Ohm's Law + Power Formula for Trades Reference
Why this matters
Ohm's Law + power formula are the foundation of electrical work. Every multimeter reading, every breaker sizing, every motor diagnosis traces back to these relationships. The tech who's fluent in them diagnoses faster + sizes equipment correctly. This is the field card.
The four basic quantities
Voltage (V OR E): electrical pressure (volts)
- Pushes current through circuit
- Measured between two points
- Single-phase residential: 120V OR 240V
Current (I): electrical flow rate (amperes / amps)
- Quantity of electrons moving
- Measured in series (clamp meter most common)
- Determines wire size + breaker
Resistance (R): opposition to current (ohms / Ω)
- All conductors have some
- Heating elements + motor windings have specific resistance
- Measured with circuit de-energized
Power (P): rate of energy use (watts / W)
- Volts × amps for resistive loads
- Determines equipment capacity + utility bill
- Often kilowatts (kW = 1,000W)
Ohm's Law
E = I × R
Voltage equals current times resistance.
Rearranged:
- I = E ÷ R (current = voltage divided by resistance)
- R = E ÷ I (resistance = voltage divided by current)
Memorize one form; derive others.
Practical Ohm's Law examples
Heater element:
- Element resistance: 12Ω
- Supply voltage: 120V
- Current: I = 120 ÷ 12 = 10A
Verify reading:
- Element should draw 10A
- Reading higher = shorted (less resistance)
- Reading lower = open (more resistance)
- Reading zero = open circuit
Motor winding:
- Spec'd resistance: 8Ω
- Measured: 8.2Ω
- Within tolerance; winding OK
Spec'd: 8Ω; measured: 0.1Ω:
- Short circuit between windings
- Motor failure
- Replace
Spec'd: 8Ω; measured: infinite:
- Open winding
- Motor failure
- Replace
Power formula
P = E × I (for DC + resistive AC)
Power equals voltage times current.
For motors + reactive loads:
- P = E × I × PF (power factor)
- PF typically 0.7-0.95 for motors
- Reactive power separate calculation
- Most service calculation: E × I sufficient estimate
Rearranged:
- I = P ÷ E (current = power divided by voltage)
- E = P ÷ I (rarely used)
Practical power examples
Refrigerator:
- Nameplate 700W running
- Voltage: 120V
- Current: I = 700 ÷ 120 = ~5.8A
Verify amp draw:
- Clamp meter on cord
- Should read ~5.8A running
- Higher = motor straining (failing)
Water heater:
- 4500W element
- Voltage: 240V
- Current: I = 4500 ÷ 240 = 18.75A
Breaker size:
- 18.75A continuous load × 125% (NEC) = 23.4A
- 30A breaker minimum
- Code-compliant
AC condenser:
- Nameplate 3500W
- Voltage: 240V
- Current: I = 3500 ÷ 240 = ~14.6A
Combining Ohm's Law + Power
P = E² ÷ R OR P = I² × R
Useful for:
- Heater element wattage from resistance
- Heat generated in wire (I²R losses)
- Voltage drop calculations
Example: I²R losses:
- 50A current through wire
- 0.05Ω wire resistance (long run small gauge)
- Power lost as heat: 50² × 0.05 = 125W
- Significant; wire warms + voltage drops
Series circuit math
Resistors in series:
- Total resistance = sum
- R_total = R1 + R2 + R3
Current:
- Same through all (series circuit)
Voltage:
- Divides across each resistor proportionally
Example: Series christmas lights:
- 50 bulbs, each 2.4V
- Total = 120V
- One out = all out (open circuit)
Parallel circuit math
Resistors in parallel:
- Total resistance = 1 / (1/R1 + 1/R2 + 1/R3...)
- Always LESS than smallest resistor
Voltage:
- Same across all (parallel)
Current:
- Divides through branches inverse to resistance
- Lower resistance branch = more current
Example: Household circuit:
- All outlets parallel
- Each device sees 120V
- Each draws different current
Practical parallel example
Bedroom circuit:
- Lamp: 100W = 0.83A
- Phone charger: 5W = 0.04A
- TV: 80W = 0.67A
- Heater: 1500W = 12.5A
Total current: 0.83 + 0.04 + 0.67 + 12.5 = ~14A
On 15A circuit: near max; one more device could trip. On 20A circuit: comfortable margin.
AC vs DC
Ohm's Law applies to both:
- DC: straightforward
- AC: same for resistive loads (heaters)
- AC: more complex for motors / transformers (reactance)
For practical residential service:
- Resistive: heaters, incandescent, electric stove elements
- Inductive: motors, transformers, fluorescent ballasts
- Capacitive: certain loads, capacitor banks
Reactance + impedance
For inductive loads (motors):
- Inductive reactance (X_L) at AC frequency
- Adds to resistance
- Impedance (Z) = total opposition
- Z = √(R² + X_L²)
Practical implication:
- Motor draws more current than R alone suggests
- Power factor < 1
- Account in sizing
For typical residential: most service tech uses simple V × I × PF approximation.
Voltage drop
Common diagnostic + design concern:
Long wire run:
- Wire has resistance per foot
- Voltage drops along length
- At end of run, less voltage than source
Calculation:
- V_drop = I × R_wire
- R_wire = ρ × L ÷ A (resistivity × length / cross-section)
Practical rule:
- 3% maximum voltage drop typical
- 5% maximum branch + feeder combined
Example:
- 240V circuit, 50A load
- 100 ft run, 8 AWG copper
- 8 AWG: ~0.78Ω per 1000 ft
- Round trip 200 ft: 0.156Ω
- V_drop = 50 × 0.156 = 7.8V
- 7.8 ÷ 240 = 3.25% (just over limit)
- Consider larger wire
Wire ampacity
NEC tables:
- 14 AWG copper: 15A (60°C column for branch)
- 12 AWG: 20A
- 10 AWG: 30A
- 8 AWG: 40A (60°C); 50A (75°C)
- 6 AWG: 55A (60°C); 65A (75°C)
- 4 AWG: 70A (60°C); 85A (75°C)
Sizing logic:
- Calculate load
- Apply 125% for continuous
- Match wire to current
- Match breaker to wire
Heat dissipation
Power dissipated in wire = I² × R
- Heats wire
- Reduces ampacity if cluster
- Sometimes derate per NEC
Loose / corroded connection:
- Higher resistance
- More heat
- Eventually arc
- Fire risk
Tight connections matter.
Calculating amp draw for diagnosis
Compare measured to expected:
For element:
- Nameplate watts ÷ voltage = expected amps
- Measure with clamp meter
- Within ±10% = OK
- Higher = short (replace)
- Lower = open partial (replace)
- Zero = open complete (replace)
For motor:
- Nameplate FLA (Full Load Amps)
- Measure under load
- Higher than FLA = motor straining
- Lower = light load (sometimes OK)
Three-phase calculations
For commercial three-phase:
- P = E × I × 1.732 × PF
- 1.732 = √3 (square root of 3)
- Three-phase relationship
Example: 480V three-phase motor:
- 10 HP motor
- ~7,460W (746W per HP)
- Current per leg: I = 7460 ÷ (480 × 1.732 × 0.85) ≈ 10.5A
- Nameplate confirms
Power factor
For motors:
- Real power (kW) - useful work
- Apparent power (kVA) - what utility supplies
- Power factor = kW ÷ kVA
Lower PF:
- More current for same useful work
- Inefficient
- Commercial customers pay penalty
References
- NEC (National Electrical Code) ampacity tables
- IEEE electrical engineering references
- Standard electrical theory textbooks
- Manuall internal: Reading Wiring Schematics, Common Motor Diagnosis