Ohm's Law, Rearranged Three Ways: Solving for Voltage, Current, or Resistance

New to Ohm’s law, it’s easy to come away thinking there are three formulas to memorize: V = I × R, I = V ÷ R, and R = V ÷ I. There’s really only one. It’s the same relationship between the same three quantities, and which “formula” you reach for depends entirely on which two quantities you already know. This is a companion to Ohm’s Law Explained, focused specifically on drilling all three directions until picking the right rearrangement stops requiring any thought at all.

One relationship, three starting points

Every Ohm’s law problem hands you exactly two of the three quantities and asks for the third. There are only three possible starting pairs, which means there are only three possible things to solve for:

• Know voltage and resistance → solve for current: I = V ÷ R
• Know voltage and current → solve for resistance: R = V ÷ I
• Know current and resistance → solve for voltage: V = I × R

The triangle trick (V on top, I and R on the bottom) exists purely to make this lookup instant: cover the letter you want, and whatever’s left tells you whether to multiply or divide. But the triangle is a memory aid, not the actual understanding — the real skill is recognizing which two quantities a problem hands you, since that alone tells you which rearrangement to use.

Direction 1: solving for current

This is the direction you reach for when you know your supply and the resistance in the circuit, and want to know what current will actually flow — the most common real-world question, since it’s how you check whether a circuit is safe before powering it up.

• 12V across a 4Ω resistor: I = 12 ÷ 4 = 3A.
• A 5V supply across a 220Ω resistor: I = 5 ÷ 220 ≈ 0.0227A (about 22.7mA).
• A single 1.5V AA cell across a 68Ω resistor: I = 1.5 ÷ 68 ≈ 0.0221A (about 22.1mA).
• A 6V supply across a 330Ω resistor: I = 6 ÷ 330 ≈ 0.0182A (about 18.2mA).

Notice the last three all land in the same rough neighborhood — roughly 18 to 23mA — despite using different voltages and different resistors. That’s a reminder that it’s the ratio of voltage to resistance that sets the current, not either number on its own: a bigger resistor paired with a bigger voltage can land close to the same current as a smaller resistor paired with a smaller voltage, even without landing on exactly the same number.

Direction 2: solving for resistance

This direction answers “what resistor do I need?” — you know your supply voltage and the current you’re targeting, and you need the resistance that gets you there. It’s the exact calculation behind sizing a current-limiting resistor for an LED, just without yet subtracting the LED’s forward voltage first (see Choosing the Right Resistor for an LED for that fuller version).

• A 9V supply, targeting 0.02A (20mA): R = 9 ÷ 0.02 = 450Ω.
• A 3V supply, targeting 0.015A (15mA): R = 3 ÷ 0.015 = 200Ω.
• A 10V supply, targeting 0.1A (100mA): rearranged from I = V ÷ R, R = V ÷ I = 10 ÷ 0.1 = 100Ω.

The pattern to notice: a higher target current always calls for a smaller resistance at the same voltage, and a lower target current calls for a larger one — more resistance always means less current, for a fixed voltage. That inverse relationship is worth having as intuition, not just as an equation, because it lets you sanity-check an answer before you even finish the arithmetic: if you want less current and your calculated resistance came out smaller than what you started with, something went wrong.

Direction 3: solving for voltage

This direction answers “what voltage will this drop, or what voltage do I need to supply?” — you know the current flowing and the resistance it’s flowing through, and you want the voltage across that resistance. It shows up constantly in circuits with more than one resistor, where you need to know how much of the total supply voltage lands across each individual part (the subject of Series vs Parallel: What Each Wiring Actually Does to Current).

• 0.5A through a 10Ω resistor: V = 0.5 × 10 = 5V.
• 0.1A through a 100Ω resistor: V = 0.1 × 100 = 10V.
• 0.02A (20mA) through a 68Ω resistor: V = 0.02 × 68 = 1.36V.

This is also the direction that answers “how much voltage did this component actually use up?” — useful when you’re trying to account for where every volt of a supply went across several components in a row.

Power comes free once you have any two

Whichever direction you solve, power is the same follow-up step: P = V × I. Once a problem hands you two of the three basic quantities, you can find the third and the power without any extra formula to memorize, because power only ever needs voltage and current, and by the time you’ve solved for the missing quantity, you have both.

• Solving for current gave 12V and 3A above → power is 12 × 3 = 36W.
• Solving for resistance gave 9V and 0.02A above → power is 9 × 0.02 = 0.18W.
• Solving for voltage gave 5V and 0.5A above → power is 5 × 0.5 = 2.5W.

There are two other forms of the power formula — P = I² × R and P = V² ÷ R — that skip straight to power without finding the missing quantity first, useful when a problem only ever asks for power and you don’t need the third basic quantity for anything else. But all three power formulas are just P = V × I with one of V or I swapped out using Ohm’s law itself, not new physics.

Why the triangle can mislead if you lean on it too hard

The triangle trick is genuinely useful for beginners, but it has a blind spot: it tells you whether to multiply or divide, but not which order to divide in, and it’s easy to misremember which letter goes on top. R = V ÷ I and R = I ÷ V look similarly plausible if you’re recalling the triangle from memory under pressure, but only one of them is dimensionally sensible — resistance has to get larger as voltage increases (for fixed current) and smaller as current increases (for fixed voltage), which only R = V ÷ I satisfies. Checking a rearrangement against that kind of common-sense direction — does the answer move the way it should when one input goes up? — catches a flipped formula faster than trying to re-picture the triangle from scratch.

A mixed drill: identify the direction first

The hardest part of Ohm’s law usually isn’t the arithmetic — it’s correctly identifying which two quantities a problem actually gives you before doing any math. Try sorting these by direction before solving:

• “A 220Ω resistor is connected across a 5V supply. What current flows?” (Given voltage and resistance → solve for current: I = 5 ÷ 220 ≈ 0.0227A, about 22.7mA.)
• “A circuit needs to draw exactly 0.1A from a 10V supply. What resistance achieves that?” (Given voltage and current → solve for resistance: R = 10 ÷ 0.1 = 100Ω.)
• “0.5A is measured flowing through a 10Ω resistor. What voltage is across it?” (Given current and resistance → solve for voltage: V = 0.5 × 10 = 5V.)

Once identifying the direction becomes automatic, the arithmetic is the easy part — and the Ohm’s Law Calculator handles that arithmetic instantly once you know which two values to type in.

A second round, without the direction labeled

The real test is doing this without the direction spelled out first — that’s the skill a calculator can’t practice for you. Three more, in no particular order:

• A resistor measures 100Ω on a multimeter, and 0.1A is flowing through it in the circuit. What’s the voltage across it? (Given current and resistance: V = 0.1 × 100 = 10V.)
• A project runs from a 3V coin-cell supply and needs to hold its current to 15mA (0.015A). What resistor value achieves that? (Given voltage and current: R = 3 ÷ 0.015 = 200Ω.)
• A 200Ω resistor sits across a 3V supply. What current does it draw? (Given voltage and resistance: I = 3 ÷ 200 = 0.015A, exactly 15mA — the reverse of the previous problem, confirming the two directions are consistent with each other.)

That last pair is worth sitting with: the same three numbers (3V, 200Ω, 15mA) answer two different questions depending on which one is treated as unknown. That’s the entire point of thinking of Ohm’s law as one relationship rather than three separate formulas — the same three quantities, the same fixed relationship between them, just a different one singled out as the question each time. (One practical note: because the Ohm’s Law Calculator displays current rounded to the nearest hundredth of an amp — fine resolution for amp-scale circuits, coarser at milliamp scale — a milliamp-level result like this 15mA is easiest to read precisely by entering resistance in kilohms instead of ohms and reading the current field as milliamps directly, since volts ÷ kilohms equals milliamps by the same underlying formula.)

Where this becomes second nature

Rearranging Ohm’s law stops feeling like a lookup once you’ve used all three directions on real projects rather than just practice problems: solving for current to check a circuit is safe before powering it up, solving for resistance to size a component before you buy it, and solving for voltage to understand where the supply’s voltage actually goes once a circuit has more than one part in it. All three are the same short equation, asked three different ways.