Ohm's Law & DC/AC Electrical Power Calculator

Calculate exact Voltage (V), Current (I), Resistance (R), and Power (P) across DC and single-phase AC circuits using the 12-formula Ohm's Law Wheel with thermal copper resistance shift.

1.00 (Pure DC/Resistive)
20°C (Standard Room Temp)
Engineering & Appliance Circuit Presets
⚡ Estimate-First Circuit Intuition Challenge

Before checking the table below, can you guess the resulting Current (Amps) for this exact circuit setup?

Voltage (V)
12 V
Current (I) - DC/Resistive
3 A
Resistance (R) @ 20°C Base
4 Ω
Active Power (P)
36 W
AC Impedance & Copper Thermal Shift Analysis
AC Apparent Power (S = P / PF)36 VA
AC RMS Current (at PF 1.00)3 A
Hot Copper Resistance (20°C)4 Ω
Hot Conductor Current (20°C)3 A
Step-by-Step Mathematical Walkthrough
Step 1: Determine all 4 primary Ohm/Joule parameters from known pair
I = V / R = 12 / 4 = 3 A; P = V × I = 36 W
Step 2: Single-Phase AC Power Factor Adjustment (PF = cos θ = 1.00)
PF = 1.0 (Purely resistive or DC load) → True Power P = Apparent Power S = 36 VA; AC RMS Current = 3 A
Step 3: Copper Conductor Thermal Resistance Shift at 20°C (Base 20°C, α = 0.00393 /°C)
R_T = 4 Ω × [1 + 0.00393 × (20 - 20)] = 4 Ω → Hot Current I_hot = 3 A

About the Ohm's Law Calculator

Ohm's Law defines the relationship between voltage, current, and resistance in an electrical circuit. Formulated by the German physicist Georg Ohm, it is the foundational rule of electrical engineering, electronics design, and circuit diagnostics. The practical value of the law is that these quantities are not independent. Fix any two and the third is determined, and once power is included, knowing any two of the four fixes all four. That is why this calculator accepts whichever pair you happen to have measured rather than demanding a particular one: a multimeter reading of voltage and current is as good a starting point as the resistance printed on a component. Ohm published the relation in 1827 to a hostile reception — it was thought too simple to be a discovery — and it now underpins everything from sizing a resistor in a hobby circuit to deciding what voltage a national grid should run at. Both of those cases are worked through below, because they are the same equation applied at scales that differ by six orders of magnitude.

Mathematical Formula & Logic

The primary mathematical expression is: V = I × R Where: - V = Voltage in Volts (V) - I = Current in Amperes (A) - R = Resistance in Ohms (Ω) Additionally, Power (P) in Watts (W) can be calculated using: P = V × I or P = I² × R or P = V² / R

Step-by-Step Example

Calculate the current flowing through a 10 Ω resistor when connected to a 12V battery: I = V / R I = 12V / 10 Ω = 1.2 Amperes. Finishing the same circuit Knowing two quantities fixes all four, so the example above is only a quarter done: 1. Current: I = V / R = 12 / 10 = 1.2 A. 2. Power: P = V × I = 12 × 1.2 = 14.4 W, or equivalently V² / R = 144 / 10 = 14.4 W. That second figure is the one beginners omit, and it is the one that sets fire to things. A resistor is sold with two ratings, not one: its resistance and the power it can dissipate. The common quarter-watt resistor in a hobby kit would be asked to shed 14.4 W here — nearly sixty times its rating. It would char within seconds. The resistance value was never the constraint. A useful mental model Think of voltage as water pressure, current as the rate of flow, and resistance as the narrowness of the pipe. Raise the pressure and more flows; narrow the pipe and less does. Power is then the rate at which the flow does work, which is why it depends on both pressure and flow rather than either alone. Why the grid runs at high voltage The relation P = I² × R has a consequence that shaped the entire electricity network. Losses in a cable depend on the square of the current, not on the voltage, so pushing the same power at a higher voltage draws proportionally less current and wastes dramatically less. Send 100 kW down a line with 0.5 Ω of resistance: At 240 V the current is 416.67 A, and the line dissipates 416.67² × 0.5 = 86.8 kW. You would lose almost 87% of what you sent. At 24,000 V the current is 4.17 A, and the line dissipates 0.0087 kW. Essentially nothing. A hundredfold rise in voltage cuts the current a hundredfold and the losses ten-thousandfold, because the current term is squared. This single inequality is why transmission runs at hundreds of kilovolts and why transformers sit between the grid and your house. Where the law stops working Ohm's Law is a description of a particular kind of material, not a universal law of nature. It holds for metals and standard resistors at steady temperature. It fails for diodes, transistors, filament lamps and gas discharge tubes, whose resistance changes with the voltage applied — a filament bulb can be ten times more resistive when hot than when cold, which is why bulbs almost always fail at the moment you switch them on.

Reference Data & Values

metricunitformuladescription
Voltage (V)Volts (V)V = I × RElectrical potential difference
Current (I)Amperes (A)I = V / RFlow of electric charge carriers
Resistance (R)Ohms (Ω)R = V / IOpposition to charge flow
Power (P)Watts (W)P = V × IRate of electrical energy dissipation

Frequently Asked Questions

Ohm's Law states that the current passing through a conductor between two points is directly proportional to the voltage across the two points, and inversely proportional to the resistance between them.
No. Ohm's law only applies to linear, ohmic conductors (like standard metal resistors) under constant temperature. It does not apply to non-ohmic components like diodes, transistors, or gas discharge tubes where resistance changes with voltage.
In most metals, electrical resistance increases as temperature rises because thermal agitation increases electron collisions. In semiconductors, resistance decreases as temperature rises because more charge carriers are freed.
Work out the dissipation with P = V² / R, or P = I² × R, then choose a part rated comfortably above it. A 10 Ω resistor across 12 V dissipates 14.4 W, so a common quarter-watt component would be overloaded roughly sixty times and would char within seconds. Specifying at least double the calculated figure keeps the component running cool.
Because line losses follow P = I² × R, depending on the square of the current and not at all on the voltage. Sending 100 kW through a 0.5 Ω line at 240 V draws 416.67 A and wastes 86.8 kW — almost everything. The same power at 24,000 V draws 4.17 A and wastes 0.0087 kW. Raising voltage a hundredfold cuts losses ten-thousandfold, which is the whole reason for transformers and pylons.
P = V × I, P = I² × R and P = V² / R are algebraically identical once Ohm's Law is substituted in, so they always agree. Use whichever matches the two quantities you already know — that avoids computing an intermediate value and rounding it twice.
In this form, only for purely resistive loads such as a heater or an incandescent bulb. Once capacitance or inductance is present — motors, transformers, fluorescent lighting — resistance becomes impedance, current and voltage fall out of step, and real power is voltage × current × power factor. A power factor below one means the supply carries more current than the delivered power alone would suggest.
A tungsten filament is far less resistive when cold, sometimes a tenth of its hot resistance. The instant the switch closes it therefore draws a large inrush current before it heats and settles. That surge is the point of greatest stress, which is why a failing bulb almost always dies at switch-on rather than during use.