Calculate power consumption of household devices:
Example: A refrigerator drawing 2A at 120V consumes P = 120V × 2A = 240W
Power calculations for industrial equipment:
Example: A 3-phase motor with line voltage 400V, current 12A, and power factor 0.85 consumes P = √3 × 400V × 12A × 0.85 = 7.06kW
Power calculations for renewable energy systems:
Example: A solar panel with 20V output and 5A current produces P = 20V × 5A = 100W
Power calculations for electronic circuits:
Example: A resistor with 10Ω resistance and 2A current dissipates P = I²R = 2²A × 10Ω = 40W
Electrical power (P) is the rate at which energy is used or produced in a circuit. It can be calculated using three equations depending on which quantities you know:
An electric bulb operates at 220 V and draws a current of 0.5 A. What is the power consumption?
Solution: P = 220 × 0.5 = 110 W
A resistor has a resistance of 10 Ω and a voltage of 20 V across it. What is the power dissipated?
Solution: P = 20² / 10 = 400 / 10 = 40 W
A resistor has a resistance of 8 Ω and a current of 2 A flows through it. What is the power dissipated?
Solution: P = 2² × 8 = 4 × 8 = 32 W
An AC device operates at 230 V, draws 5 A, and has a power factor of 0.85. What is the real power?
Solution: P = 230 × 5 × 0.85 = 977.5 W
In a DC circuit power is simply P = VI. In AC there are three distinct quantities, and conflating them is the source of most confusion about electrical loading.
| Quantity | Symbol | Unit | Meaning |
|---|---|---|---|
| Real power | P | watt (W) | Energy actually converted to work or heat |
| Reactive power | Q | volt-ampere reactive (var) | Energy shuttled to and from magnetic or electric fields |
| Apparent power | S | volt-ampere (VA) | The product of RMS voltage and RMS current |
They relate as a right triangle: S = √(P2 + Q2), and the power factor is P/S. Reactive power does no net work, but it is not fictional — the current carrying it is real, and it heats every conductor it passes through.
Transformers, generators and UPS units are rated in VA or kVA because their limits are thermal and depend on current, not on how much of that current is useful. Motors and heaters are rated in watts because what matters is the work they do. A 5 kVA transformer supplying a load at power factor 0.7 can deliver only 3.5 kW — which is why sizing a supply from the load’s wattage alone leads to under-specified equipment.
P = VI, P = I2R and P = V2/R are all equivalent, derived by substituting Ohm’s law. The squared forms are worth internalising because they explain why small changes matter so much: doubling the current through a resistance quadruples the heat. This is why high-current connections are where thermal problems appear first, and why a slightly loose terminal — adding a fraction of an ohm of contact resistance — can become hot enough to be a fire risk.
Watts measure real power that does work. Volt-amperes measure apparent power, the simple product of RMS voltage and current. Var measures reactive power, which is shuttled back and forth without doing net work.
Because their limit is thermal and depends on current magnitude, which is unaffected by power factor. A 5 kVA transformer delivers only 3.5 kW into a load at power factor 0.7.
P = VI, P = I²R and P = V²/R. All three are equivalent, obtained by substituting Ohm's law, and which to use depends on which quantities are known.
Because dissipation in a resistance follows P = I²R, so power rises with the square of current. This is why high-current joints fail first and why a loose terminal can become dangerously hot.
It does no net work, but the current carrying it is real and heats every conductor it flows through. That is why power factor correction is worthwhile: it keeps reactive current local rather than drawing it through the whole supply.
P = √3 × VLL × IL × PF for a balanced load, where VLL is the line-to-line voltage and IL the line current.