Watt, Volt, Amp Calculator

Ohm's LawPower Law
Calculator Input
Enter any two values to calculate the third
W

Power in watts

V

Voltage in volts

A

Current in amperes

Ω

Resistance in ohms

Frequently Asked Questions

Common questions about this calculator

Power (Watts) = Voltage (Volts) × Current (Amps). For DC: P = V × I. For AC: P = V × I × PF. Knowing any two values lets you calculate the third. This is the fundamental power equation in electrical engineering.

Watts = Volts × Amps. Volts = Watts / Amps. Amps = Watts / Volts. For AC, include power factor: W = V × A × PF. These relationships apply to all electrical systems.

In AC circuits: Real Power (W) = Apparent Power (VA) × PF. When PF<1, current (amps) is higher than needed for the real power. At PF=0.8, a 1000W load draws 1250VA or 10.4A at 120V instead of 8.3A at PF=1.

Real Power² + Reactive Power² = Apparent Power². In symbols: W² + VAR² = VA². Power factor = W/VA = cos(θ) where θ is phase angle between voltage and current. This helps visualize the relationship between all power types.

DC: P = V × I (always). AC: P = V × I × PF (for single-phase) or P = √3 × V × I × PF (three-phase). DC has no reactive power, so power factor concept does not apply. AC calculations must account for phase relationships.

P = √3 × V_line × I_line × PF. For 480V, 50A, PF=0.85: P = 1.732 × 480 × 50 × 0.85 = 35,341W. Use line-to-line voltage and line current. The √3 factor accounts for three-phase power relationships.

Learn More

The relationship between electrical power (watts), voltage (volts), and current (amperes) forms the foundation of electrical engineering analysis and system design. These three fundamental quantities, governed by Ohm's Law and the definition of electrical power, enable engineers to analyze circuits, size components, troubleshoot systems, and design reliable electrical installations. Understanding how power, voltage, and current interact allows calculation of any unknown quantity when two are measured or specified, supporting diverse applications from residential circuit design to industrial power distribution and renewable energy systems throughout the electrical engineering profession.

Basic Power Relationships in DC and AC Circuits: The fundamental power equation P=V×IP = V \times I applies universally to DC circuits and resistive AC loads, defining electrical power as the product of voltage and current. This simple equation enables immediate calculation: a 120-volt circuit carrying 10 amperes dissipates 1,200 watts. These calculations extend throughout electrical engineering—sizing conductors, selecting circuit breakers, specifying power supplies, and verifying equipment nameplates. The relationship holds constant across voltage levels: 1,000 watts at 120V draws 8.33A, while the same power at 240V draws only 4.17A, demonstrating the current-reducing advantage of higher voltages.

Power Factor and Reactive Power Considerations: Alternating current systems introduce power factor considerations that distinguish real power (watts) from apparent power (volt-amperes). Real power performs useful work—rotating motors, heating elements, illuminating lamps. Reactive power magnetizes inductive loads and charges capacitive components, oscillating between source and load without performing net work but consuming current capacity. The power factor, ranging from 0 to 1, quantifies the phase relationship between voltage and current. Inductive loads exhibit lagging power factor typical of motors and transformers, while capacitive loads show leading power factor.

Three-Phase Power Systems and Efficiency: Three-phase power systems dominate commercial and industrial applications due to superior efficiency and power density. The mathematical relationship incorporates the 3\sqrt{3} factor (1.732): P = 3\sqrt{3} × V × I × PF, where voltage represents line-to-line measurement and current represents line current. This geometry, arising from 120-degree phase displacement between three phases, delivers more power per conductor than equivalent single-phase systems. A 100 kW three-phase load at 480V with 0.90 PF draws 134A per conductor, compared to 417A for equivalent 240V single-phase system—68% current reduction enabling smaller conductors and transformers.

Equipment Sizing and System Design Applications: Equipment sizing requires careful analysis of power-voltage-current relationships to prevent overload while avoiding oversizing waste. Circuit breakers protect conductors from overcurrent damage, sized at 125% of continuous load current per NEC requirements. Transformers match source voltage to utilization voltage, sized in kVA (apparent power) rather than kW to account for reactive loads. Generators must supply both real and reactive power, specified in kVA with power factor rating. Understanding apparent versus real power prevents undersizing disasters and costly oversizing throughout electrical system design.

Voltage Drop Analysis and Power Quality: Voltage drop analysis critically depends on current-voltage-power relationships throughout electrical distribution systems. Conductor resistance generates voltage drop proportional to current according to Ohm's Law. For constant-power loads including switched-mode power supplies, voltage drop causes proportional current increase, partially offsetting voltage reduction effects. This behavior affects voltage drop predictions and conductor sizing, particularly for modern facilities with predominantly electronic loading. Excessive voltage drop reduces motor torque proportional to voltage squared, dims lighting, and may cause electronic equipment malfunction requiring careful design consideration.

Standards Reference: NEC Article 210 specifies circuit design requirements including conductor sizing and overcurrent protection based on voltage-current relationships. IEEE 141 (Red Book) provides comprehensive guidance on industrial and commercial power systems design. IEC 60364 establishes international standards for electrical installations. NFPA 70E addresses electrical safety in the workplace including proper calculation and verification procedures.

Electric Heater - Resistance Calculation

Calculate current and resistance from power and voltage for electric heater

1
Power: 1,500 W
2
Voltage: 120 V

Result

Current:
**12
5 A** (1500W / 120V). Resistance: 9.6 Ω (120V ÷ 12.5A or 120² ÷ 1500). Heat output: 5,118 BTU/hr.
Wiring: 12 AWG minimum, 20A breaker required. Actual measured resistance: Cold element reads 20-30% lower (positive temperature coefficient).

Additional Notes

Power relationships: P=V2/R=I2R=V×IP = V^2/R = I^2 R = V \times I. Heating elements: Nichrome wire (80% Ni, 20% Cr), resistance increases with temperature. Safety: Never exceed 80% breaker rating continuous (1500W ÷ 120V ÷ 0.8 = 15.6A, needs 20A circuit). Energy cost: 0.18 USD/hour at 0.12 USD/kWh. Efficiency: 100% conversion to heat, but expensive vs natural gas (1.20 USD/therm = 0.04 USD/kWh equivalent). Sizing: 10W per sq ft for supplemental heat, 20W for primary.

Motor Power Triangle. Apparent Power

Calculate motor current and apparent power from real power and voltage

1
Power: 11,190 W (15 HP × 746 W/HP)
2
Voltage: 480 V (three-phase)
3
Power Factor: 0.85

Result

Current (3-phase):
**15
8 A** (11,190W/(3×480V×0.85)=15.8A11,190W / (\sqrt{3} \times 480V \times 0.85) = 15.8A). Apparent power: 13,165 VA or 13.2 kVA (11,190W ÷ 0.85). Reactive power: 6,950 VAR (13,165211,1902\sqrt{13,165^2 - 11,190^2}). VFD sizing: 20 HP minimum. Wire: 10 AWG (30A ampacity with 75°C terminations per NEC 310.16).

Additional Notes

Three-phase power: P=3×VL×I×PFP = \sqrt{3} \times V_{L} \times I \times PF where VLV_{L} is line-to-line voltage. Power triangle: Real (kW), Reactive (kVAR), Apparent (kVA) form right triangle. Power factor correction: Add 6.95 kVAR capacitor to achieve PF 0.95, reduces current to 14.1A (10% savings). VFD benefits: Soft start, speed control, energy savings 20-50% for variable loads. Motor efficiency: Premium efficiency (IE3) 93-95%, standard 88-91%. Full load amps (FLA): Nameplate specifies actual, NEC Table 430.250 for design. Overload protection: 115-125% FLA per NEC 430.32.