Enter any two of voltage, current, resistance and power, typed with prefixes such as 4k7, 10m or 2.2µ, and this Ohm’s law calculator gives the other two. The result also shows the smallest chip resistor that carries the power with a 2× margin.
Resistance R
1 kΩ
Chip ratings are Yageo RC_L standard-power values at 70 °C ambient, falling linearly to zero at 155 °C (125 °C for 0201). Other series rate the same size differently: check the datasheet of the part you buy.
Chip resistor sizes for this load (Yageo RC_L ratings at 70 °C)
| Size | Rated power | Max voltage | Load | Verdict |
|---|---|---|---|---|
| 0201 (0603) | 1/20 W | 25 V | 50 % of rating | OK with 2× margin |
| 0402 (1005) | 1/16 W | 50 V | 40 % of rating | OK with 2× margin |
| 0603 (1608) | 1/10 W | 75 V | 25 % of rating | OK with 2× margin |
| 0805 (2012) | 1/8 W | 150 V | 20 % of rating | OK with 2× margin |
| 1206 (3216) | 1/4 W | 200 V | 10 % of rating | OK with 2× margin |
| 1210 (3225) | 1/2 W | 200 V | 5 % of rating | OK with 2× margin |
| 2010 (5025) | 3/4 W | 200 V | 3.33 % of rating | OK with 2× margin |
| 2512 (6332) | 1 W | 200 V | 2.5 % of rating | OK with 2× margin |
Ohm’s law is V = I × R. Combined with power, P = V × I, any two of the four quantities give the other two. For example, 12 V across 1 kΩ drives I = 12 ÷ 1000 = 12 mA and dissipates P = 12² ÷ 1000 = 144 mW, so a 1210 (3225 metric), 1/2 W resistor or larger is needed for a 2× margin.
Two relations connect the four quantities. The Ohm’s law equation links voltage, current and resistance; the power law links power to any two of them.
Substituting one into the other gives the Ohm’s law chart below: twelve forms, three for each quantity, as on an Ohm’s law wheel. Use RMS values for AC through a resistor; the power is then the average power.
| To find | Form 1 | Form 2 | Form 3 |
|---|---|---|---|
| Voltage V | V = I × R | V = P ÷ I | V = √(P × R) |
| Current I | I = V ÷ R | I = P ÷ V | I = √(P ÷ R) |
| Resistance R | R = V ÷ I | R = V² ÷ P | R = P ÷ I² |
| Power P | P = V × I | P = I² × R | P = V² ÷ R |
A resistor across a 9.00 V battery draws 3.00 mA. R = V ÷ I = 9 ÷ 0.003 = 3 kΩ, and it dissipates P = V × I = 27 mW. With a 2× margin that needs 54 mW of rating, so the smallest chip that fits is 0402 (1005 metric), 1/16 W.
V = I × R = 3 × 0.1 = 300 mV, and P = I² × R = 9 × 0.1 = 900 mW. That is within a 2512 chip’s 1 W rating but leaves almost no margin, so no standard-power chip size has the 2× margin. Use a 2 W or 3 W current-sense resistor, or two 200 mΩ parts in parallel to share the heat.
I = P ÷ V = 1500 ÷ 120 = 12.5 A, and R = V² ÷ P = 120² ÷ 1500 = 9.6 Ω. That is the element’s resistance when hot: heating wire has a positive temperature coefficient, so its cold resistance is lower and the switch-on current is higher than the running current.
I = V ÷ R and P = V² ÷ R at the usual supply rails. Look for the power column before you pick a package.
| R | I at 3.3 V | P at 3.3 V | I at 5 V | P at 5 V | I at 12 V | P at 12 V | I at 24 V | P at 24 V |
|---|---|---|---|---|---|---|---|---|
| 10 Ω | 330 mA | 1.09 W | 500 mA | 2.5 W | 1.2 A | 14.4 W | 2.4 A | 57.6 W |
| 100 Ω | 33 mA | 109 mW | 50 mA | 250 mW | 120 mA | 1.44 W | 240 mA | 5.76 W |
| 220 Ω | 15 mA | 49.5 mW | 22.7 mA | 114 mW | 54.5 mA | 655 mW | 109 mA | 2.62 W |
| 330 Ω | 10 mA | 33 mW | 15.2 mA | 75.8 mW | 36.4 mA | 436 mW | 72.7 mA | 1.75 W |
| 470 Ω | 7.02 mA | 23.2 mW | 10.6 mA | 53.2 mW | 25.5 mA | 306 mW | 51.1 mA | 1.23 W |
| 1 kΩ | 3.3 mA | 10.9 mW | 5 mA | 25 mW | 12 mA | 144 mW | 24 mA | 576 mW |
| 4.7 kΩ | 702 µA | 2.32 mW | 1.06 mA | 5.32 mW | 2.55 mA | 30.6 mW | 5.11 mA | 123 mW |
| 10 kΩ | 330 µA | 1.09 mW | 500 µA | 2.5 mW | 1.2 mA | 14.4 mW | 2.4 mA | 57.6 mW |
| 100 kΩ | 33 µA | 109 µW | 50 µA | 250 µW | 120 µA | 1.44 mW | 240 µA | 5.76 mW |
Standard-power ratings of a common thick-film family (Yageo RC_L) at 70 °C ambient. Above the critical resistance, Vmax² ÷ P, the working voltage limits the power before the power rating does.
| Size (imperial) | Metric | Rated power | Max working voltage | Critical resistance |
|---|---|---|---|---|
| 0201 | 0603 | 1/20 W (50 mW) | 25 V | 12.5 kΩ |
| 0402 | 1005 | 1/16 W (62.5 mW) | 50 V | 40 kΩ |
| 0603 | 1608 | 1/10 W (100 mW) | 75 V | 56.3 kΩ |
| 0805 | 2012 | 1/8 W (125 mW) | 150 V | 180 kΩ |
| 1206 | 3216 | 1/4 W (250 mW) | 200 V | 160 kΩ |
| 1210 | 3225 | 1/2 W (500 mW) | 200 V | 80 kΩ |
| 2010 | 5025 | 3/4 W (750 mW) | 200 V | 53.3 kΩ |
| 2512 | 6332 | 1 W (1 W) | 200 V | 40 kΩ |
Ohm’s law describes the resistor, not the circuit around it. A real source has its own resistance, so the voltage at the resistor sags as the current rises; a meter or the next stage loads the node too. Resistance also changes with temperature: a thick-film chip has a temperature coefficient of ±100 to ±200 ppm/K, copper about 0.39 %/K, and a lamp filament changes many times over between cold and hot. LEDs and diodes are not ohmic at all, which is why they need the LED resistor calculator rather than R = V ÷ I. Finally, the power a resistor can carry depends on how hot its surroundings are: the ratings above fall linearly to zero between 70 °C and 155 °C (125 °C for 0201), and a part on a small pad or in a closed box runs hotter than the datasheet’s test board.
The voltage across a resistor equals the current through it times its resistance: V = I × R. Rearranged, I = V ÷ R and R = V ÷ I. It holds for resistors and other ohmic materials whose resistance does not depend on the current; diodes, LEDs and lamps are not ohmic.
Divide the voltage across the part by the current through it: R = V ÷ I. A resistor across a 9.00 V battery that draws 3.00 mA is 3 kΩ. To do it here, pick V and I above and enter the two values.
Volts are the potential difference across a part, amps the current through it, ohms its resistance and watts the power it draws, which a resistor turns into heat. One volt across one ohm passes one ampere and dissipates one watt. Enter any two in the calculator above to get the other two.
P = V² ÷ R. For example 12 V across 1 kΩ dissipates 144 mW (and draws 12 mA). The same power is also I² × R and V × I; use whichever pair you know.
I = 3.3 V ÷ 10 kΩ = 330 µA, and the resistor dissipates 1.09 mW. That is the current a 10 kΩ pull-up draws when its line is held low.
V = √(P × R) = √(0.25 × 10,000) = 50 V, at 5 mA. In practice keep the dissipation to half the rating, and check the part’s maximum working voltage too: a chip resistor can hit its voltage limit before its power limit.
At least twice the power it dissipates, which is the margin Broadcom’s LED application note AN 1005 uses. Ratings apply at 70 °C ambient and fall to zero at 155 °C for common thick-film chips (125 °C for 0201), so a hot enclosure needs more margin.
Because of its voltage rating. A 0603 chip in the Yageo RC_L series is rated 1/10 W but only 75 V, so a 1 MΩ part reaches its voltage limit at 5.63 mW. Above about 56.3 kΩ the voltage limit, not the power, sets what a 0603 can take.
For a resistor, yes: use RMS voltage and current and the power comes out as the average power. For capacitors and inductors, replace R with the impedance Z, which depends on frequency and adds a phase shift between voltage and current.
See all free electronics tools →
ProtoFlow is free AI PCB design software: describe a circuit and get a schematic with real parts, PCB layout with an autorouter, SPICE simulation and Gerber files. Native KiCad, Altium and EAGLE import and export. Free on macOS, Windows and Linux.
Free to start. No card required.
Download ProtoFlow
