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Ohm's Law Calculator

Ohm's Law — V = I × R — is the first thing every circuit problem needs. Enter any two of voltage (V), current (I) and resistance (R), and this calculator solves the third, then derives power dissipation (P) from the result. Unit prefixes (kΩ, mA, µF-style) are supported so lab values like 4.7 kΩ work directly.

Voltage (V)
Current (I)
Resistance (R)

Solved circuit values

Solved

Voltage
9 V
Current
19.15 m A
Resistance
470 Ω
Power
172.34 m W

How it works

  1. Enter any two of the three quantities: voltage, current, resistance.
  2. Pick a unit prefix for each (none, k, M, m, µ, n) — values are converted automatically.
  3. The calculator solves the missing quantity from V = I × R.
  4. Power is derived from the solved values: P = V × I.
  5. Every field updates as you type; entering fewer than two values shows a helpful hint.

The formula

Ohm's law

V = I × R

Voltage (volts) = current (amps) × resistance (ohms). Rearranged: I = V ÷ R and R = V ÷ I.

Power

P = V × I = V² ÷ R = I² × R

Power in watts, derived once two electrical quantities are known.

Worked example

Worked example — a 9 V battery and a 470 Ω resistor

  1. V = 9 V, R = 470 Ω → I = 9 ÷ 470
  2. I = 0.0191 A = 19.1 mA
  3. P = 9 × 0.0191 = 0.172 W = 172 mW

The circuit draws 19.1 mA and dissipates 172 mW — a ¼ W resistor handles it comfortably.

Common mistakes

  • Mixing prefixes — 4.7 kΩ is 4700 Ω, not 4.7 Ω. Select the k prefix in the dropdown.
  • Forgetting that R must be positive — a zero or negative resistance is invalid.
  • Using AC peak values as if they were RMS — this calculator assumes DC or RMS values.

Frequently asked questions

What if I only know one value?

Ohm's law needs two of the three to solve the rest — the calculator shows a hint asking for a second value.

Does this work for AC circuits?

For resistive (non-reactive) AC circuits, V, I and R can be RMS values and the relationship still holds. For circuits with capacitors or inductors, use impedance (Z) instead of resistance.

How do I know if a resistor can handle the power?

Compare the calculated power to the resistor's rating (1/8 W, 1/4 W, 1 W…). A good rule: use a rating at least twice the calculated dissipation.

Can it calculate power from voltage and resistance alone?

Yes — enter V and R, and the calculator derives both I and P (using P = V² ÷ R).

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