All 12 formula permutations for voltage (V), current (I), resistance (R) and power (P). Use this at the bench to find any unknown from any two known values.
V = voltage (volts, V) · I = current (amperes, A) · R = resistance (ohms, Ω) · P = power (watts, W)
| Configuration | Total resistance |
|---|---|
| 2 resistors in series | Rtotal = R1 + R2 |
| 3+ resistors in series | Rtotal = R1 + R2 + R3 ... |
| 2 equal resistors in parallel | Rtotal = R ÷ 2 |
| 2 resistors in parallel | Rtotal = (R1 × R2) ÷ (R1 + R2) |
| 3+ resistors in parallel | 1/Rt = 1/R1 + 1/R2 + 1/R3 |
Series adds resistances. Parallel always gives a total lower than the smallest individual resistor.
| Prefix | Symbol | Factor | Example |
|---|---|---|---|
| Mega | M | 10&sup6; | 1 MΩ = 1,000,000 Ω |
| Kilo | k | 10³ | 4.7 kΩ = 4,700 Ω |
| (base) | 1 | 220 Ω | |
| Milli | m | 10-3 | 100 mA = 0.1 A |
| Micro | µ | 10-6 | 470 µF = 0.00047 F |
| Nano | n | 10-9 | 100 nF = 0.0001 mF |
| Pico | p | 10-12 | 22 pF ceramic cap |
| Scenario | Known values | Formula | Result |
|---|---|---|---|
| LED resistor for 5V supply, 20 mA LED (2V Vf) | V = 3V (supply minus Vf), I = 0.02A | R = V ÷ I | R = 3 ÷ 0.02 = 150 Ω (use 180 Ω from E12) |
| Current draw of a 60W bulb on 230V | P = 60W, V = 230V | I = P ÷ V | I = 60 ÷ 230 = 0.26 A |
| Power dissipated by a 470 Ω resistor at 12V | V = 12V, R = 470 Ω | P = V² ÷ R | P = 144 ÷ 470 = 0.31 W (use 0.5 W rated) |
| Voltage drop across a 10 Ω resistor at 50 mA | I = 0.05A, R = 10 Ω | V = I × R | V = 0.05 × 10 = 0.5 V |
| Unknown resistance from 9V supply drawing 18 mA | V = 9V, I = 0.018A | R = V ÷ I | R = 9 ÷ 0.018 = 500 Ω |