What does this calculator do?
Ohm’s law links voltage, current and resistance in a resistive element. It is one of the most used tools for sizing resistors, estimating currents and checking voltage drops in DC circuits.
V = I × R
If you know the current through a resistor and its value, multiply them to get the voltage. The same relation can be rearranged to find current or resistance.
Worked example
With I = 10 mA and R = 1 kΩ, the voltage drop is V = 0.01 × 1000 = 10 V.
Practical tips
- Convert milliamps to amps before applying the formula.
- Check the resistor’s power rating with P = I²R when the current is high.
- For non-linear components, simple Ohm’s law does not describe the whole behaviour.
Common mistake to avoid
Mixing up milliamps and amps multiplies the result by a thousand. Always check the unit before sizing a resistor.
Complete Ohm’s law formulas
Ohm’s law relates three quantities: if you know two, you can find the third. It is often remembered with the triangle of V over I and R:
- Voltage: V = I · R
- Current: I = V / R
- Resistance: R = V / I
- Power dissipated: P = V · I = I² · R = V² / R
- Consistent units: V in volts, I in amps, R in ohms. If you use kΩ and mA together, V comes out directly in volts (1 kΩ × 1 mA = 1 V).
This calculator finds the voltage drop V from the current and the resistance. To solve for the other quantities, rearrange the formula as shown above.
Typical values
| Voltage | Resistance | Current | Power | Typical case |
|---|---|---|---|---|
| 3.3 V | 10 kΩ | 0.33 mA | 1.1 mW | Pull-up resistor |
| 5 V | 220 Ω | 22.7 mA | 114 mW | Load on a 5 V pin |
| 9 V | 470 Ω | 19.1 mA | 172 mW | 9 V battery circuit |
| 12 V | 1 kΩ | 12 mA | 144 mW | Biasing at 12 V |
| 24 V | 4.7 kΩ | 5.1 mA | 123 mW | PLC input or indicator |
| 120 V | 14.4 Ω | 8.33 A | 1000 W | 1 kW heating element (120 V) |
| 230 V | 52.9 Ω | 4.35 A | 1000 W | 1 kW heating element (230 V) |
Most electronics cases stay below 1/4 W, but some come close: always check the resistor’s power rating.
Worked examples: voltage drop in a resistor and in a cable
Case 1. 10 mA flows through a 1 kΩ resistor: V = 0.01 × 1000 = 10 V. The power is P = 10 × 0.01 = 0.1 W, fine for a 1/4 W resistor.
Case 2. A supply cable has 0.15 Ω round trip and carries 8 A to a 12 V load. The drop is V = 8 × 0.15 = 1.2 V, so the load only gets 10.8 V (10 % less) and the cable dissipates 8² × 0.15 = 9.6 W. The fix is a thicker or shorter cable.
Common mistakes
- Mixing units: 10 mA × 1 kΩ is not 10 000 V; in SI units, 0.01 A × 1000 Ω = 10 V.
- Applying it to non-linear parts: diodes, LEDs and transistors have no fixed resistance; calculate the series resistor, not the semiconductor.
- Measuring resistance in a powered circuit: the ohmmeter injects its own current; with external voltage present the reading is wrong and the meter can be damaged.
- Forgetting power: the right resistance with too low a power rating ends in a burnt resistor.
- Ignoring temperature: the resistance of metals rises as they heat up.
Frequently asked questions
Does Ohm’s law work for AC?
Yes, with RMS values and resistive loads. With coils or capacitors, replace R by the impedance Z; see the RLC impedance calculator.
How do I find the current if I know voltage and resistance?
Divide: I = V / R. For example, 5 V across 220 Ω gives 5 / 220 ≈ 22.7 mA.
How do I combine several resistors before applying the law?
First find the equivalent resistance with the series resistance or parallel resistance calculator, then apply V = I · R.
For the history of the law, the resistivity of materials and why some components don’t follow it, read the Ohm’s law guide.
Educational tool. Check tolerances, datasheets, standards and real operating conditions before using the results in critical applications.