Ohm's Law Calculator

solve V = I · R for any variable

Voltage V = I × R

Current I = V ÷ R

Resistance R = V ÷ I

Worked example — Ohm's Law Calculator

I — Current	0.015 A
plug in the numbers	0.015A

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

Lesson

The theory — Ohm's Law Calculator

Two different statements hide inside V = IR. Resistance is defined as the ratio V/I, and that ratio can be measured for any component at any operating point — a diode, a lamp, a length of wet string. Ohm’s law is the further, separate claim that for some materials the ratio does not move when you change the voltage. That is what turns the graph into a straight line through the origin, and it is why a single number can stand in for the whole component.

What each symbol means

V: voltage, in volts — energy per unit charge. One volt hands one joule to every coulomb that passes. That unit does real work below: joules per coulomb times coulombs per second is joules per second, so V and I multiply into watts without any new physics being added.
I: current, in amperes — charge per second, one coulomb each second. Around a single loop it is the same at every point, including on both sides of the resistor. A resistor does not use up current; it uses up energy.
R: resistance, in ohms — one ohm is one volt per ampere. Push it down and nothing holds the current back: at 0.001 Ω the readout reports 12000 A from 12 V. At exactly 0 it prints — rather than a number, which is the honest response to a question with no finite answer.

Where the formula comes from

Start from the units rather than from a formula. V is joules per coulomb and I is coulombs per second, so V × I is joules per second — watts. The power a resistor turns into heat is not an extra law to memorise; it is what those two units give when multiplied.
Now substitute the law. Putting V = IR into P = VI gives P = (IR)I = I²R. At a fixed resistance the heat follows the square of the current, so doubling the current quadruples the dissipation.
Substitute the other way instead. I = V/R in P = VI gives P = V(V/R) = V²/R. Three expressions, one quantity — use whichever names the two things you already know.
Put it on the short-circuit preset: 12 V across 0.1 Ω, and the readout says 120 A. Then P = I²R = 120² × 0.1 = 1440 W. That is a kilowatt and a half inside a piece of wire, and it is why the note attached to that preset warns the current can overheat wires. The wattage that justifies the warning is never printed anywhere on the page.

How to read what you see

Read down the panel and the units shift under you. The large readout is in amperes — the LED preset shows 0.015 A — while the sentence directly beneath it switches to milliamperes and says 15 mA. The same number twice, in two units, on one screen. The rest is a worked line you can check instead of trust: the formula shows the relation rearranged for whatever you asked for, I = V ÷ R or R = V ÷ I, and plug in the numbers repeats that line with your own values in place.

Assumes: One ohmic element, at a steady temperature, on direct current, with the entire supply voltage across it. Nothing else in the loop: no internal resistance in the battery, no second component, no signal that changes with time. The three boxes describe a single resistor rather than a circuit, and on a real bench that is the assumption most likely to be false.
Breaks when: The trap is this page’s own first preset. LED circuit loads 3.3 V and 220 Ω and returns 15 mA, which is exactly right for a resistor with 3.3 V across it — but an LED is not ohmic. It holds a roughly fixed forward drop, near 2 V for a red one, almost regardless of the current through it, so in a real circuit the LED takes its 2 V first and the resistor is left with 1.3 V. Type 1.3 into the voltage box and the readout returns 0.005909 A — under 6 mA, less than half the preset’s answer. Ohm’s law did not fail here; it was applied to the supply instead of to the resistor.

V = IR is an approximation, not a law 🖖

While V = IR is a convenient macroscopic relation, Ohm's law is not a fundamental law of nature, but an empirical approximation. Microscopically, it is expressed as the vector relation J = σE, where J is current density, σ is conductivity, and E is the electric field. In anisotropic materials, conductivity σ becomes a second-rank tensor, meaning an applied electric field can push current in a different direction. Under the classical Drude model, this linearity arises because the electric force on charge carriers is balanced by scattering against lattice vibrations (phonons) and impurities. However, this approximation breaks down under extreme conditions: in high electric fields, carrier drift velocity saturates; in superconductors, Cooper pairs undergo macroscopic quantum coherence to bypass scattering entirely (R = 0); and at the nanoscale, transport becomes ballistic, where conductance is quantized in integer steps of 2e²/h (Landauer formula), rendering the traditional concept of localized resistance obsolete.

Voltage, current, and resistance in balance 🖖

Think of a circuit like water in a pipe: voltage is the pressure pushing charge, current is how much flows, and resistance is how much the pipe restricts it. Ohm's law V = IR ties them together, so knowing any two values gives you the third. The practical takeaway: at a fixed voltage, doubling the resistance halves the current.

Ohm borrowed his law from heat 🖖

Georg Ohm built his 1827 relation by direct analogy to Fourier's law of heat conduction, swapping temperature for voltage and heat flow for electric current. The same linear form reappears across physics — Fourier's heat law, Fick's law of diffusion, and Darcy's law for groundwater all share it. German academics first dismissed Ohm's work as fanciful, and he resigned his teaching post before finally winning recognition.

See how Ohm's Law works like a water pipe - animated 🖖

Water analogy: imagine the circuit as a closed pipe loop. The battery is a pump that builds up 9 V of pressure. The resistor (10 Ω) is a narrow section of pipe - the higher the resistance, the tighter the constriction. That pressure, fighting through that restriction, forces a steady flow of water: 0.9 A (900 mA) of current. Double the pressure - double the flow. Double the pipe width - double the flow. That linearity is Ohm's Law.

Problem solved in full

A red LED on a 5 V supply with a 220 Ω resistor 5 steps
A red LED on a 5 V supply, with the classic 220 Ω resistor. Find the current, then find where the power goes — because most of it does not go to the LED.
1. Ohm's law gives the current directly from the voltage across the resistor and its resistance.
  
  $I = \frac{V}{R} = \frac{3.3}{220} = 0.015 A = 15 mA$
2. But that 3.3 V is not the supply. An LED holds a roughly fixed forward voltage — about 1.7 V for a red one — and the resistor gets whatever is left of the 5 V. That subtraction is why 220 Ω became the default value in every beginner's kit.
  
  $V_{R} = V_{supply} - V_{f} = 5.0 - 1.7 = 3.3 V$
3. The resistor's power dissipation follows from the current it carries. At 0.05 W, a quarter-watt resistor is running at a fifth of its rating and will not even feel warm.
  
  $P_{R} = I^{2} R = (0.015)^{2} \times 220 = 0.0495 W$
4. The LED's share is its forward voltage times the same current, since the two are in series and carry the same current by definition. The total is the supply voltage times that current, and the two parts add up to it exactly.
  
  $P_{LED} = V_{f} I = 1.7 \times 0.015 = 0.0255 W, P_{total} = 5.0 \times 0.015 = 0.075 W$
5. So a third of the power reaches the LED and two thirds is turned into heat in the resistor.
  
  $η = \frac{0.0255}{0.075} = 34%$
Answer

The tool prints 0.015 A. The efficiency in step 5 is not a coincidence to be computed — it is the voltage ratio 1.7/5.0 read back as a fraction, because both components carry the same current and power is voltage times current. A series resistor is therefore a divider whose loss is fixed by the supply voltage you chose, and no resistor value can improve it. That is the whole argument for switching regulators in lighting: they change the current without standing in its way, and a torch that burned two thirds of its battery in a resistor would be a poor torch.

$\frac{V _{f}}{V _{supply}} = \frac{1.7}{5.0} = 34% : the divider IS the efficiency$

Learning path

Resistance, then reactance

Leads to RC circuit resistance as a ratio, V/I, rather than a substance a component contains.

References (3)

The lesson's failure case — an LED holds a roughly fixed forward drop, so the resistor sees the supply voltage minus that drop: P. Horowitz & W. Hill, The Art of Electronics, 3rd ed. Cambridge University Press, 2015. ISBN 978-0-521-80926-9.
The three forms of resistive power, P = VI = I²R = V²/R, as set out in a standard circuits text: J. W. Nilsson & S. A. Riedel, Electric Circuits, 10th ed. Pearson/Prentice Hall, 2015. ISBN 978-0-13-376003-3.
Insight block 3 — the analogy Ohm built the law on: G. S. Ohm, Die galvanische Kette, mathematisch bearbeitet. Riemann, Berlin, 1827 — modelled directly on Fourier's Théorie analytique de la chaleur (1822).

By the 42Math team · Updated 2026-08-17

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

LED circuit - Typical LED current-limiting setup: with 3.3 V and 220 Ω, current is about 15 mA, which is in a safe range for many indicator LEDs.
light bulb - Mains-bulb example: at 230 V and 0.5 A, the equivalent resistance is 460 Ω, corresponding to roughly 115 W power draw.
short circuit - Near-short warning case: 12 V across 0.1 Ω gives about 120 A. This is dangerous current and can overheat wires or damage power supplies.
USB charger - USB-style output case: 2 A through 2.5 Ω implies 5 V, matching common USB power rails.

YouTube

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