Voltage Divider Calculator
Enter the input voltage and both resistors to get the divider's output voltage, the current it draws and the power each resistor dissipates.
How it works
Two resistors in series across a supply carry the same current, so each drops voltage in proportion to its own resistance. Tapping the point between them gives a fraction of the input — R2 divided by the total.
Only the ratio sets the output voltage. 1 kΩ and 2 kΩ give the same 2/3 output as 10 kΩ and 20 kΩ. What the absolute values change is the current wasted in the divider and how much load it can tolerate.
The catch that catches everyone: a voltage divider is not a power supply. Anything you connect to the output sits in parallel with R2 and pulls the voltage down. The rule of thumb is to make the divider draw at least ten times the load current, or buffer the output with an op-amp.
The formula
Output voltage
V_out = V_in × R2 ÷ (R1 + R2)
Divider current
I = V_in ÷ (R1 + R2)
Power in each resistor
P_n = I² × R_n
With a load attached
replace R2 with (R2 ∥ R_load)
Which is why a loaded divider reads lower than the unloaded figure.
Worked examples
| Scenario | Working | Result |
|---|---|---|
| 9 V, R1 = 1 kΩ, R2 = 2 kΩ | 9 × 2000 ÷ 3000 | 6 V out, 3 mA |
| 5 V halved with two 10 kΩ | 5 × 10000 ÷ 20000 | 2.5 V out, 0.25 mA |
| 12 V to 3.3 V for a logic input | R1 = 8.2 kΩ, R2 = 3.3 kΩ | ≈ 3.44 V — close enough for a 3.3 V input |
When you'd use it
- Scaling a voltage down to something a microcontroller ADC can read
- Setting a reference or bias point
- Reading a battery voltage safely on a 3.3 V pin
- Shifting a 5 V signal down to 3.3 V logic
Common questions
Why is my measured output lower than the calculated one?
Almost always loading. Whatever you connected sits in parallel with R2, lowering the effective resistance and the output with it. Use much smaller divider resistors relative to the load, or buffer the output with an op-amp follower.
Should I use large or small resistor values?
It is a trade-off. Large values waste less current but are more affected by load and by noise pickup. Small values hold their voltage better under load but burn power continuously. For a battery-powered ADC divider, tens to hundreds of kΩ is usual.
Can I power something from a voltage divider?
No. It cannot supply meaningful current — the output collapses as soon as a real load draws from it, and the resistors turn the difference into heat. Use a regulator or a buck converter for anything that draws current.
Which resistor is R1 and which is R2?
R1 sits between the input and the tap; R2 sits between the tap and ground. The output is measured across R2, which is why R2 appears on the top of the fraction.
Is anything I enter sent to a server?
No. Every calculation runs in your browser, so component values, circuit figures and measurements are never uploaded, logged or stored. There is no account and no record of what you typed.