminum | 2.82 × 10-8 | Common lightweight conductor |
| Tungsten | 5.60 × 10-8 | Higher resistance than copper |
| Nichrome | 1.10 × 10-6 | Often used as a resistance wire |
Unit Conversions Used in the Calculation
| Quantity | Unit | Base-unit conversion |
|---|---|---|
| Voltage | mV | 1 mV = 0.001 V |
| Voltage | kV | 1 kV = 1000 V |
| Area | ft² | 1 ft² = 0.092903 m² |
| Area | yd² | 1 yd² = 0.836127 m² |
| Length | ft | 1 ft = 0.3048 m |
| Length | yd | 1 yd = 0.9144 m |
| Current | mA | 1 mA = 0.001 A |
| Current | kA | 1 kA = 1000 A |
Example
Example 1: Calculate current
A copper conductor has a voltage of 12 V across it, a cross-sectional area of 0.000001 m², a resistivity of 1.68 × 10-8 Ω·m, and a length of 10 m.
I = (V*A)/(\rho*L)
I = (12*0.000001)/(0.0000000168*10) = 71.428571 A
The current is about 71.43 A.
Example 2: Calculate required voltage
A conductor carries 5 A, has a resistivity of 2.82 × 10-8 Ω·m, a length of 25 m, and a cross-sectional area of 0.000002 m².
V = (I*\rho*L)/A
V = (5*0.0000000282*25)/0.000002 = 1.7625 V
The required voltage is 1.7625 V.
FAQ
Why does current decrease when the conductor is longer?
A longer conductor has more resistance because charge has to travel through more material. Since resistance is proportional to length, increasing length reduces current when voltage, area, and material stay the same.
Why does a larger cross-sectional area increase current?
A larger cross-sectional area gives charge more space to move through the conductor. This lowers resistance. With the same voltage and material, lower resistance means higher current.
Can this calculation be used for any wire?
It works best for a uniform conductor with constant cross-sectional area and known resistivity. Real wires can heat up, and resistivity changes with temperature. For high-current or safety-critical electrical work, temperature rating, insulation, installation conditions, and electrical code limits also matter.
