Amps to Volts Calculator

Ohm's LawIEC 60364
Amps to Volts Calculator
Convert current (amps) to voltage (volts) using power or resistance
A

Current in amperes (A)

W

Power in watts (W)

Frequently Asked Questions

Common questions about this calculator

You need additional information—either resistance or power. Using Ohm's Law: Volts = Amps × Ohms (V = I × R). Using power: Volts = Watts / Amps (V = P / I). Example: 5A through 10Ω resistance = 5 × 10 = 50V. Or: 100W at 5A = 100 / 5 = 20V.

Amps and volts measure different electrical properties—current and electrical potential. They are related through Ohm's Law (V = I × R) or power equation (P = V × I). Like speed and distance, you need additional information (time for speed/distance, resistance or power for amps/volts) to convert between them.

Ohm's Law states V = I × R (Voltage = Current × Resistance). Rearranged: I = V/R and R = V/I. If a circuit has 2A flowing through 100Ω, voltage is 2 × 100 = 200V. This applies to DC and purely resistive AC circuits. For reactive AC loads, impedance replaces resistance.

Use the power formula: V = W / A (Volts = Watts / Amps). For a 1500W heater drawing 12.5A: Voltage = 1500 / 12.5 = 120V. This works for DC and resistive AC loads. For AC with power factor, use: V = W / (A × PF). Our calculator handles both scenarios.

Ohm's Law defines: V = I × R. Higher resistance with same current means higher voltage. Lower resistance with same voltage means higher current. A 240V circuit pushing current through 12Ω has: I = 240/12 = 20A. These relationships are fundamental to all electrical calculations.

Voltage drop = Current × Wire Resistance. Wire resistance depends on material, length, and gauge: R = ρL/A. For 50ft of 12 AWG copper carrying 15A: R ≈ 0.16Ω, voltage drop = 15 × 0.16 = 2.4V. NEC limits voltage drop to 3-5% for efficient operation. Our voltage drop calculator automates this.

Learn More

Converting current to voltage through Ohm's Law (V=I×RV = I \times R) enables calculation of voltage drop across circuit elements, essential for power supply design, conductor sizing, and instrumentation circuits. This fundamental relationship determines terminal voltages under load, verifies power supply adequacy, and predicts conductor losses in distribution systems. Understanding voltage drop prevents equipment malfunction, motor torque reduction, and electronic device failure. Proper voltage calculations ensure system reliability and compliance with NEC voltage drop limits for branch circuits and feeders.

Ohm's Law and Resistance Fundamentals: Ohm's Law establishes V=I×RV = I \times R for resistive circuits, where voltage drop equals current times resistance. A conductor carrying 10A through 5Ω develops 50V drop, representing I²R power dissipation as heat. Conductor resistance depends on material resistivity, length, cross-sectional area, and temperature: R = ρ × L / A. Copper exhibits 0.01724 Ω·mm²/m resistivity at 20°C; aluminum 0.02826 Ω·mm²/m. Temperature increases resistance approximately 0.4% per °C, requiring correction factors for elevated operating temperatures in thermal calculations.

Voltage Drop in Conductor Sizing: NEC 210.19(A) recommends limiting voltage drop to 3% for branch circuits, 2% for feeders, and 5% total from service to load. A 100m run of 10mm² copper carrying 50A produces 8.6V drop (3.7% at 230V nominal), approaching maximum limits. Motor circuits demand special attention—terminal voltage affects starting torque proportionally to V², so 10% voltage drop reduces starting torque 19%. Long cable runs or high currents require larger conductors to maintain acceptable voltage levels and prevent equipment performance degradation.

Power-Based Voltage Calculation: When power is known but resistance varies with operating conditions, voltage calculation uses V=P/IV = P / I for DC and resistive AC loads. AC systems with reactive components require power factor consideration: P=V×I×PFP = V \times I \times \text{PF}. This method determines supply voltage requirements from specified load power and current, particularly useful for unknown impedances. Current-output sensors (4-20 mA) develop voltage across precision burden resistors for signal conditioning, with 250Ω producing 1-5V range for analog-to-digital conversion.

Voltage Division and Measurement Circuits: Series-connected resistors distribute voltage proportionally: V₁ = VtotalV_{\text{total}} × (R₁ / RtotalR_{\text{total}}). This principle underlies voltage divider networks in analog circuits, sensor interfaces, and precision references. Potentiometers use variable voltage division for controls. Long signal cable runs require Kelvin (4-wire) connections to eliminate lead resistance errors in precision measurements. Proper burden resistor sizing balances signal-to-noise ratio against power consumption and self-heating effects that degrade measurement accuracy.

Three-Phase and Neutral Conductor Considerations: Balanced three-phase loads theoretically produce zero neutral current as phase currents sum vectorially to zero. Unbalanced loading or harmonic currents create neutral current potentially exceeding phase currents. Triplen harmonics (3rd, 9th, 15th orders) from non-linear loads add arithmetically in neutral conductors rather than canceling. Modern facilities with significant switching power supply loading often require neutral conductors sized equal to or larger than phase conductors, contrary to traditional reduced neutral sizing practices.

Standards Reference: NEC 210.19(A) establishes voltage drop limits: 3% branch circuits, 2% feeders, 5% total maximum. NEC Article 310 covers conductor sizing and ampacity tables. NEC 430 addresses motor circuit voltage considerations and starting torque effects. IEC 60364 provides international installation standards for voltage drop and conductor sizing. IEEE standards govern instrumentation accuracy and measurement system design practices for industrial applications.

Extension Cord Voltage Drop - Portable Power Tool

Calculate voltage drop across extension cord for portable power tool application

1
Current: 15 A
2
Resistance: 0.52 Ω
3
System Voltage: 120 V

Result

Voltage Drop:
7
8 V (15A × 0.52Ω). Tool receiving 112.2V at end of cord (120V - 7.8V = 112.2V). Voltage drop = 6.5% (7.8V ÷ 120V).
Per NEC 210.19(A)(1), branch circuits should limit voltage drop to 3% maximum, 5% for feeders and branch circuits combined. 6.5% drop exceeds recommendations.

Additional Notes

Per Ohm's Law, V=I×RV = I \times R relates voltage, current, and resistance. Applications: calculate voltage drop across components, design voltage dividers, size current-sense resistors. Power dissipation: P = I² × R. For AC circuits, consider impedance Z (includes reactance). NEC voltage drop limits: 3% feeders, 5% total.

Solar Array String Voltage - Rooftop PV System

Calculate string voltage for solar panel array from current and power measurements

1
Current: 9.2 A
2
Power: 4,830 W
3
System Type: DC

Result

String Voltage:
525 V

Calculations

  • String voltage: 4,830 W / 9.2 A = 525 V
  • Per panel: 525 V / 15 panels = 35 V per panel

Verification

  • Matches expected Vmp (voltage at maximum power) for 72-cell panels
  • String is properly connected in series

Additional Notes

Per NEC 210.19, voltage drop should not exceed 3% for branch circuits. Calculate: Vdrop=I×RwireV_{\text{drop}} = I \times R_{\text{wire}}. Wire resistance varies with size, material, and temperature. Size conductors to maintain voltage within equipment operating range. Long runs or high currents require larger wire. Consider aluminum conductors for cost savings (60% more resistance).

Industrial Busbar Voltage Drop - Switchgear Distribution

Calculate voltage drop across industrial busbar for switchgear distribution system

1
Current: 400 A
2
Resistance: 0.0021 Ω
3
System Voltage: 480 V

Result

Busbar Voltage Drop:
0.84 V per phase (400A × 0.0021Ω = 0.84V). For three-phase system, line-to-line voltage drop = 0.84V. Source voltage 480V, load voltage = 479.16V. Voltage drop = 0.175% (0.84V ÷ 480V). Well within acceptable limits.

Resistance Calculation

Busbar resistance: R=ρ×L/AR = \rho \times L / A. Where: ρ=0.0175Ωmm2/m\rho = 0.0175 \Omega \cdot \text{mm}^2/\text{m} (copper at 20°C), L = 50m length × 2 (round trip for single-phase. Per-phase for 3-phase), A = 100mm × 10mm = 1,000 mm². R = 0.0175 × 50 / 1,000 = 0.000875Ω per conductor. For AC circuits, add 15% for skin effect and proximity effect: RAC=0.000875×1.15=0.001ΩR_{\text{AC}} = 0.000875 \times 1.15 = 0.001\Omega per phase. Two-way resistance (out and back for voltage drop): 0.001 × 2 = 0.002Ω (using 0.0021Ω including temperature rise to 40°C).

Temperature Effects

Copper resistance increases with temperature: Rhot=Rcold×(1+α×ΔT)R_{\text{hot}} = R_{\text{cold}} \times (1 + \alpha \times \Delta T), where α\alpha = 0.00393/°C for copper. At 20°C: R = 0.002Ω. At 60°C (busbar under load): R = 0.002 × (1 + 0.00393 × 40°C) = 0.00231Ω. Voltage drop increases proportionally: 400A × 0.00231Ω = 0.924V.

Power Loss in Busbar

P = I² × R = 400² × 0.0021 = 336W per phase. Three-phase total: 1,008W (1kW) heat dissipation. Busbar temperature rise: ΔT=P/(h×Asurface)\Delta T = P / (h \times A_{\text{surface}}). For natural convection, h ≈ 10 W/m²·K. Surface area = 2 × (100mm × 50m) = 10m². ΔT = 1,000W / (10 × 10) = 10°C rise above ambient. At 30°C ambient, busbar reaches 40°C (acceptable, <70°C limit per IEC 61439-1).

Per NEC 408.30 & IEC 61439-1

  • Voltage drop limits: 2% for feeders, 3% for branch circuits, 5% total
  • Busbar temperature rise: <65K for enclosed equipment, <70K for open equipment
  • Short-circuit withstand: Verify busbar can handle fault current—for 50kA fault, I²t calculation required
  • Support spacing: 100×10mm busbar needs support every 1.

Motor Starting Considerations

400A inrush during motor starting represents ~6× full load current (FLC67 A\text{FLC} \approx 67 \text{ A} continuous). Starting voltage dip affects other equipment: 0.84V drop during start is minimal. Per IEC 60034-1, motors rated for \pm10% voltage variation (432-528V). At 479V, all motors operate normally. For voltage-sensitive equipment, calculate steady-state voltage drop at full continuous load (67A): Drop = 67A × 0.0021Ω = 0.14V (0.03%)—negligible.

Additional Notes

For instrumentation: minimize voltage drop in sensor supply circuits (<0.5%). Use twisted-pair shielded cable to reduce noise pickup. Kelvin (4-wire) connection eliminates lead resistance errors in precision measurements. Current-sense resistors: select low-value (<0.1Ω) with precision tolerance (\pm0.1%) and low TCR (<50ppm/°C) for accurate current measurement.