Current and voltage drop: formulas and examples
Calculate resistive voltage drop from the complete circuit resistance and load current. The result is a circuit estimate, not a cable-size or protection approval. Start with the supply voltage, actual load and conductor data at the relevant temperature.
Reviewed:
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Ohm's law: voltage, current and resistance
V = I × R; I = V / R; R = V / I; P = V × I
For a resistive element, 12 V across 6 Ω gives 2 A and 24 W. Choose two known positive quantities to find the third. This is a DC or purely resistive circuit calculation; it does not select cable sizes, protection or account for AC impedance.
Calculator — Ohm's law: voltage, current and resistance — I = V / R
Calculator — Ohm's law: voltage, current and resistance — V = I × R
Calculator — Ohm's law: voltage, current and resistance — R = V / I
Voltage drop and percentage
ΔV = I × Rloop; drop [%] = 100 × ΔV / Vsupply
Rloop includes both outgoing and return conductors plus relevant connections. For I = 10 A and Rloop = 0.10 Ω, the drop is 1.0 V: 8.33% at 12 V or 0.435% at 230 V. Do not confuse one-way cable resistance with the full loop. A drop greater than the supply indicates inconsistent assumptions for a fixed-current resistive model, not a valid negative load voltage.
Current from power
DC/resistive: I = P / V; single-phase: I = P / (V × PF); balanced three-phase: I = P / (√3 × Vline × PF)
Use watts, not kilowatts, in these equations. For an illustrative resistive load 2300 W / 230 V = 10 A. For AC use the actual power factor; for balanced three-phase use line-to-line voltage. If P is shaft output rather than electrical input, include efficiency in the denominator. Starting current and harmonics need separate treatment.
Calculator — Current from power
Cable data and model limits
If the manufacturer gives resistance per metre, multiply by the appropriate total conductor length. Resistance changes with conductor area, material and temperature. This simple resistive method excludes AC reactance and does not check current-carrying capacity, fault protection, installation conditions or locally permitted voltage drop. Verify those separately with the product documentation and an applicable electrical design method.
How many amps is 5 kW?
I1 = P / (V × PF × η); I3 = P / (√3 × Vline × PF × η)
For 5 kW electrical input, PF = 1 and efficiency input η = 1, the single-phase current at 230 V is 21.74 A; balanced three-phase at 400 V gives 7.22 A. Use η below 1 only when the power is mechanical output. Specify PF and the correct voltage; this does not select cable or protection.
Calculator — How many amps is 5 kW?
How many amps is 10 kW?
I1 = P / (V × PF × η); I3 = P / (√3 × Vline × PF × η)
For 10 kW electrical input, PF = 1 and efficiency input η = 1, the single-phase current at 230 V is 43.48 A; balanced three-phase at 400 V gives 14.43 A. Use η below 1 only when the power is mechanical output. Specify PF and the correct voltage; this does not select cable or protection.
Calculator — How many amps is 10 kW?
Wire cross-section from resistive voltage drop
Smin [mm²] = 2 × ρ [Ω·mm²/m] × Lone-way [m] × I [A] / ΔVallowed [V]
This two-conductor resistive estimate solves the loop voltage-drop equation for cross-section. With illustrative ρ = 0.0237, one-way length 20 m, current 10 A and allowed drop 6.9 V, Smin = 1.374 mm². The Schneider source uses 0.0237 Ω·mm²/m for copper in its stated operating conditions; enter product resistance at the appropriate temperature. This lower bound does not check ampacity, fault protection, installation derating or AC reactance and does not approve a cable size. The allowed drop is your input, not a universal limit.
Calculator — Wire cross-section from resistive voltage drop
Electricity use and your unit price
E [kWh] = P [W] / 1000 × hours/day × days; cost = E × price/kWh
A 1000 W appliance used 4 hours/day for 30 days consumes 120 kWh. At your example price 0.25 per kWh, the variable charge is 30 currency units. Enter your own tariff: the example is not a current electricity price. Fixed fees, taxes and time-of-use rates must be added separately; actual cycling changes operating time.
Calculator — Electricity use and your unit price
Voltage drop from one-way length and Ω/km
Rloop = 2 × length [m] × resistance [Ω/km] / 1000; ΔV = I × Rloop
For a two-conductor resistive circuit, 16 A, 25 m one-way and 7.41 Ω/km give 5.928 V or 2.577% at 230 V. Enter the product resistance at operating temperature. This model excludes AC reactance and does not approve three-phase design, ampacity or protection.
Calculator — Voltage drop from one-way length and Ω/km
Worked examples — illustrative inputs
| I [A] | Rloop [Ω] | ΔV [V] | 12 V [%] | 230 V [%] |
|---|---|---|---|---|
| 10 | 0.10 | 1.00 | 8.33 | 0.435 |
| 5 | 0.10 | 0.50 | 4.17 | 0.217 |