Ohm's Law for Electricians: Formulas, Examples + Free Calculator

Ohm's Law is the foundation of electrical calculations. It states that voltage equals current times resistance: V = I × R. This single relationship unlocks twelve formulas that let you solve for any unknown in a circuit—voltage, current, resistance, or power—when you know any two values. If you're studying for the Texas journeyman electrician exam or troubleshooting circuits on the job, mastering Ohm's Law means you can answer nearly every calculation question with confidence.

Quick tool: Use our free Ohm's Law Calculator to solve any V, I, R, P combination instantly.

The Four Core Formulas

Ohm's Law is built on two foundation equations and two power equations. Master these four, and you can derive the rest:

Basic Resistance Formula

V = I × R    (Voltage = Current × Resistance)
I = V ÷ R    (Current = Voltage ÷ Resistance)
R = V ÷ I    (Resistance = Voltage ÷ Current)

Power Formulas

P = V × I    (Power = Voltage × Current)
P = I² × R   (Power = Current squared × Resistance)
P = V² ÷ R   (Power = Voltage squared ÷ Resistance)

What each variable means:

• V = Voltage (measured in volts, V)
• I = Current (measured in amps, A)
• R = Resistance (measured in ohms, Ω)
• P = Power (measured in watts, W)

These four formulas are all you need. The rest are rearrangements of the same relationships. Write them down now. Memorize them before test day.

The Ohm's Law Wheel

Electricians use a visual tool called the Ohm's Law wheel or power wheel to quickly find any formula. Imagine a circle divided into twelve sections:

• The top half contains the four basic formulas (V = IR, I = V/R, R = V/I, P = VI)
• The bottom half adds the power variants (P = I²R, P = V²/R, and derived forms)

Here's how it works: pick the variable you're solving for, cover it with your finger, and follow the rule. For example:

• Looking for current (I) when you know V and R? Find the I section, cover it, and the remaining formula is I = V ÷ R.
• Looking for power (P) when you know V and I? Find the P section, cover it, and you get P = V × I.

You don't need to memorize the entire wheel—just the four core formulas above. The rest follow logically.

Worked Examples with NEC References

Example 1: Find Current for a Resistive Load

Problem: A 120V baseboard heater has a resistance of 8 ohms. How many amps does it draw?

Given:

• V = 120V
• R = 8Ω
• I = ?

Solution:

I = V ÷ R
I = 120 ÷ 8
I = 15 amps

The heater draws 15 amps. You would protect this circuit with a 20-amp breaker (the next standard size up, per NEC Article 210.21).

Example 2: Find Resistance and Power

Problem: A 240V circuit has a measured current of 20 amps. What is the resistance, and how much power is the load consuming?

Given:

• V = 240V
• I = 20A
• R = ?
• P = ?

Solution:

R = V ÷ I
R = 240 ÷ 20
R = 12 ohms

P = V × I
P = 240 × 20
P = 4,800 watts

The resistance is 12 ohms and the load draws 4,800 watts (or 4.8 kW).

Example 3: Find Voltage from Power and Current

Problem: An electric furnace is rated 1,800 watts and draws 15 amps. What is the supply voltage?

Given:

• P = 1,800W
• I = 15A
• V = ?

Solution:

P = V × I
1,800 = V × 15
V = 1,800 ÷ 15
V = 120V

The furnace is supplied at 120 volts.

Example 4: Exam-Style Question with NEC Compliance

Problem: A commercial kitchen is being equipped with a 4,800-watt baseboard heater rated for 240V. The electrician needs to determine the circuit current and the minimum breaker size. This is a continuous load (defined as operating at 80% or more capacity for 3+ hours). What breaker size is required?

Given:

• P = 4,800W
• V = 240V
• I = ?
• Continuous load factor per NEC Article 215.3

Solution (Step 1: Find current):

P = V × I
4,800 = 240 × I
I = 4,800 ÷ 240
I = 20 amps

Solution (Step 2: Account for continuous load):

Per NEC Article 215.3, a continuous load must be multiplied by 1.25 when selecting breaker and conductor sizes:

Breaker size = 20 × 1.25 = 25 amps

Per NEC Article 310.16, you would select a conductor based on the continuous current (20A) but protect it with a 25-amp breaker. The circuit requires a 25-amp breaker and appropriately sized copper conductor (typically 10 AWG for 25A at 60°C or 8 AWG for 25A at 75°C, depending on the installation and wire type).

This type of question—combining Ohm's Law with NEC requirements—appears frequently on the Texas journeyman exam.

When Ohm's Law Breaks Down

Ohm's Law works perfectly for resistive loads: heaters, incandescent lights, ovens, and similar equipment where the load acts as simple resistance. However, it needs adjustment for:

Inductive and Capacitive Loads

Motors, transformers, fluorescent ballasts, and other equipment with windings or capacitors introduce inductance or capacitance. These create impedance (Z), not simple resistance.

For these loads:

• Use Z = V ÷ I instead of R = V ÷ I
• Account for power factor (PF), which ranges from 0 to 1 (100%)
• The actual power consumed is P = V × I × PF

For example, a 120V motor drawing 10 amps is not consuming 1,200 watts if its power factor is 0.8—it's consuming only 960 watts.

Why You Still Need Ohm's Law

On the exam, all AC problems for journeyman-level studies use resistive loads or are simplified to assume unity power factor (PF = 1). Use Ohm's Law directly. If the question introduces power factor, it will state it explicitly.

Common Exam Traps

Trap 1: Unit Confusion (kW vs. W)

Problem: A 240V heater is rated 4.8 kW. How many amps?

Wrong approach:

I = P ÷ V = 4.8 ÷ 240 = 0.02A  ← WRONG (forgot to convert kW to W)

Right approach:

4.8 kW = 4,800 W
I = P ÷ V = 4,800 ÷ 240 = 20A  ← CORRECT

Always convert kilowatts to watts (multiply by 1,000) before using the formula.

Trap 2: Forgetting the Continuous Load Multiplier

The exam loves questions about heaters and other continuous loads. Remember: per NEC Article 215.3, multiply continuous currents by 1.25 when selecting breaker and conductor sizes.

Trap 3: Series vs. Parallel Resistance Confusion

Ohm's Law solves for individual circuit elements. If you have multiple resistances:

• Series: Total R = R₁ + R₂ + R₃
• Parallel: 1/R_total = 1/R₁ + 1/R₂ + 1/R₃

Identify the circuit topology first, then apply Ohm's Law to the individual loads or the total.

Trap 4: Mixing Up Current Formulas

You can find current two ways:

• From resistance: I = V ÷ R
• From power: I = P ÷ V

Both are correct. Use whichever values you're given. The exam will provide enough information to solve the problem if you use the right formula.

Quick Reference Table

| Find | Formula 1 | Formula 2 | Formula 3 | |----------|---------------|---------------|---------------| | Voltage (V) | V = I × R | V = P ÷ I | V = √(P × R) | | Current (I) | I = V ÷ R | I = P ÷ V | I = √(P ÷ R) | | Resistance (R) | R = V ÷ I | R = V² ÷ P | R = P ÷ I² | | Power (P) | P = V × I | P = I² × R | P = V² ÷ R |

Choose the formula that matches the values you have.

Related Study Guides

Strengthen your understanding of circuits and NEC requirements with these resources:

• Texas Electrician Exam Calculations: Complete Breakdown — Master all calculation types, from wire sizing to load calculations.
• Conductor Sizing: NEC Table 310.16 Explained — Learn how to select the correct wire gauge using amperage and temperature ratings.
• Free Ohm's Law Calculator — Instant answers for voltage, current, resistance, and power.

Final Thoughts

Ohm's Law is simple, but the exam will test it in dozens of ways. Master the four core formulas, practice until you can solve problems in under 30 seconds, and pay attention to NEC compliance rules (continuous loads, breaker sizing, conductor ampacity). These two skills—Ohm's Law and NEC knowledge—are what separate passing journeyman electricians from those who struggle with calculations.

Start with our Ohm's Law Calculator to verify your work, then solve practice problems by hand. Speed and accuracy on calculation questions will boost your exam score significantly.