Ohm's Law Calculator — V=IR Voltage, Current & Resistance

Calculate voltage, current, resistance, and power using Ohm's Law

Ohm's Law Calculator: Voltage, Current, Resistance, and Power

Table of Contents - Ohms Law


How to Use This Calculator - Ohms Law

Enter any two of the four electrical values:

  • Voltage (V) in volts
  • Current (I) in amperes (use decimals for milliamps: 15mA = 0.015A)
  • Resistance (R) in ohms
  • Power (P) in watts

Leave the unknown values blank or at zero.

Click "Calculate" to see results. The calculator determines the missing values using Ohm's Law and power equations:

  • All four values displayed
  • The formulas used for each calculation
  • Verification that V = I × R and P = V × I

For resistor combinations, select series or parallel mode and enter individual resistor values.


The Core Principle: Ohm's Law

Ohm's Law describes the fundamental relationship between voltage, current, and resistance in electrical circuits:

V = I × R

Voltage (V) is the electrical pressure pushing electrons through a circuit. Current (I) is the flow rate of electrons. Resistance (R) opposes that flow.

This relationship means:

  • Higher voltage pushes more current through a given resistance
  • Higher resistance reduces current for a given voltage
  • Current equals voltage divided by resistance

Power extends these relationships:

  • P = V × I (power equals voltage times current)
  • P = I²R (power equals current squared times resistance)
  • P = V²/R (power equals voltage squared divided by resistance)

These equations are the foundation of all circuit analysis, from LED circuits to power grids.


How to Calculate Electrical Values Manually

Finding current from voltage and resistance: I = V / R

Example: 12V battery, 100Ω resistor I = 12 / 100 = 0.12A = 120mA

Finding resistance from voltage and current: R = V / I

Example: 9V across a component drawing 0.02A R = 9 / 0.02 = 450Ω

Finding voltage from current and resistance: V = I × R

Example: 0.05A through 220Ω V = 0.05 × 220 = 11V

Calculating power: P = V × I = I²R = V²/R

Example: 12V, 0.12A P = 12 × 0.12 = 1.44W

Series resistors: R_total = R₁ + R₂ + R₃ + ...

Example: 100Ω + 220Ω + 330Ω = 650Ω

Parallel resistors: 1/R_total = 1/R₁ + 1/R₂ + 1/R₃ + ...

Example: 100Ω || 100Ω 1/R_total = 1/100 + 1/100 = 2/100 R_total = 50Ω


Real-World Applications

LED current limiting. An LED needs 20mA at 2V forward voltage. From a 5V supply: Resistor voltage = 5 - 2 = 3V R = 3 / 0.020 = 150Ω

Resistor power rating. That 150Ω resistor dissipates: P = 0.020² × 150 = 0.06W A ¼W (0.25W) resistor provides adequate margin.

Wire sizing. Long wire runs have resistance. Calculate voltage drop to ensure equipment receives adequate voltage.

Fuse selection. Calculate maximum expected current for a circuit, add safety margin, and select appropriate fuse rating.

Battery drain estimation. Device draws 0.5A from 12V battery. Power = 6W. A 12Ah battery provides approximately 24 hours of operation.

Motor starting current. Motors draw high current at startup. Calculate to ensure power supply and wiring can handle the surge.


Scenarios People Actually Run Into

The hot resistor. Your resistor gets uncomfortably hot. You calculated the resistance correctly but chose a ¼W resistor for a circuit dissipating 0.23W. Solution: use ½W or 1W resistor for adequate derating.

The dim LED. You're running an LED from 12V with a 1kΩ resistor. Current = 10mA—too low for many LEDs. Solution: decrease resistance to increase current (330Ω for 30mA).

The voltage drop problem. Your 12V power supply feeds equipment 50 feet away through thin wire. Equipment measures only 11V. Solution: use heavier gauge wire or higher supply voltage.

The parallel mystery. Two 100Ω resistors in parallel create 50Ω, not 200Ω. Parallel resistance is always less than the smallest individual resistor.

The power supply limitation. Your 12V supply is rated for 1A. Your circuit calculates to 1.2A. The supply may overheat, shut down, or fail. Solution: use adequately rated supply.


Trade-Offs and Decisions People Underestimate

Resistor power derating. A resistor rated for ½W should run at ¼W for longevity. Higher operating temperature reduces lifespan and can affect nearby components.

Standard resistor values. Calculated 453Ω? That doesn't exist. Use 470Ω (next standard value up). Slight current reduction is usually acceptable.

Tolerance matters. A 100Ω ±5% resistor could be 95-105Ω. In precision applications, use 1% or 0.1% tolerance.

Temperature effects. Resistance changes with temperature. Carbon composition resistors vary more than metal film. Consider temperature coefficient for precision circuits.

AC versus DC. Ohm's Law applies directly to DC and resistive AC loads. Inductive and capacitive loads require impedance calculations.


Common Mistakes and How to Recover

Unit confusion. Milliamps (mA) versus amps (A). 1mA = 0.001A. Enter 15mA as 0.015A.

Forgetting power dissipation. Calculating resistance without checking power rating leads to burned components. Always verify P = I²R is within component limits.

Ignoring internal resistance. Batteries and power supplies have internal resistance. Under load, output voltage is lower than no-load voltage.

Using ideal values. Real components have tolerance. A "100Ω" resistor might be 95-105Ω. Design with margins.

Measuring resistance in-circuit. Other parallel paths affect readings. Measure components out of circuit for accurate values.


Related Topics

Kirchhoff's Laws. Conservation of charge (current into a node equals current out) and conservation of energy (voltages around a loop sum to zero).

Impedance. The AC equivalent of resistance, including reactive effects from inductors and capacitors.

Power factor. In AC circuits, the ratio of real power to apparent power. Resistive loads have power factor of 1.

Voltage dividers. Two resistors in series create a voltage proportionally divided between them.

Current dividers. Two resistors in parallel divide current inversely proportional to resistance.


How This Calculator Works

Core relationship:

V = I × R

Derived equations:

I = V / R
R = V / I
P = V × I = I² × R = V² / R

Given two values, calculate others:

if V and I known:
  R = V / I
  P = V × I
if V and R known:
  I = V / R
  P = V² / R
if I and R known:
  V = I × R
  P = I² × R
if P and R known:
  V = √(P × R)
  I = √(P / R)
if P and V known:
  I = P / V
  R = V² / P
if P and I known:
  V = P / I
  R = P / I²

Resistor combinations:

Series: R_total = R₁ + R₂ + R₃ + ...
Parallel: 1/R_total = 1/R₁ + 1/R₂ + 1/R₃ + ...

All calculations happen locally in your browser.


FAQs

What units should I use?

Voltage in volts (V), current in amperes (A), resistance in ohms (Ω), power in watts (W). Convert milliamps to amps (divide by 1000) before entering.

Can this handle AC circuits?

For purely resistive AC loads (heaters, incandescent bulbs), yes—use RMS values. For inductive or capacitive loads, these calculations are approximate.

What if I only know power and resistance?

The calculator uses derived formulas: V = √(P × R) and I = √(P / R).

Why does my resistor get hot?

It's dissipating power as heat. If power approaches the resistor's rating, it runs hot. Use a higher-wattage resistor or redesign the circuit.

How do I combine resistors?

Series: add values directly. Parallel: reciprocal of sum of reciprocals. Two equal resistors in parallel have half the value of one.

Is this suitable for high-voltage circuits?

The math is valid, but never work on mains voltage without proper training and safety equipment. This tool is primarily for low-voltage DC projects.

How accurate are the results?

Mathematically exact for ideal components. Real-world tolerances, temperature effects, and measurement accuracy affect actual circuit behavior.

What about variable resistors (potentiometers)?

Treat them as resistors with current value. The calculation shows behavior at that setting; adjust and recalculate for other positions.

How do I calculate power supply requirements?

Sum the current draw of all components in your circuit. Add 20-30% margin for safety. Verify the power supply can deliver that current at the required voltage.

What's the relationship between wire gauge and resistance?

Thinner wire has higher resistance per foot. For long runs or high current, calculate voltage drop: V_drop = I × R_wire. Use heavier gauge to reduce drop.

How do I size fuses and circuit breakers?

Calculate maximum expected current, add 25% margin, and select the next standard size up. The protection device should blow before wire overheats.

Why does AC power factor matter?

For inductive loads (motors, transformers), current and voltage are out of phase. Real power (watts) is less than apparent power (volt-amps). Power factor = watts / VA.

How do I measure resistance accurately?

Use a multimeter in resistance mode. Disconnect power, isolate the component from parallel paths, and check that your meter leads don't add significant resistance.

What about temperature effects on resistance?

Most conductors increase resistance with temperature. This can affect circuit behavior in hot environments or with components that heat during operation.

How do I handle mixed AC and DC circuits?

Analyze separately. DC components see only DC values. AC components see peak, RMS, or instantaneous values depending on the analysis. Ohm's Law applies to each independently.

What is a voltage divider and how do I calculate it?

Two resistors in series divide voltage proportionally: V_out = V_in × R2 / (R1 + R2). Useful for creating reference voltages or sensor interfaces.

How do I calculate current limiting for LEDs?

R = (V_supply - V_forward) / I_desired. A 5V supply with 2V LED at 20mA needs: (5-2)/0.020 = 150Ω. Use next higher standard value (180Ω).

What is Kirchhoff's Current Law?

Current entering any junction equals current leaving. This lets you analyze complex circuits by tracking current flow at every node.

What is Kirchhoff's Voltage Law?

Voltages around any closed loop sum to zero. This lets you write equations for complex circuits and solve for unknown values.

How do I calculate power loss in cables?

P_loss = I² × R_cable. For a 10A current through 0.5Ω cable resistance: P_loss = 100 × 0.5 = 50W. That's heat in your wires!

Additional Notes

Understanding Ohm's Law is fundamental to all electrical work. These calculations form the foundation for circuit design, troubleshooting, and component selection in every electronics project. Mastering these fundamentals enables you to tackle increasingly complex circuit analysis challenges with confidence. Safety should always be your first priority when working with electricity. Always double-check calculations before applying power.