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Ohm's Law & Power Calculator

Enter any two of voltage, current, resistance and power — the solver derives the other two from V = I·R and P = V·I, with the full power wheel and the I–V line.

Calculator

Enter exactly two values; the other two are calculated.

volts (V)
amperes (A)
ohms (Ω)
watts (W)
Solved circuit
R = 6 Ω · P = 24 W
Derived from voltage and current.
Voltage
12 V
Current
2 A
Resistance
6 Ω
Power
24 W

Power wheel

Voltage (V)
12 V
given
Current (I)
2 A
given
Resistance (R)
6 Ω
calculated
Power (P)
24 W
calculated
Formula used: R = V / I · P = V · I
I–V characteristic

Straight line of current against voltage with slope 1/R through the origin, and the operating point (V, I) marked.

At this operating point

24 W

A current of 2 A flows through 6 Ω, dissipating 24 W of heat at 12 V.

Assumes an ideal ohmic resistor in a steady DC circuit (constant R, positive magnitudes). For the resistance-and-power pair the positive roots are returned.

A reference and planning tool — double-check important dates, figures and official requirements before you rely on them.

About this calculator

This is an Ohm's-law and DC-power solver. Supply any two of the four quantities — voltage V (volts), current I (amperes), resistance R (ohms) and power P (watts) — and it derives the remaining two, then shows the full "power wheel" and the resistor's current-versus-voltage line. It is the everyday tool for sizing a resistor, checking how much current a load will draw, or finding the heat a component dissipates, without re-deriving the formula each time.

How to read your results

The result panel shows all four quantities at once: the two you entered are flagged "given" and the two the calculator found are flagged "calculated". The formula line tells you exactly which rearrangement was used (for example, from V and I it computes R = V / I and P = V·I). The I–V chart plots the straight line I = V / R — its slope is 1 / R, so a steeper line means a smaller resistance — and marks the operating point (V, I) where your circuit actually sits.

How it's calculated

The solver works in SI base units throughout. From the two values you supply it picks the matching closed form: V & I → R = V/I, P = V·I; V & R → I = V/R, P = V²/R; V & P → I = P/V, R = V²/P; I & R → V = I·R, P = I²·R; I & P → V = P/I, R = P/I²; and R & P → I = √(P/R), V = √(P·R). Every quantity must be a positive, finite number, which rules out any divide-by-zero; a result that still came out non-finite (from an extreme combination) is flagged rather than shown. The I–V chart is the line I = V/R sampled from zero to 1.6× the operating voltage, with the operating point (V, I) marked. All four relations are exact algebraic rearrangements of Ohm's law and Joule's power law, so there is no iteration or approximation.

Worked example

A 12 V supply drives a 6 Ω resistor. Enter 12 in the Voltage box and 6 in the Resistance box; leave Current and Power blank.

The solver returns I = V / R = 12 / 6 = 2 A and P = V² / R = 144 / 6 = 24 W. The power wheel shows V = 12 V and R = 6 Ω as given, with I = 2 A and P = 24 W calculated, and the operating point (12 V, 2 A) is marked on the I–V line.

Frequently asked questions

What is Ohm's law in one line?

Ohm's law states that the voltage across an ohmic resistor equals its current times its resistance: V = I·R. Rearranged, that is I = V / R and R = V / I. It holds for an ideal resistor whose resistance does not change with voltage or current.

How are voltage, current, resistance and power related?

Three relations connect them. Ohm's law links voltage, current and resistance as V = I·R. DC power is P = V·I, which substitution turns into P = I²·R and P = V² / R. Together these let you find any two of V, I, R and P once the other two are known.

Why do I have to enter exactly two values?

Two known quantities fix the operating point of an ohmic DC circuit completely — the other two follow from a single formula with no ambiguity. One value is not enough to solve, and three or more would over-constrain the circuit (and usually contradict each other), so the calculator asks for exactly two.

What happens when I only know resistance and power?

Resistance and power give P = I²·R and P = V² / R, so the calculator takes the positive square roots: I = √(P / R) and V = √(P·R). For example, R = 100 Ω with P = 1 W gives I = √(1/100) = 0.1 A and V = √(100) = 10 V. Only the positive (physical) magnitudes are reported.

Does this work for AC circuits?

It is a DC (resistive) solver. It assumes an ideal ohmic resistor with constant resistance and no reactance, so it does not model the phase, impedance or power factor of an AC circuit. For purely resistive AC values it still gives the right RMS magnitudes, but reactive (inductive or capacitive) loads need an impedance calculation instead.

Popular scenarios

Popular scenarios

Sources

Reviewed by the YouCalc Team · Last reviewed

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