Ohm's Law Examples: 10 Real-World Problems

Basic DC Circuit Examples

These first four examples use the fundamental Ohm's Law formula V = IR and power formula P = VI with simple DC circuits. They are the building blocks for all more complex problems.

Example 1: Find Current Through a Resistor

Scenario: A 12V battery is connected to a 220Ω resistor. What current flows through the resistor?

Known: V = 12V, R = 220Ω

Find: I = ?

Solution:

• Apply I = V / R
• I = 12 / 220 = 0.0545 A (54.5 mA)

Power: P = V × I = 12 × 0.0545 = 0.654 W (use a 1W resistor for safety margin)

Application: This is exactly how you calculate the current-limiting resistor for an LED circuit.

Example 2: Find Voltage Across a Load

Scenario: A DC motor draws 4.5A through its 8Ω winding. What voltage is applied?

Known: I = 4.5A, R = 8Ω

Find: V = ?

Solution:

• Apply V = I × R
• V = 4.5 × 8 = 36 V

Power: P = I²R = 4.5² × 8 = 162 W

Application: Used when diagnosing motor windings — if measured resistance gives a different voltage than applied, there's a problem.

Example 3: Find Resistance of a Heating Element

Scenario: A 240V electric water heater draws 18.75A. What is the element resistance?

Known: V = 240V, I = 18.75A

Find: R = ?

Solution:

• Apply R = V / I
• R = 240 / 18.75 = 12.8 Ω

Power: P = V × I = 240 × 18.75 = 4,500 W (4.5 kW)

Application: HVAC technicians measure element resistance to verify it matches nameplate specs. A reading of 0Ω means shorted; ∞Ω means open/burned out.

Example 4: LED Current-Limiting Resistor

Scenario: A white LED requires 20mA at 3.2V forward voltage. The supply is 5V. What series resistor is needed?

Known: V_supply = 5V, V_LED = 3.2V, I = 20mA = 0.020A

Find: R = ?

Solution:

• Voltage across resistor: V_R = 5 − 3.2 = 1.8V
• R = V_R / I = 1.8 / 0.020 = 90 Ω
• Nearest standard value: 100 Ω (use the next higher standard value)

Power in resistor: P = I²R = 0.020² × 100 = 0.04 W (a ⅛W resistor is fine)

Application: Every LED circuit in HVAC control panels needs this calculation.

Series & Parallel Circuit Examples

These examples combine Ohm's Law with series and parallel resistance rules. Understanding these patterns is essential for troubleshooting real circuits.

Example 5: Series Resistors — Voltage Divider

Scenario: Three resistors in series (100Ω, 220Ω, 330Ω) are connected to a 24V supply. Find the total current and voltage drop across each resistor.

Known: R1 = 100Ω, R2 = 220Ω, R3 = 330Ω, V = 24V

Solution:

• Step 1: Total resistance: R_total = 100 + 220 + 330 = 650 Ω
• Step 2: Total current: I = V / R_total = 24 / 650 = 0.0369 A (36.9 mA)
• Step 3: Voltage across each (V = IR):
  • V1 = 0.0369 × 100 = 3.69V
  • V2 = 0.0369 × 220 = 8.12V
  • V3 = 0.0369 × 330 = 12.18V
• Verification: 3.69 + 8.12 + 12.18 = 23.99V ≈ 24V ✓

Application: Series voltage dividers are used in HVAC thermostat circuits and sensor signal conditioning.

Example 6: Parallel Resistors — Current Divider

Scenario: Two heating elements (48Ω and 72Ω) are connected in parallel across a 240V supply. Find branch currents and total current.

Known: R1 = 48Ω, R2 = 72Ω, V = 240V

Solution:

• Step 1: Branch currents:
  • I1 = V / R1 = 240 / 48 = 5.0 A
  • I2 = V / R2 = 240 / 72 = 3.33 A
• Step 2: Total current: I_total = 5.0 + 3.33 = 8.33 A
• Step 3: Total resistance: R_total = V / I_total = 240 / 8.33 = 28.8 Ω
• Verification (parallel formula): 1/R_total = 1/48 + 1/72 = 0.02083 + 0.01389 = 0.03472 → R_total = 28.8Ω ✓

Power: P_total = V × I_total = 240 × 8.33 = 2,000 W

Application: Multiple heating stages in HVAC air handlers are wired in parallel to provide variable heat output.

AC Circuit Examples

AC circuits introduce impedance (Z) and power factor (cos φ). These examples use single-phase and three-phase AC formulas.

Example 7: Single-Phase AC Motor

Scenario: A 230V single-phase HVAC blower motor has a nameplate rating of 8.5A and 0.82 power factor. Find real power, apparent power, and impedance.

Known: V = 230V, I = 8.5A, cos φ = 0.82

Solution:

• Real power: P = V × I × cos φ = 230 × 8.5 × 0.82 = 1,603 W (1.6 kW)
• Apparent power: S = V × I = 230 × 8.5 = 1,955 VA
• Reactive power: Q = √(S² − P²) = √(1955² − 1603²) = √(3,822,025 − 2,569,609) = 1,119 VAR
• Impedance: Z = V / I = 230 / 8.5 = 27.06 Ω
• Winding resistance (from nameplate): R = Z × cos φ = 27.06 × 0.82 = 22.19 Ω

Application: This is how you verify a blower motor is operating within spec. Measure current — if it exceeds 8.5A significantly, the bearings may be failing.

Example 8: Three-Phase Compressor Motor

Scenario: A 460V three-phase scroll compressor draws 22A per line with a power factor of 0.88. Find real power and efficiency if output is 12.5 HP.

Known: V_LL = 460V, I_L = 22A, cos φ = 0.88, output = 12.5 HP

Solution:

• Real power input: P = √3 × V_LL × I_L × cos φ = 1.732 × 460 × 22 × 0.88 = 15,553 W (15.55 kW)
• Output power: 12.5 HP × 746 W/HP = 9,325 W (9.33 kW)
• Efficiency: η = P_out / P_in = 9,325 / 15,553 = 0.599 (59.9%)

Application: Low efficiency suggests the compressor may be oversized, the system may have high head pressure, or the motor windings may be deteriorating. Typical scroll compressor efficiency is 65–80%.

Power Calculation Examples

These examples focus on power-related calculations common in HVAC equipment sizing and energy analysis.

Example 9: HVAC Electric Furnace Heating Stages

Scenario: A 240V electric furnace has three heating stages: Stage 1 = 5kW, Stage 2 = 5kW, Stage 3 = 10kW. Find resistance, current, and total power for each stage and all stages combined.

Known: V = 240V, P1 = 5000W, P2 = 5000W, P3 = 10,000W

Solution for each stage:

Wire sizing: At 83.33A total, use 4 AWG copper wire (rated 85A in 75°C column per NEC Table 310.16) with a 100A breaker.

Application: This is the standard calculation for sizing furnace heating elements and verifying wire/breaker sizes for residential electric furnaces.

Example 10: Voltage Drop in HVAC Control Wiring

Scenario: A 24V thermostat signal wire runs 75 feet from the furnace control board to the thermostat. The wire is 18 AWG (6.385 Ω/1000 ft). The thermostat relay draws 0.5A. Calculate the voltage drop and verify the signal is adequate.

Known: V_source = 24V, L = 75 ft (one way), I = 0.5A, wire resistance = 6.385 Ω/1000 ft

Solution:

• Step 1: Total wire resistance (round trip): R = 2 × 75 × 6.385 / 1000 = 0.958 Ω
• Step 2: Voltage drop: V_drop = I × R = 0.5 × 0.958 = 0.479 V
• Step 3: Voltage at thermostat: V_load = 24 − 0.479 = 23.52 V
• Step 4: Percentage drop: 0.479 / 24 × 100 = 2.0%

Result: The 2.0% voltage drop is well within the 5% maximum recommended by NEC for branch circuits. The thermostat will operate reliably.

When to worry: If the wire run exceeds 200 ft, or if using 22 AWG wire, voltage drop can exceed 5% and cause thermostat malfunctions. Use our Voltage Drop Calculator to check specific scenarios.

Summary of Formulas Used

Common Mistakes in Ohm's Law Problems

• Forgetting to convert units: mA must be converted to A (divide by 1000), kΩ to Ω (multiply by 1000), kW to W (multiply by 1000) before using formulas.
• Using line voltage in three-phase formulas: In three-phase circuits, P = √3 × V_LL × I uses line-to-line voltage, not phase voltage. V_phase = V_LL / √3.
• Ignoring wire resistance: In low-voltage circuits (24V control, 12V DC), wire resistance can cause significant voltage drop. Always check with the Voltage Drop Calculator.
• Confusing resistance and impedance: In AC circuits, the measured resistance (with an ohmmeter) is less than the impedance. A motor with 5Ω resistance may have 20Ω impedance due to inductive reactance.
• Not accounting for power factor: P = VI only works for DC or purely resistive AC loads. For motors and inductive loads, you must include cos φ.

Standards Reference

• IEC 60071 — Insulation Co-ordination (voltage levels and clearances)
• IEEE 141 — Recommended Practice for Electric Power Distribution for Industrial Plants
• NEC (NFPA 70) — National Electrical Code, Article 210 (branch circuits), Article 310 (conductors)
• NFPA 70E — Electrical Safety in the Workplace
• IEC 60364 — Low-Voltage Electrical Installations

Frequently Asked Questions

What are some real-world Ohm's Law examples?

Real-world Ohm's Law examples include: calculating current through a resistor (I=V/R), sizing a heating element (R=V²/P), finding voltage drop in wire (V=IR), determining motor current draw (I=P/V), and calculating LED resistor values (R=(V_supply−V_LED)/I_LED). These apply V=IR and P=VI to everyday electrical problems.

How do you solve series circuit problems with Ohm's Law?

In a series circuit, total resistance is the sum of individual resistances: R_total = R1 + R2 + R3. The same current flows through all components. Apply Ohm's Law to find the total current (I = V/R_total), then find individual voltage drops (V_n = I × R_n). The sum of all voltage drops equals the source voltage.

How do you solve parallel circuit problems with Ohm's Law?

In a parallel circuit, total resistance is found by: 1/R_total = 1/R1 + 1/R2 + 1/R3. The same voltage appears across all branches. Total current is the sum of branch currents (I_total = I1 + I2 + I3). Each branch current is found by I_n = V/R_n. Parallel resistance is always less than the smallest individual resistance.

How is Ohm's Law used in HVAC systems?

HVAC technicians use Ohm's Law to: diagnose compressor windings (measure R, calculate expected I), size heating elements (R=V²/P), calculate voltage drop in control wiring, verify contactor coil operation, and select proper wire sizes. A 240V/5kW heater needs R=11.52Ω and draws I=20.8A.

What is the difference between Ohm's Law in DC and AC examples?

DC examples use V=IR and P=VI directly. AC examples must account for impedance (Z) instead of resistance (R): V=IZ, and real power includes power factor: P=VI×cos φ. Three-phase AC adds a √3 factor: P=√3×V_LL×I×cos φ. DC problems are simpler; AC problems require understanding of reactance and phase angle.