The formula
- Ohm's law
V = IR- Solved for current or resistance
I = V / R and R = V / I- Electrical power dissipated
P = VI = I²R = V² / R
What the symbols mean
| Symbol | Meaning | Unit |
|---|---|---|
V | Potential difference across the component | V |
I | Current through the component | A |
R | Resistance of the component | Ω |
P | Power dissipated as heat | W |
When it applies
- Ohmic components, such as resistors and metal wires held at a steady temperature.
- V is the potential difference across that one component, not the supply voltage of the whole circuit.
- Nonohmic parts, including diodes, transistors and filament lamps as they heat up, do not follow it.
Worked example
Problem. A 9.0 V supply is connected across a 330 Ω resistor. What current flows, and how much power does the resistor dissipate?
- Rearrange for current: I = V / R = 9.0 V / 330 Ω.
- I = 0.02727 A, which is 27 mA to two significant figures.
- Power: P = I²R = (0.02727 A)²(330 Ω).
- (0.02727)² = 7.44 × 10⁻⁴, so P = (7.44 × 10⁻⁴)(330) = 0.25 W. A quarter-watt resistor would be right at its limit here.
Answer. 27 mA flows, and the resistor dissipates about 0.25 W.
Common mistakes
- Using the battery voltage instead of the voltage across the individual component in a series circuit.
- Substituting milliamps and kilohms without converting to amps and ohms first.
- Applying it to a filament lamp, whose resistance climbs sharply as it heats.
- Treating resistance as a property of the circuit rather than of the component.
Related formulas
- Coulomb's law:
F = k·|q₁q₂| / r² - Kirchhoff's laws:
ΣI_in = ΣI_out - Faraday's law of induction:
ε = -N·ΔΦ / Δt