What is Ohm's Law?
Ohm's Law is one of the most fundamental principles in electrical engineering and electronics. Discovered by German physicist Georg Ohm in 1827, this law describes the relationship between voltage, current, and resistance in electrical circuits.
The Fundamental Principle
Ohm's Law states that the current through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance between them. This relationship is expressed mathematically as:
V = I × RWhere V is voltage (volts), I is current (amperes), and R is resistance (ohms)
Understanding the Components
Voltage (V) - Volts
Voltage is the electrical pressure or potential difference that pushes electrons through a circuit. It's similar to water pressure in a pipe. Higher voltage means more electrical "pressure" pushing current through the circuit.
Current (I) - Amperes
Current is the flow rate of electrical charge through a conductor. It's analogous to the flow rate of water through a pipe. One ampere equals one coulomb of charge passing through a point per second.
Resistance (R) - Ohms
Resistance is the opposition to current flow in a conductor. It's like friction in a pipe that restricts water flow. Different materials have different resistances - copper has low resistance (good conductor), while rubber has high resistance (good insulator).
The Water Analogy
A helpful way to understand Ohm's Law is through a water pipe analogy:
- Voltage is like water pressure - higher pressure pushes more water
- Current is like water flow rate - how much water flows per second
- Resistance is like pipe narrowness - narrow pipes restrict flow
- Just as more pressure or wider pipes increase water flow, more voltage or less resistance increases current
Historical Note: Georg Ohm published his findings in 1827, but his work was initially met with skepticism. Today, Ohm's Law is recognized as one of the cornerstones of electrical engineering, and the unit of resistance (ohm, Ω) is named in his honor.
Ohm's Law Formulas
Ohm's Law can be rearranged to solve for any of the three variables (voltage, current, or resistance) when you know the other two. Additionally, power calculations extend Ohm's Law to include energy consumption.
Core Ohm's Law Formulas
Calculate Voltage (V)
V = I × RVoltage equals current multiplied by resistance
Example: If current is 2 amps and resistance is 6 ohms:
V = 2 A × 6 Ω = 12 volts
Calculate Current (I)
I = V ÷ RCurrent equals voltage divided by resistance
Example: If voltage is 12 volts and resistance is 6 ohms:
I = 12 V ÷ 6 Ω = 2 amps
Calculate Resistance (R)
R = V ÷ IResistance equals voltage divided by current
Example: If voltage is 12 volts and current is 2 amps:
R = 12 V ÷ 2 A = 6 ohms
Power Formulas
Power (measured in watts) represents the rate of energy conversion. Combining Ohm's Law with the power equation P = V × I gives us twelve useful formulas:
Calculate Power (P)
P = V × IPower equals voltage times current
P = I² × RPower equals current squared times resistance
P = V² ÷ RPower equals voltage squared divided by resistance
Calculate Voltage from Power
V = P ÷ IVoltage equals power divided by current
V = √(P × R)Voltage equals square root of (power times resistance)
Calculate Current from Power
I = P ÷ VCurrent equals power divided by voltage
I = √(P ÷ R)Current equals square root of (power divided by resistance)
Calculate Resistance from Power
R = V² ÷ PResistance equals voltage squared divided by power
R = P ÷ I²Resistance equals power divided by current squared
Memory Aid: The "Ohm's Law Wheel" or "Ohm's Law Triangle" helps remember these formulas. Place V at the top, I and R at the bottom. Cover the value you want to find, and the remaining values show the formula. For example, cover V to see I × R.
Practical Applications
Ohm's Law is used daily by electricians, electronics technicians, and engineers for designing circuits, troubleshooting problems, and selecting components. Understanding these applications makes electrical work safer and more effective.
LED Circuit Design
One of the most common applications is calculating the current-limiting resistor for LEDs:
Example: Power a 2V LED from a 9V battery
- Supply voltage: 9V
- LED forward voltage: 2V
- Desired LED current: 20mA (0.02A)
- Voltage across resistor: 9V - 2V = 7V
Calculate resistor value:
R = V ÷ I = 7V ÷ 0.02A = 350ΩUse standard 360Ω or 390Ω resistor (next higher value)
Calculate power rating:
P = I² × R = (0.02)² × 350 = 0.14WUse 1/4 watt (0.25W) resistor for safety margin
Electrical Troubleshooting
Ohm's Law helps diagnose circuit problems by identifying abnormal readings:
- Open Circuit: Infinite resistance, zero current - indicates broken wire or disconnection
- Short Circuit: Near-zero resistance, excessive current - dangerous condition requiring immediate attention
- High Resistance: Lower than expected current - may indicate corrosion, loose connections, or degraded components
- Load Testing: Measure voltage and current to calculate actual resistance and compare to specifications
Battery Life Estimation
Calculate how long a battery will power a device:
Example: Battery-powered device runtime
- Battery: 9V with 500mAh capacity
- Device resistance: 450Ω
I = V ÷ R = 9V ÷ 450Ω = 0.02A = 20mARuntime = Capacity ÷ Current = 500mAh ÷ 20mA = 25 hoursHeater Element Sizing
Design heating elements for specific power output:
Example: 120V, 1500W space heater
I = P ÷ V = 1500W ÷ 120V = 12.5AR = V² ÷ P = (120)² ÷ 1500 = 9.6ΩHeating element must be 9.6Ω and handle 12.5A continuously
Wire Ampacity Verification
Ensure wire size is adequate for current draw:
Example: 240V, 5kW water heater
I = P ÷ V = 5000W ÷ 240V = 20.83ANEC requires 125% safety factor for continuous loads:
20.83A × 1.25 = 26.04A minimum wire ampacity
Use 10 AWG copper (rated 30A at 60°C) or larger
Circuit Design Examples
Real-world circuit design requires applying Ohm's Law with practical considerations like component tolerances, temperature effects, and safety margins.
Example 1: Arduino Sensor Circuit
Design a pull-up resistor for a digital input
- Supply voltage: 5V
- Input pin leakage: 1µA (0.000001A) maximum
- Desired high-level voltage: 4.5V minimum
- Maximum current budget: 100µA
Minimum resistance (prevent excess current):
R_min = V ÷ I_max = 5V ÷ 0.0001A = 50,000Ω = 50kΩMaximum resistance (ensure high level):
Voltage drop allowed = 5V - 4.5V = 0.5V
R_max = 0.5V ÷ 0.000001A = 500,000Ω = 500kΩSolution: Use standard 10kΩ resistor (well within range, commonly available)
Example 2: Series Resistor Voltage Divider
Create 3.3V from 5V supply for sensor
- Input voltage: 5V
- Output voltage: 3.3V
- Load current: 10mA maximum
- Divider current: 1mA (10× load for stability)
Total resistance:
R_total = V ÷ I = 5V ÷ 0.001A = 5,000Ω = 5kΩResistor ratio for voltage division:
V_out = V_in × (R2 ÷ (R1 + R2))
3.3V = 5V × (R2 ÷ 5kΩ)
R2 = 3.3kΩ, R1 = 1.7kΩSolution: Use 1.8kΩ (R1) and 3.3kΩ (R2) standard values
Actual output: 5V × (3.3kΩ ÷ 5.1kΩ) = 3.24V (within tolerance)
Example 3: Motor Current Limiting
Limit inrush current for DC motor
- Motor rated voltage: 12V
- Motor running current: 500mA
- Motor winding resistance: 2Ω
- Supply voltage: 12V
- Desired startup current limit: 1A
Calculate startup current without limiting:
I_startup = V ÷ R_motor = 12V ÷ 2Ω = 6A (too high!)Required series resistance:
R_total = V ÷ I_limit = 12V ÷ 1A = 12Ω
R_limit = R_total - R_motor = 12Ω - 2Ω = 10ΩPower dissipation in limiting resistor:
P = I² × R = (1A)² × 10Ω = 10W during startupSolution: Use 10Ω, 20W power resistor or soft-start circuit with timer relay
Example 4: Fuse Selection
Select appropriate fuse for 12V automotive circuit
- Load: 12V LED light bar
- Power rating: 120W
- Operating voltage: 12V
Calculate operating current:
I = P ÷ V = 120W ÷ 12V = 10ASolution: Use 15A fuse (125-150% of operating current)
10A fuse would be too close to operating current and might blow during normal operation
20A fuse would be too large and might not protect against shorts adequately
How We Calculate
Our Ohm's Law calculator uses scientifically validated formulas to provide accurate results for electrical circuit calculations. Understanding the methodology helps you apply these principles correctly.
Calculation Methodology
1. Input Validation
The calculator first validates that at least two values are known and non-zero. Ohm's Law requires two known values to solve for a third. Zero or negative values are physically meaningless for these calculations.
2. Formula Selection
Based on which variable you're solving for and which values you've provided, the calculator automatically selects the appropriate formula from the twelve Ohm's Law and power formulas.
Solving for Voltage (V):
- If I and R are known: V = I × R
- If P and I are known: V = P ÷ I
- If P and R are known: V = √(P × R)
3. Computation
All calculations use double-precision floating-point arithmetic for accuracy. Results are computed to full precision, then rounded to three decimal places for display.
4. Dependent Value Updates
When solving for one variable, the calculator also computes the fourth variable to show complete circuit parameters. For example, when solving for voltage with known current and resistance, power is automatically calculated.
Accuracy Considerations
The calculator provides mathematically exact results based on ideal Ohm's Law. Real-world accuracy depends on several factors:
- Temperature Effects: Resistance changes with temperature (typical metals increase ~0.4% per °C)
- Component Tolerance: Resistors typically have ±1%, ±5%, or ±10% tolerance
- Non-Ohmic Devices: Diodes, transistors, and batteries don't follow Ohm's Law exactly
- Measurement Precision: Multimeter accuracy is typically ±0.5% to ±2%
- AC Impedance: For AC circuits with reactive components, use impedance instead of resistance
Interactive Circuit Diagram
The calculator includes an SVG-based circuit diagram that updates in real-time to show:
- Voltage source (battery symbol) with current value
- Resistor (zigzag symbol) with resistance value
- Current flow direction with arrow and magnitude
- Power dissipation across the circuit
Engineering Note: This calculator assumes ideal DC circuits. For AC circuits with inductors or capacitors, replace R with impedance Z, where Z = √(R² + X²) and X is reactance. For three-phase power systems, additional factors apply.