Current through a 4.7kΩ resistor at 9V
Worked answer for 9V applied across a 4.7kΩ resistor, using Ohm's law (I = V/R) and power (P = V²/R).
Current 1.9 mA 9V ÷ 4.7kΩ · dissipates 17.2 mW
| Voltage (V) | 9 V |
| Resistance (R) | 4.7kΩ |
| Current I = V / R | 1.9 mA |
| Power P = V² / R | 17.2 mW |
Driving 9V through 4.7kΩ gives 1.9 mA of current (I = V/R) and the resistor must dissipate 17.2 mW of heat (P = V²/R). Pick a resistor power rating comfortably above that figure.
Different values? Enter any two of voltage, current, resistance or power in the interactive tool:
Open the Ohm's Law Calculator →Disclaimer: This is a reference estimate using ideal Ohm's law. Real components have tolerance and self-heating; verify against your part's datasheet and power rating before building.