How this instrument works
Current is the rate at which charge moves past a point in a circuit, and Ohm's law states that it is proportional to the push behind it and inversely proportional to what resists it: I = V ⁄ R. Georg Simon Ohm established this relationship experimentally in 1827, testing wires of different lengths and thicknesses against a fixed voltage source and finding the ratio held steady for a given conductor. Divide the potential difference across a component, in volts, by its resistance, in ohms, and the result is the current, in amps, flowing through it.
The formula is a rearrangement of V = IR, which itself falls out of the microscopic relation J = σE — current density is proportional to electric field through a material's conductivity. Solve that macroscopic version for current and division is unavoidable: double the resistance at a fixed voltage and only half the current gets through, because a single ohm is defined as exactly the opposition that lets one volt push one amp through a conductor. The shape of the equation is not a convention chosen for convenience; it is what resistance means.
The formula only holds for ohmic conductors — materials whose resistance stays essentially constant regardless of the current flowing or the voltage applied, which covers resistors, copper wire, and simple heating elements across their normal operating range. It breaks down for diodes, LEDs, transistors, and thermistors, whose resistance shifts with current, voltage, or temperature, and it does not directly apply to AC circuits with capacitance or inductance, where impedance replaces plain resistance in the ratio.
- Enter the potential difference across the component in the Voltage field, in volts.
- Enter the component's resistance in the Resistance field, in ohms — this value must stay constant for the answer to be exact.
- Read the result in the Current field: the instrument divides voltage by resistance automatically to return amps.
- Sanity-check the instrument yourself first: 12 V into 4 Ω should return exactly 3 A, the simplest ratio in electronics.
Worked example — a 12 V accessory circuit through 4 Ω
A 12 V car battery feeds a resistive accessory — say a seat-heater element — with 4 Ω of resistance measured cold. Apply the formula directly: I = V ⁄ R = 12 ⁄ 4 = 3.0 A. The circuit draws exactly 3 amps, a figure worth checking against the fuse rating before the element is ever wired in, since a 2 A fuse would blow the first time the seat warmer is switched on.
Halve the resistance to 2 Ω on the same 12 V line and the draw rises to 6 A — half the resistance, double the current, the inverse relationship working exactly as the formula says. Raise the voltage instead, to a 24 V truck electrical system across the same 4 Ω element, and the current also climbs to 6 A; either change moves the current by the same ratio it moved the numerator or the denominator.
Questions
Why is Ohm's law written as I = V ÷ R instead of V = IR?
Both describe the same relationship; the form used depends on which quantity is unknown. Solving for voltage, V = IR is the natural shape; solving for current, given a known voltage and resistance, I = V ÷ R is the rearrangement that isolates it. This instrument treats current as the unknown, so voltage divided by resistance is the version it applies.
Does Ohm's law apply to every electrical component?
No. It only holds for ohmic materials — resistors, copper wire, and similar conductors whose resistance stays constant across the voltages and currents they normally see. Diodes, LEDs, transistors, and thermistors are non-ohmic: their resistance changes with current or temperature, so I = V ⁄ R gives at best a rough, instantaneous approximation rather than an exact answer for them.
What happens to the current if resistance doubles?
It halves, for a fixed voltage. The relationship is inversely proportional: I = V ⁄ R means current and resistance move in opposite directions once voltage is held steady. At 12 V, a 4 Ω load draws 3 A; the same 12 V across 8 Ω draws 1.5 A — doubling the resistance exactly halves the current.
Where does Ohm's law actually come from?
From Georg Simon Ohm's 1827 experiments relating the voltage across a conductor to the current through it, published in his book on the galvanic circuit. Microscopically it follows from current density being proportional to electric field within a material, J = σE; integrated over a real conductor's length and cross-section, that relation becomes the familiar V = IR.
Why might a real circuit draw less current than I = V ÷ R predicts?
Usually because resistance was measured cold. Most conductors heat up as current flows, and resistance in metals rises with temperature, so the actual resistance under load is higher than the value measured at room temperature, pulling the real current below the calculated figure. Incandescent filaments show this dramatically; a wirewound resistor stays far more stable.
Can this calculator be used on an AC circuit?
Only where the load is purely resistive and reactance is negligible — a heating element or incandescent bulb, for instance. Where a circuit includes a capacitor or inductor, impedance Z replaces R and carries a phase relationship this instrument does not model; that situation calls for an AC impedance calculation instead.