Ohm's Law Solver (V = IR)
Solve Ohm's law for voltage, current or resistance: pick the unknown, enter the other two values and get V = I × R rearranged for you.
Last updated: October 2026
Result
20 V
Formula below · 3 sources (britannica.com, allaboutcircuits.com, electronics-tutorials.ws) · Updated Oct 2026
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About this calculator
Ohm's Law is the cornerstone of electronics: the voltage across a resistor equals the current through it multiplied by its resistance, V = I × R. Rearranged, it also gives current as I = V ÷ R and resistance as R = V ÷ I. This solver lets you pick which quantity is unknown and enter the other two. Voltage (V) is measured in volts and represents the electrical "pressure" pushing charge through a circuit; current (I) in amperes is the rate of charge flow; and resistance (R) in ohms (Ω) is how strongly a component opposes that flow. The three are locked together — fix any two and the third is determined. For example, 2 amps flowing through a 10-ohm resistor must have 20 volts across it (2 × 10). The law holds exactly for ideal "ohmic" components like resistors at constant temperature, and approximately for many real components. It does not apply directly to non-ohmic devices such as diodes, transistors, and incandescent bulbs, whose resistance changes with voltage, current, or temperature. To use the solver, choose the quantity you want to find and enter the two known values; the field for the unknown is ignored, so whatever it holds does not matter. Watch your units: amperes, not milliamps; ohms, not kilo-ohms — mixing prefixes is the most common source of wrong answers. Ohm's Law pairs naturally with the power equations (P = V × I = I²R = V²/R), which let you find power dissipation once any two of voltage, current, and resistance are known.
How to use
Pick the quantity you want in "Solve for" and enter the other two. The field for the unknown is ignored. Example (the defaults): solving for voltage with 2 A through a 10 Ω resistor gives V = 2 × 10 = 20 V. Solving for current with 12 V across 6 Ω gives I = 12 ÷ 6 = 2 A. Solving for resistance with 9 V and 0.03 A (30 mA) gives R = 9 ÷ 0.03 = 300 Ω.
Frequently asked questions
What do I put in the field I am solving for?
Anything; the calculator ignores it. Choose the unknown in "Solve for" and fill in the other two. For example, to find voltage, enter the current and the resistance.
What units should I use?
Use base SI units: volts for voltage, amperes for current, and ohms for resistance. The most common mistake is entering milliamps instead of amps (1000 mA = 1 A) or kilo-ohms instead of ohms (1 kΩ = 1000 Ω), which throws the answer off by orders of magnitude. If your component is rated in milliamps or kilo-ohms, convert first: 50 mA becomes 0.05 A, and 4.7 kΩ becomes 4700 Ω. Keeping everything in base units guarantees the result comes out in volts, amps, or ohms as expected. When in doubt, write out the powers of ten before entering values.
Does Ohm's Law work for all electrical components?
No. Ohm's Law applies exactly only to "ohmic" components — chiefly resistors — whose resistance stays constant regardless of the voltage or current applied. Many real-world devices are non-ohmic: diodes conduct only above a threshold voltage and then barely increase in resistance; incandescent bulbs have a resistance that rises sharply as the filament heats; and transistors and LEDs have highly non-linear behaviour. For those, you cannot simply plug values into V = IR and trust the result across their operating range. Ohm's Law also assumes steady (DC) conditions; in AC circuits with capacitors and inductors you need impedance, not plain resistance.
What are common mistakes with Ohm's Law calculations?
The number-one mistake is unit mismatch — mixing milliamps with ohms, or kilo-ohms with volts — which produces answers off by factors of 1000. Another is forgetting that the law applies to a single resistive element or an equivalent resistance; for networks you must first combine resistors in series (add them) or parallel (reciprocal sum) before applying it. People also misuse it on non-ohmic parts like diodes. A subtler error is confusing source voltage with the voltage across one component in a circuit with several — Ohm's Law relates the voltage across a resistor to the current through that same resistor, not to the whole supply.
When should I use the power equations instead?
Use the power equations when you care about energy dissipation, heat, or component ratings rather than just the voltage-current-resistance relationship. Power is P = V × I, and by substituting Ohm's Law you also get P = I²R and P = V²/R. These tell you how many watts a resistor must dissipate — critical for choosing a component that will not overheat. For example, a resistor passing 0.5 A at 10 V dissipates 5 W and needs an adequately rated part. Ohm's Law finds the missing electrical quantity; the power equations tell you whether your components can handle it safely.
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