Why do I need Ohm's law?

Ohm's law, V = IR, links the push of a battery (voltage), how much a part resists (resistance), and how much electricity flows (current). With any two you can find the third. That is how you choose a resistor so an LED doesn't burn out, size a fuse, or check whether a wire will overheat.

9 V+R = 10 ΩI = 900 mAP = 8.1 Wslope = 1/R
Current
900 mA
Power (heat in the resistor)
8.1 W

Challenge: Get exactly 20 mA flowing, a typical LED current.

The push from the battery or power supply.
How hard the resistor squeezes the flow.

Play

Turn up the voltage to push more current. Turn up the resistance to squeeze it down.

Challenge: Get exactly 20 mA flowing, a typical LED current. The box under the picture turns green when you get it.

Stuck? Pick one of the examples from the “Try an example” menu, or press “New example.”

Understand

I=VRI = \frac{V}{R}

Ohm's law is a linear equation. Hold the resistance steady, and current is proportional to voltage:

I=1R VI = \frac{1}{R}\,V

That is a straight line through zero on the graph, with slope 1/R1/R. A bigger resistor makes a flatter line: the same voltage pushes less current. The dot shows your circuit on that line.

In the circuit drawing, the arrow thickness follows the current, and the glow around the resistor follows the power P=VIP = VI it turns into heat.

Set the resistance to zero and the formula tries to divide by zero. There is no answer, so the site shows #DIV/0!. In a real circuit, that is a short.

Use

Every input has a unit menu, so you can type values in the units you already have. Results follow your units.

Show the work

  1. Ohm’s lawV = IR ;\Rightarrow; I = \frac{V}{R}
  2. With your numbersI = \frac{9\,\mathrm{V}}{10\,\Omega} = 0.9\,\mathrm{A}
  3. PowerP = VI = (9\,\mathrm{V})(0.9\,\mathrm{A}) = 8.1\,\mathrm{W}

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  • LED resistor: R=(Vsupply−VLED)/ILEDR = (V_{\text{supply}} - V_{\text{LED}}) / I_{\text{LED}}. For a 5 V supply, a 2 V red LED, and 20 mA, that is 3/0.02=150 Ω3 / 0.02 = 150\,\Omega.
  • Resistor wattage: check P=I2RP = I^2R and choose a part rated for at least twice that.
  • Fault finding: measure V and I and compute R to spot a failing heater element or a corroded joint.

Real parts have tolerances (±1% to ±10%), and resistance changes with temperature, so leave margin.

For learning and estimation. Verify with applicable codes, standards, and a qualified professional before using in design, construction, or safety-critical work.

Cheat card

V=IRV = IR
I=VRI = \frac{V}{R}
R=VIR = \frac{V}{I}
P=VI=I2R=V2RP = VI = I^2 R = \frac{V^2}{R}
SymbolMeaningUnit
VVvoltage, the electrical pushV
IIcurrent, how much charge flowsA
RRresistance, how hard the part pushes backΩ
PPpower turned into heatW
  • Double the voltage, double the current. Double the resistance, half the current.
  • LED resistor: R = (supply voltage − LED voltage) ÷ LED current.
  • Pick a resistor rated for at least twice the power it will actually burn.

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Where it’s used

  • Electrical & Electronics
    Electricians and engineers use V = IR to size fuses, breakers, and wires, and to find faults with a meter.
  • Hobbies & Crafts
    Hobbyists pick the resistor that keeps an LED at a safe current on an Arduino or a battery pack.

Questions people ask

What does Ohm's law say in plain words?

Current equals voltage divided by resistance. More push means more flow, and more resistance means less flow.

What happens if resistance is zero?

The formula would divide by zero, so it has no answer. In real life that is a short circuit, and the current is limited only by the battery and wires, often dangerously. The site shows this as

How do I find the power a resistor uses?

Multiply voltage by current (P = VI). You can also use P = I²R or P = V²/R. All three give the same answer.