Ohm's Law Calculator
Enter any 2 values — solves the rest
Try an example
Solved
12 V
2 A
6 Ω
24 W
V = IR P = VI
About the Ohm's Law Calculator
Ohm's law states that the current through a conductor is proportional to the voltage across it and inversely proportional to its resistance. Expressed as V = IR, it is the most-used relationship in electrical work: three quantities, any two of which give the third.
A useful mental model treats voltage as pressure, current as flow rate, and resistance as pipe narrowness. Increase the pressure and more flows; narrow the pipe and less does. The analogy is imperfect — electrons do not accumulate the way water does — but it correctly predicts the direction of every change, which is most of what you need when troubleshooting.
In practice Ohm's law is almost always used alongside the power relationship P = VI. Together they answer the questions that actually matter on a bench or in a panel: what resistor do I need, how much heat will this dissipate, is this wire thick enough, and why is that component running hot. Knowing any two of voltage, current, resistance and power determines the other two.
Formula
V = I × R | I = V / R | R = V / I | P = V × I = I²R = V²/R
- V
- Voltage, or potential difference, in volts (V)
- I
- Current in amperes (A)
- R
- Resistance in ohms (Ω)
- P
- Power in watts (W)
The three forms of V = IR are the same equation rearranged; there is nothing to memorise beyond the original. The power forms are derived by substituting Ohm's law into P = VI — replacing V with IR gives P = I²R, and replacing I with V/R gives P = V²/R.
Which power form to use depends on what you know. P = I²R is the one to reach for when analysing losses in wiring, because current is what flows through the whole run and resistance is a property of the conductor. P = V²/R suits fixed-voltage loads like heaters and resistive dividers.
Units must be base units. Milliamps must become amps (divide by 1,000) and kilohms must become ohms (multiply by 1,000) before the numbers go in. A convenient shortcut for electronics: volts divided by kilohms gives milliamps directly, which avoids most of the decimal-point errors.
Worked Examples
A basic resistive load
A 4 Ω resistor across a 12 V supply. Find the current and the power dissipated.
Result: 3 A flowing, 36 W dissipated as heat
All three power formulas must agree — a fast check that the arithmetic is right. Note that 36 W in a small resistor is a lot of heat; a component rated below that will fail, often within seconds.
Sizing a current-limiting resistor for an LED
Driving a red LED with a 2 V forward voltage at 15 mA from a 5 V supply. The classic first electronics calculation.
Result: Use 220 Ω; a standard 1/4 W resistor is ample at 41 mW
Subtracting the LED forward voltage first is the step most often skipped. Using the full 5 V would suggest 333 Ω and run the LED at only 9 mA — visibly dim. Always round resistance up, since that reduces current and protects the LED.
Voltage drop in a long cable run
A 12 V circuit drawing 10 A through 20 metres of cable with a total round-trip resistance of 0.16 Ω.
Result: Only 10.4 V reaches the load and 16 W is lost heating the cable
Well beyond the 3% drop normally considered acceptable, so this run needs thicker cable. Because loss goes as I², low-voltage high-current systems suffer far worse than mains-voltage ones delivering the same power — the reason transmission grids run at hundreds of kilovolts.
How to Use the Result
Series and parallel resistance
Ohm's law applies to a single resistance, so multi-component circuits need combining first:
- —In series, resistances add: R_total = R₁ + R₂ + R₃. Current is identical through every component, while voltage divides in proportion to each resistance.
- —In parallel, conductances add: 1/R_total = 1/R₁ + 1/R₂ + 1/R₃. Voltage is identical across every branch, while current divides inversely with resistance.
- —For exactly two resistors in parallel, the shortcut R = (R₁ × R₂) / (R₁ + R₂) is quicker than reciprocals.
- —Parallel resistance is always lower than the smallest individual resistor. If your parallel result is higher than any branch, you have inverted something.
- —Equal resistors in parallel divide simply: n identical resistors of value R give R/n.
Choosing a resistor power rating
Resistance sets the current; power rating determines whether the component survives. A resistor dissipating more than its rating will drift in value, discolour, and eventually fail open or short.
Standard through-hole ratings are 1/8 W, 1/4 W, 1/2 W, 1 W and 2 W. Common practice is to specify at least twice the calculated dissipation, giving headroom for supply variation and poor airflow. In sealed enclosures or above 50 °C ambient, derate further — manufacturer curves typically require 50% derating by 100 °C.
The 41 mW in the LED example is comfortable on a 1/4 W part. The 36 W in the first example needs a wirewound or aluminium-clad resistor on a heatsink, not anything from a standard assortment kit.
Practical troubleshooting patterns
Most faults show a recognisable signature once you measure two of the three quantities:
- —Correct voltage present but no current means an open circuit — a broken conductor, blown fuse or bad connection presents effectively infinite resistance.
- —Excessive current with normal voltage means resistance has fallen: a short circuit, or a component that has failed shorted.
- —Voltage sagging under load points at excessive source resistance — a tired battery, undersized cable, or a corroded connection adding resistance in series.
- —A component running hotter than expected is dissipating more power than intended. Measure the actual voltage across it and current through it rather than trusting nominal values.
Edge Cases and Common Mistakes
Many components are not ohmic
Ohm's law describes a linear relationship, and plenty of components do not have one. Diodes and LEDs have an exponential voltage-current curve with a roughly constant forward voltage. Incandescent bulbs increase resistance dramatically when hot — a 60 W bulb measures around 20 Ω cold and near 240 Ω at operating temperature, which is why they draw a large inrush current and usually fail at switch-on.
Transistors, thermistors, varistors and any semiconductor junction are similarly non-linear. For these, V = IR gives the resistance only at one specific operating point, not a property of the device.
AC circuits need impedance rather than resistance
With alternating current, capacitors and inductors oppose current in a frequency-dependent way that also shifts the phase between voltage and current. The generalisation is V = IZ, where impedance Z combines resistance with reactance.
A consequence is that P = VI overstates real power in AC circuits with reactive loads. True power is P = VI × cos(φ), where cos(φ) is the power factor. For a purely resistive load the power factor is 1 and the simple form is correct; for motors and switch-mode supplies it can be 0.6 or lower.
Resistance changes with temperature
Copper resistance rises roughly 0.4% per °C. A cable run calculated at 20 °C will have about 8% more resistance at 40 °C, which increases voltage drop and heating — a feedback loop that matters in tightly loaded circuits and in attics or engine bays.
A zero-ohm path is a short circuit
Setting R = 0 in I = V/R gives infinite current. Reality limits it to whatever the source and wiring resistance allow, which is typically hundreds or thousands of amps from a mains supply or a lithium battery. This is what fuses and breakers exist to interrupt, and why a calculated result of near-zero resistance should be treated as a design error rather than a valid answer.
Frequently Asked Questions
Which formula should I use for power?
Use whichever matches what you already know. P = VI when you have measured both voltage and current. P = I²R for losses in wiring, since current is common to the run. P = V²/R for fixed-voltage resistive loads. All three give identical answers on a resistive circuit, so they double as a check on each other.
Why does my LED need a resistor at all?
Because an LED is not ohmic. Its forward voltage barely rises as current climbs, so with no resistance limiting it, current runs away until the junction overheats — usually within a fraction of a second. The resistor supplies the missing current limiting, which is why the calculation starts by subtracting the forward voltage.
How do I find current if I only know power and voltage?
Rearrange P = VI to I = P / V. A 1,200 W appliance on a 120 V supply draws 10 A. Then Ohm's law gives the effective resistance: R = V / I = 12 Ω. This is the standard route for sizing breakers and cables from an appliance nameplate.
Does Ohm's law apply to AC?
In modified form. Replace resistance with impedance: V = IZ. For purely resistive AC loads such as heaters and incandescent lamps, the ordinary form works fine using RMS values. For anything with significant capacitance or inductance, you need impedance and a power factor.
What resistor value should I pick if my calculation gives a non-standard number?
Round to the nearest standard E12 or E24 value, and prefer rounding up for current-limiting applications since that lowers current. Then recalculate the actual current at the chosen value to confirm it is still acceptable — as in the LED example, where 220 Ω gave 13.6 mA against a 15 mA target.
Why do transmission lines use such high voltages?
Because loss is I²R. Delivering a fixed amount of power at ten times the voltage requires one tenth the current, which cuts resistive loss to one hundredth. The resistance of the conductor is unchanged; only the current matters, and it enters squared.