ASVAB · Electricity Basics

How does electrical power relate to voltage and current?

Correct answer

Power = Voltage × Current (P = VI); in a DC circuit or RMS values for AC

  1. A Power = Voltage × Current (P = VI); in a DC circuit or RMS values for AC
  2. B Power = Voltage / Current
  3. C Power = Voltage + Current
  4. D Power is unrelated to voltage and current

Why this is the answer

Electrical power: P = V × I (DC, or AC with RMS values). Units: WATTS (W) = volts × amperes. P = V × I × cos(φ) for AC with phase shift (φ = phase angle between V and I); cos(φ) is the POWER FACTOR. Combined with Ohm's Law (V = IR): P = I²R (current-squared times resistance) and P = V²/R (voltage-squared divided by resistance). Choose the formula based on what's known. Examples: (1) 120V × 10A = 1200W (typical microwave); (2) 12V × 5A = 60W (laptop power supply); (3) 5V × 2A = 10W (phone charger); (4) 240V × 30A = 7200W (electric dryer). Power consumption examples: LED bulb 9-15W; incandescent equivalent 60W (LEDs 4-6x more efficient); laptop 50-100W; refrigerator 100-400W average; washer/dryer 1000-5000W; air conditioner 1000-3500W; electric oven 2000-5000W; electric vehicle charging 6-22 kW for Level 2, 50-350 kW for fast charging. Energy = Power × Time. Common energy unit: kWh (kilowatt-hour) = 1000W × 1hour = 3,600,000 J. Electric bills typically quoted in kWh. Cost = kWh × rate (US average ~$0.16/kWh). A 100W bulb running 10 hours = 1 kWh; runs every day = 365 kWh/year ≈ $58/year. LED equivalent (15W) saves ~$50/year per bulb. RMS values for AC: 120V AC household is RMS — peak is about 170V; 240V AC is RMS, peak 340V. RMS is used because it gives the equivalent DC voltage that delivers the same power; P = V_RMS × I_RMS for resistive loads. POWER FACTOR: for AC loads with inductive or capacitive components (motors, transformers, fluorescent lights), current is OUT OF PHASE with voltage; real power is less than apparent power (VA). Power factor = real power / apparent power. Resistive loads (heaters, incandescent lights) have PF = 1.0. Motors and electronics may have PF 0.5-0.95. Utilities charge industrial customers based on apparent power; power factor correction (capacitors) reduces costs.
Source: ASVAB EI, Power Equations

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