How this calculator works
A voltage divider taps a fraction of Vin at the node between R1 and R2. That fraction only holds while nothing draws current from the tap. The moment you connect a load — a sensor, an ADC input, anything with finite resistance — it forms a parallel path with R2, so the divider is really only sourcing R1 and (R2 ∥ Rload) together.
- Vin — input voltage
- R1 — top resistor, between Vin and the tap
- R2 — bottom resistor, between the tap and ground
- Rload — resistance of whatever you connect to the tap
- R2' — effective R2 once Rload is in parallel with it
More detail
A divider is not a power supply
The Vout you calculate for an unloaded divider is only a promise until something actually draws current. Op-amp inputs and most ADCs pull negligible current, so the unloaded number holds close enough. But drive an LED, a relay coil, or a low-impedance sensor straight off the tap and R2' collapses toward Rload, dragging Vout down with it — sometimes by more than half. Leave Load resistance blank to see the ideal case, then fill it in with what you're actually driving to see the real number.
Design tip. If your load resistance isn't at least 10× R2, don't trust the unloaded Vout — compute the loaded number instead, or add a buffer (op-amp follower) so the divider only has to feed a high-impedance input.
Frequently asked questions
5V into a 1kΩ/1kΩ divider with no load — what's Vout?
Vout = 5 × 1000/(1000+1000) = 2.5 V, with 2.5 mA flowing through the divider and 12.5 mW dissipated.
Same divider, but now a 1kΩ load is attached — how far does Vout drop?
R2' = (1000×1000)/(1000+1000) = 500 Ω, so Vout = 5 × 500/(1000+500) = 1.667 V — down from 2.5 V, a 33% drop from a load equal to R2.
Why did my divider's output collapse when I connected my circuit?
Your circuit's input resistance is acting as Rload in parallel with R2. If it's low compared to R2, R2' shrinks toward it and Vout follows. Enter your circuit's actual input resistance as the load above to see the real Vout instead of the ideal one.