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Voltage Divider Calculator (with Loading Effect)

Calculate a two-resistor voltage divider's output voltage, current, and power dissipation — including how much the output sags once a real load is connected.

Output voltage (unloaded)
6 V
Divider current
6 mA
Power dissipated — R1
36 mW
Power dissipated — R2
36 mW
Total power dissipated
72 mW

How it works

A voltage divider's output is Vout = Vin × R2 ÷ (R1 + R2), which follows directly from Ohm's law: the same current I = Vin ÷ (R1 + R2) flows through both series resistors, and Vout is simply that current times R2. Power dissipated in each resistor is I² × R. Verified via web search against SparkFun's and Circuit Digest's own voltage-divider references rather than relied on from memory.

The output most divider-only calculators skip is what happens once you actually connect something to the output: any real load resistor sits electrically in parallel with R2, combining to R2 ∥ RL = (R2 × RL) ÷ (R2 + RL) — a smaller resistance than R2 alone, which pulls the output voltage down below the 'ideal' unloaded figure. This calculator reports that loaded output voltage and the resulting percentage sag whenever a load resistance is entered, so you can see whether a divider you're designing will actually deliver close to its calculated voltage once it's driving something, or needs a buffer/op-amp instead.

FAQ

How low does the load resistance need to be before sag becomes a problem?

A commonly cited rule of thumb is to keep the load at least 10 times R2 — at that ratio the sag stays under about 5%, which this calculator will show directly for your own values. Below that, especially when the load resistance gets close to R2, the sag grows quickly and the divider stops being a reliable voltage reference for that load.

Why does a lower R1/R2 pair reduce sag compared to higher values at the same ratio?

Sag depends on how R2 compares to the load, not on the absolute size of R1 and R2 alone — but for a fixed load, smaller R1/R2 values (keeping the same ratio) mean R2 is smaller relative to that load, so the parallel combination R2 ∥ RL barely changes from R2 alone. That's also why very low resistor values draw more current and dissipate more power, a real design trade-off against the lower sag.

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