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Ohm’s Law Calculator

Set any two of voltage, resistance and current and read the third — with the power, the heating in the resistor and the safety headroom shown at the same time, because those are what decide whether the part survives.

Live circuit
+12.0 V6.0 ΩA2.00 AP = 24.0 W
Live equations
V = I × R
12.0 V = 2.00 A × 6.0 Ω
I = V / R
2.00 A = 12.0 V / 6.0 Ω
P = V × I
24.0 W = 12.0 V × 2.00 A
Power dissipation
24.0 W
μWmWW100+ W
Safety check
ℹ️ All values within safe range
Relationship triangle
VIR×
To find V: cover V → I × R = 2.00 × 6.0 = 12.0
To find I: cover I → V / R = 12.0 / 6.0 = 2.00
To find R: cover R → V / I = 12.0 / 2.00 = 6.0

Ohm's Playground

Explore wider voltage and resistance ranges, pin multiple R curves to compare them on the V vs I graph, and log measurements in the table below.

Current
120.00 mA
Power
1.44 W
V vs I graph
010203040500167m333m500mVoltage (V)Current
Measurement log — release voltage slider to record a row

No data yet — move the voltage slider.

One law, three rearrangements

Ohm’s Law says the current through a conductor is proportional to the voltage across it: V = I × R. Everything else is that one statement rearranged. Cover the quantity you want on the V-I-R triangle and what remains is the calculation — cover V and you are left with I × R, cover I and you are left with V / R.

Power is the part that burns things

The law tells you the current; it says nothing about whether the resistor survives. That is P = V × I, and because V and I are themselves linked, it can be written as I²R or V²/R — useful when you only know one of them. Power rises with the square of current, so doubling the current quadruples the heat. A resistor passing 100 mA at 100 Ω dissipates 1 W and needs a 2 W part; the same resistor at 200 mA needs 8 W.

Where the law stops applying

Ohm’s Law describes linear (ohmic) elements — ones whose V-I graph is a straight line through the origin. Resistors qualify. Diodes, transistors, lamps and thermistors do not: their resistance changes with operating point, so no single R describes them. On an AC circuit with capacitance or inductance, R is replaced by impedance Z and the same form holds as V = I × Z, but with phase to track as well.

Learn more → Ohm’s Law — Learn

Quick experiments

  • Watch power outrun current. Hold the resistance at 6 Ω and take the voltage from 12 V to 24 V. The current doubles from 2 A to 4 A, but the power goes from 24 W to 96 W — four times, not two, because P = V × I and both terms grew.
  • Find the short-circuit point. Drag the resistance down towards 0.1 Ω. Current shoots past 100 A and the safety panel turns red. This is what a dead short looks like numerically: the only thing limiting the current is the wiring itself.
  • Approach an open circuit. Take the resistance up to 10 kΩ at 12 V. Current falls to about 1.2 mA and the resistor barely warms. Add a decade more and it is microamps — the reason a multimeter's 10 MΩ input barely disturbs the circuit it measures.
  • Size a resistor for an LED. A red LED drops about 2 V and wants 20 mA from a 5 V supply, so the resistor must drop 3 V. Set 3 V and adjust R until the current reads 20 mA — you should land on 150 Ω, and the power readout confirms 60 mW, well inside a ⅛ W part.
  • Prove the square law on heating. Fix the voltage and halve the resistance. Current doubles and power quadruples. This is why a slightly undersized resistor does not run slightly hot — it runs dramatically hot.

Formula reference

Ohm's Law
V=I×RV = I \times R

The definition. Everything below is this rearranged.

Solving for current or resistance
I=VRR=VII = \frac{V}{R} \qquad R = \frac{V}{I}

Cover the unknown on the V-I-R triangle.

Power — three equivalent forms
P=V×I=I2R=V2RP = V \times I = I^2 R = \frac{V^2}{R}

Pick whichever two quantities you already know.

Resistor power rating
Prating2×PactualP_{rating} \geq 2 \times P_{actual}

Derating headroom; more in still air or at high ambient.

Series resistor for a known drop
R=VsupplyVloadItargetR = \frac{V_{supply} - V_{load}}{I_{target}}

The LED resistor calculation.

Conductance
G=1RI=G×VG = \frac{1}{R} \qquad I = G \times V

In siemens. Convenient for parallel branches.

SymbolMeaningUnit
VVVoltage across the elementV
IICurrent through the elementA
RRResistanceΩ
PPPower dissipated as heatW
GGConductance, the reciprocal of RS

Common mistakes

  • Applying Ohm's Law to a diode, LED or transistor.

    These are non-linear — their V-I curve is not a straight line, so no single R describes them. An LED does not have a resistance; it has a forward voltage. Size the series resistor from the voltage left over after that drop, not from the LED itself.

  • Calculating current but never checking power.

    The law tells you nothing about whether the part survives. A 100 Ω resistor at 30 V passes a harmless-looking 300 mA — and dissipates 9 W, which will destroy a quarter-watt part in seconds. Always compute P = I²R afterwards.

  • Mixing units — millivolts with amps, or kilohms with milliamps.

    The formula is only valid in base units. 5 V across 2 kΩ is 2.5 mA, not 2.5 A. Convert everything to volts, amps and ohms first, or track the prefixes deliberately: V / kΩ gives mA directly.

  • Choosing a resistor rated for exactly the calculated power.

    Ratings assume free air at 25 °C. A resistor dissipating its full rating runs extremely hot and drifts in value. Pick at least double, and more if it is enclosed or near other heat sources.

  • Assuming resistance is constant as things heat up.

    It is not. A tungsten lamp filament measures a few ohms cold and ten times that when glowing, which is why lamps draw a large inrush current at switch-on. Metal-film resistors are stable enough to ignore this; hot components are not.

Frequently asked questions

How do I calculate current from voltage and resistance?

Divide the voltage by the resistance: I equals V over R. Twelve volts across six ohms gives two amps. Keep everything in base units first, because volts divided by kilohms gives milliamps, not amps.

What size resistor do I need for an LED?

Subtract the LED's forward voltage from the supply, then divide by the current you want. A red LED dropping about two volts, run at twenty milliamps from five volts, needs three volts across the resistor, so R equals 3 divided by 0.02, which is 150 ohms.

Does Ohm's Law work for AC circuits?

For a purely resistive load, yes, using RMS voltage and current. Once capacitors or inductors are involved, resistance is replaced by impedance and the relation becomes V equals I times Z, with a phase angle to track as well.

Why does Ohm's Law not apply to a diode or transistor?

They are non-linear. Their current does not rise in proportion to voltage, so their V-I curve is not a straight line and no single resistance describes them. Ohm's Law holds only for linear, or ohmic, elements such as resistors.

What power rating should I choose for a resistor?

Work out P equals I squared times R, then pick a part rated at least twice that. Ratings assume free air at room temperature, so a resistor run at its exact rating gets very hot and drifts in value.

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