Electric • Core Flagship Pillar

Ohm’s Law Calculator

Solve for Voltage ($V$), Current ($I$), Resistance ($R$), or Power ($P$) by entering any two known parameters with step-by-step formula proofs and interactive Ohm's Law identity wheel diagrams.

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Last Updated: September 2026
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IEEE Standard Electrical Circuit Verified
Quick-Select Circuit Scenarios Standard Benchmarks

Enter Any 2 Parameters

Calculation Mode
Solving from Voltage (V) and Current (I)
Ohm's Law 12-Identity Wheel Matrix
Voltage (V)
120.00 V
Electric Potential Difference
Current (I)
10.00 A
Charge Flow Rate (Coulombs/sec)
Resistance (R)
12.00 Ω
Opposition to Current Flow
Power (P)
1,200.00 W
Energy Dissipation Rate (Joules/sec)
V=IR

Step-by-Step Ohm's Law & Joule's Law Derivation

Direct Answer & Overview
Verified Educational Guide

How to Calculate Ohm's Law (V, I, R, P)

Ohm's Law defines the relationship between Voltage (V), Current (I), and Resistance (R): V = I × R. Joule's Law defines Electrical Power: P = V × I = I²R = V²/R. Enter any two known circuit values to calculate the remaining two instantly.

Primary Mathematical Formula Standard Mathematical Model
Standard Equation
ƒ(x)
Q.E.D.
V=I⋅R,I=VR,R=VI,P=V⋅I=I2R=V2RV = I \cdot R, \quad I = \frac{V}{R}, \quad R = \frac{V}{I}, \quad P = V \cdot I = I^2 R = \frac{V^2}{R}
Evaluated with exact mathematical formulation • Rigorously verified
Exact Formula
Input Parameters
Required
1
Any 2 known values from Voltage (V), Current (I), Resistance (R), or Power (P)
Expected Outputs
Calculated
All 4 solved circuit parameters (V in Volts, I in Amps, R in Ohms, P in Watts)
Step-by-step formula breakdown and interactive 4-quadrant identity wheel
Worked Numerical Example
Instant Verification
120V household voltage powering a 10A electric heater
→ Resistance R = 120V / 10A = 12 Ω; Power P = 120V × 10A = 1,200 Watts (1.2 kW)
Voltage = 120 V | Current = 10 A | Resistance = 12 Ω | Power = 1,200 W

Ohm’s Law Definition & The Hydraulic Circuit Analogy

Discovered in 1827 by German physicist Georg Simon Ohm, Ohm’s Law defines the fundamental proportional relationship governing direct current electrical circuits:

Voltage (V) Volts (V)

Electrical potential pressure driving charge

Current (I) Amperes (A)

Rate of electrical charge flow (Coulombs/sec)

Resistance (R) Ohms (Ω)

Material opposition to charge flow

The 12 Ohm’s Law & Joule’s Power Formulas

To Find Formula 1 Formula 2 Formula 3
Voltage (V)V = I · RV = P / IV = √(P · R)
Current (I)I = V / RI = P / VI = √(P / R)
Resistance (R)R = V / IR = V² / PR = P / I²
Power (P)P = V · IP = I² · RP = V² / R

Real-World Applications of Ohm’s Law

LED Current-Limiting Resistors

Circuit designers calculate ($R = (V_{\text{supply}} - V_{\text{forward}}) / I$) to prevent light-emitting diodes from burning out under excess current.

Household Breaker Sizing

Electricians sum total appliance wattage ($P/V = I$) to ensure circuit draw does not exceed 15A or 20A branch breakers.

Audio Speaker Impedance

Car and home audio engineers match 4Ω and 8Ω speaker loads to amplifier voltage rails ($P = V^2 / R$) for clean audio power delivery.

Step-by-Step Worked Numerical Solutions

Example 1: Solving Resistance and Power DC Circuit

Problem: A 12V battery delivers 2.5 Amps to a DC motor. Find the resistance and power output.

1. Resistance R = V / I = 12 V / 2.5 A = 4.8 Ω.
2. Power P = V × I = 12 V × 2.5 A = 30 Watts.
3. Verify via P = I²R: (2.5)² × 4.8 = 6.25 × 4.8 = 30 Watts.
Result: Resistance R = 4.8 Ω | Power P = 30 W

Common Pitfalls & Mistakes

Milliamps to Amps Conversion

Forgetting that 20 mA = 0.02 A results in calculations off by a factor of 1,000.

Resistor Power Dissipation Rating

Selecting the right resistance value is useless if actual power dissipation exceeds the component's 1/4W or 1/2W thermal rating.

Applying to Non-Ohmic Diodes

Semiconductor PN junctions have dynamic non-linear resistance that cannot be modeled by simple static Ohm's Law.

Fact-Checked & Verified • Computational Accuracy Standards
Updated August 2026 • Editorial Policy
Authored By
Sanjay Samanta

Lead Developer & Founder of Basic Math Tools. Specializes in browser-native computational algorithms and applied mathematics.

Reviewed & Verified By
Academic Review Board

Mathematics & curriculum specialists. Audited against standard algebraic and arithmetic principles.

Found an error or have an improvement suggestion? Report a calculation issue

Frequently Asked Questions

What is Ohm’s Law and what is its fundamental formula?
Ohm’s Law states that the electrical current (I) flowing through a conductor between two points is directly proportional to the voltage (V) across the points and inversely proportional to the resistance (R): V = I × R (Voltage = Current × Resistance).
What are the 12 mathematical formulas of the Ohm’s Law power wheel?
Voltage: V = I·R = P/I = √(P·R); Current: I = V/R = P/V = √(P/R); Resistance: R = V/I = V²/P = P/I²; Power: P = V·I = I²·R = V²/R.
How does the hydraulic water analogy explain voltage, current, and resistance?
In the hydraulic analogy: Voltage corresponds to water pressure in a pipe; Current corresponds to the volumetric flow rate of water; and Resistance corresponds to pipe constriction or friction that opposes water flow.
What is Joule’s Law of electrical heating (Power P = I²R)?
Joule’s Law states that the rate of electrical energy dissipated as thermal heat in a resistor is proportional to the resistance multiplied by the square of the current: P = I² × R. This explains why power transmission lines use ultra-high voltages to minimize current and transmission heat losses.
What is the difference between Ohmic and non-Ohmic conductors?
Ohmic conductors (like copper wires and carbon resistors) maintain a constant resistance regardless of voltage (linear V-I curve). Non-Ohmic devices (like silicon diodes, LEDs, and transistors) have dynamic, non-linear resistance that changes with applied voltage or temperature.
How do you calculate a current-limiting resistor for an LED?
Use the formula R = (V_source − V_forward) / I_desired. For example, to run a 2.0V, 20mA (0.02A) LED from a 5V supply: R = (5 − 2) / 0.02 = 3 / 0.02 = 150 Ω.