Ohm's Law Calculator
Compute voltage across a component using Ohm’s Law, V = I·R, given the current flowing through it and its resistance. The most fundamental relationship in electronics, used in every circuit design from LED dropper-resistors to industrial power systems.
Last updated: May 2026
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About this calculator
Ohm’s Law states that the voltage across a conducting component equals the current through it multiplied by its resistance: V = I·R, where V is in volts (V), I is in amperes (A), and R is in ohms (Ω). The relationship rearranges to solve for any of the three: I = V/R (current from voltage and resistance) or R = V/I (resistance from voltage and current). One ohm is the resistance that produces a current of one ampere when a voltage of one volt is applied across it. The law was published by Georg Ohm in 1827 and applies to ‘ohmic’ materials — those where the V-I relationship is linear at constant temperature — which covers most resistors, wires, and bulk conductors in normal conditions. Non-ohmic devices (diodes, transistors, light bulbs at varying temperatures, electrolytes) have V-I curves that are not straight lines, and Ohm’s Law doesn’t apply directly — use device characteristic curves or model equations instead. Edge cases: at high frequencies (radio, microwave) reactive impedances from capacitance and inductance dominate, and V = IR is replaced by V = I·Z where Z is the complex impedance. At cryogenic temperatures superconductors have zero resistance (R = 0), so any current produces zero voltage drop — Ohm’s Law still holds (V = 0) but is uninformative. At atomic-scale dimensions and very low currents, quantum effects produce conductance quantised in units of e²/h, replacing the smooth V = IR behaviour with discrete steps. Power dissipated in the component follows from Ohm’s Law: P = V·I = I²·R = V²/R, in watts.
How to use
Example 1 — Sizing a current-limiting resistor for an LED. You want to drive a red LED (forward voltage ~2.0 V, recommended current 20 mA) from a 5 V supply. The resistor must drop the excess voltage: V_resistor = 5 − 2.0 = 3.0 V. R = V/I with V = 3.0 and I = 0.020: R = 3.0/0.020 = 150 Ω. ✓ Choose a 150 Ω (or 180 Ω) ¼-watt resistor; power dissipation P = V·I = 3.0 × 0.020 = 0.060 W = 60 mW, well within a ¼ W (250 mW) rating. Example 2 — Voltage drop across a long cable. You run 5 A through a 30 m extension cable. Each metre of 14-gauge copper has about 8.3 mΩ resistance, so the cable totals 30 × 0.0083 = 0.249 Ω. V = I·R = 5 × 0.249 = 1.245 V drop. ✓ For a 120 V circuit that is ~1% of supply voltage, acceptable; for a 12 V circuit (automotive) it is ~10%, which causes noticeable performance drops in lights or motors at the far end and may indicate the need for thicker gauge wire.
Frequently asked questions
Why isn’t Ohm’s Law actually a law in the strict sense?
Modern physics distinguishes between fundamental laws (conservation of energy, Newton’s laws) and empirical relationships that hold only under specific conditions. Ohm’s Law is the latter: it describes the behaviour of a particular class of materials (ohmic conductors) at fixed temperature and over a limited range of currents and voltages. Plenty of materials violate it routinely — diodes don’t conduct below their forward voltage and then conduct heavily above it, light bulbs have resistance that varies dramatically with temperature, and electrolytes show non-linear conductivity. Even ordinary copper wire stops obeying Ohm’s Law at extreme current densities when it heats enough to vaporise. The ‘law’ was originally a description of what Ohm observed in metal wires, and the name stuck. Calling it a rule of thumb that happens to work for most practical components would be more accurate, but less catchy.
How do I calculate power from Ohm’s Law?
Electrical power dissipated in a resistor is P = V·I, in watts. Using Ohm’s Law to substitute, this can be written three equivalent ways: P = V·I, P = I²·R, and P = V²/R. Pick the form that uses the quantities you already know. For a 100 Ω resistor carrying 0.5 A: P = I²·R = 0.25 × 100 = 25 W. The same resistor across 50 V: P = V²/R = 2500/100 = 25 W. Power matters because it appears as heat — a 25 W resistor must be physically large enough to dissipate that heat without burning out. Most resistors are rated by power-handling capacity (typically ⅛ W, ¼ W, ½ W, 1 W, 5 W) and choosing one below the actual dissipation is the most common cause of resistor failure in hobbyist circuits.
What happens to Ohm’s Law in AC circuits?
At frequencies high enough that capacitance and inductance matter, the simple V = I·R is replaced by V = I·Z where Z is the complex impedance — a generalised resistance with both real (resistive) and imaginary (reactive) parts. A pure resistor has Z = R (real only). A capacitor has Z_C = 1/(jωC) — impedance shrinks at high frequencies. An inductor has Z_L = jωL — impedance grows at high frequencies. For a series RLC circuit, Z = R + j(ωL − 1/ωC), and you must use complex arithmetic to find voltage drops and currents. For DC (frequency = 0), capacitors are open circuits (infinite Z) and inductors are short circuits (zero Z), reducing back to the simple Ohm’s Law form. Practical AC-circuit analysis adds phasor diagrams, power factor (cosine of the phase angle between V and I), and the distinction between real power (W), reactive power (VAR), and apparent power (VA).
What are the most common mistakes people make with Ohm’s Law?
The first is unit confusion: confusing milliamps (mA) with amps, or kΩ with Ω. A 1 mA current through a 10 kΩ resistor produces V = 0.001 × 10,000 = 10 V, but mishandling either prefix throws the answer off by factors of 1,000. The second is applying Ohm’s Law to non-ohmic components — calculating ‘resistor equivalents’ for diodes or transistors without checking that the operating point is in a linear region. The third is forgetting that resistance varies with temperature — a tungsten light bulb’s resistance is about ten times higher when hot than when cold, so its cold resistance does not determine its operating current. The fourth is ignoring wire and contact resistance in high-current circuits where the cable itself drops a meaningful fraction of the supply voltage. The fifth is mismatching impedance in AC circuits and applying DC Ohm’s Law where complex impedances are needed. Finally, real components have tolerance — a ‘100 Ω 5%’ resistor can be 95 to 105 Ω, and designs must remain stable across that range.
When should I not use this calculator?
Skip it for non-ohmic components — diodes, transistors, LEDs above their threshold voltage, vacuum tubes, electrolytic conduction, and gas-discharge devices all have non-linear V-I characteristics that V = IR cannot describe. Look up the device’s V-I curve or use its model equation (Shockley for diodes, Ebers-Moll for BJTs). Avoid it for high-frequency AC circuits where reactive impedance matters — use V = I·Z with complex impedance instead. It is wrong for power-supply problems where you need to track multiple voltage drops, parallel paths, and Kirchhoff’s laws (current and voltage conservation around loops) — Ohm’s Law gets used inside the analysis but is not the whole answer. Do not use it as the sole tool for designing high-current circuits where wire heating, fusing, and connector resistance all interact; full circuit simulation is needed. And do not use it at cryogenic or quantum scales where superconductivity or conductance quantisation breaks the classical relationship.