What is the voltage divider formula?

Vout = Vin × R2 / (R1 + R2). The output is taken at the node between R1 (top, in series with the supply) and R2 (bottom, tied to ground). Because both resistors carry the same series current, the fraction of the total resistance that R2 occupies is exactly the fraction of Vin that appears at the output.

How do I find R1 or R2 from a target output voltage?

Rearrange the main formula. For R1: R1 = R2 × (Vin / Vout − 1). For R2: R2 = R1 × Vout / (Vin − Vout). Both rearrangements are implemented here — just choose the unknown from the "Solve for" dropdown and enter the three known values. The calculator also shows the full algebraic step that produced the answer.

Why does loading affect the output voltage?

The divider formula assumes no current flows into or out of the output node. The moment you connect a load (say, a microcontroller GPIO or ADC input), that load appears in parallel with R2 and reduces its effective resistance, pulling Vout down. The rule of thumb is to make the divider current (Vin / (R1 + R2)) at least ten times the maximum load current, so the load barely disturbs the ratio. For very sensitive references, use an op-amp voltage follower (unity-gain buffer) to isolate the divider from the load entirely.

How is divider current and power calculated?

Current through the chain: I = Vin / (R1 + R2). Power in R1: P₁ = I² × R1. Power in R2: P₂ = I² × R2. Total: P = Vin × I = Vin² / (R1 + R2). Large resistor values (kΩ–MΩ range) waste very little power — a 10 kΩ + 10 kΩ divider fed from 5 V dissipates only 1.25 mW. The calculator shows all three power figures so you can verify no resistor exceeds its rated wattage.

What resistor values should I choose for a 3.3 V output from 5 V?

You need R2 / (R1 + R2) = 3.3 / 5 = 0.66. A common choice is R1 = 1.8 kΩ and R2 = 3.3 kΩ (ratio = 3.3 / 5.1 ≈ 0.647, giving 3.24 V) or R1 = 2 kΩ and R2 = 3.9 kΩ (ratio ≈ 0.661, giving 3.3 V). Use the "Solve for R1" or "Solve for R2" mode here — enter 5 V, 3.3 V and one standard resistor value, and the calculator finds the ideal companion value, which you can then round to the nearest E24 or E96 standard.

Is a voltage divider the same as a potential divider?

Yes — the terms are interchangeable. "Potential divider" is the more common term in UK/EU educational contexts; "voltage divider" is the standard term in US industry and datasheets. Both describe the same two-resistor series circuit where the output voltage is proportional to the resistor ratio.

Can I use this for capacitors or inductors too?

The same ratio formula works for AC impedance dividers if you replace resistance with impedance (Z). For a pure resistive divider the result is frequency-independent. For RC or RL dividers the impedance of C or L varies with frequency (XC = 1/2πfC, XL = 2πfL), so the ratio changes with frequency — that is exactly how passive filters (low-pass, high-pass) work. This calculator handles only resistive dividers; use a dedicated filter calculator for reactive networks.

Voltage Divider Calculator

A voltage divider (or potential divider) is a two-resistor series circuit that produces a predictable fraction of the supply voltage at the junction between the two resistors. It is one of the most frequently used building blocks in electronics: scaling sensor outputs into an ADC range, setting bias points, producing a reference voltage for a comparator, or level-shifting a 5 V logic signal down to a 3.3 V input. This calculator is a full solve-for-any-variable tool — enter any three of Vin, R1, R2 and Vout and it finds the fourth, plus the divider current, per-resistor power dissipation and the voltage ratio. Every calculation is shown step-by-step.

How it works

The supply voltage Vin is applied to the top of R1. The bottom of R1 connects to the top of R2, and R2’s bottom connects to ground. The output Vout is measured at the node between R1 and R2. Because R1 and R2 carry the same series current, the output voltage is simply the fraction of the total resistance that R2 contributes:

Vout = Vin × R2 / (R1 + R2)

Three further quantities follow automatically:

Divider current: I = Vin / (R1 + R2)
Power in R1: P₁ = I² × R1
Power in R2: P₂ = I² × R2
Total power: P = Vin × I = Vin² / (R1 + R2)

Solving for a resistor

Rearranging the main formula gives two inverse forms:

Find R1: R1 = R2 × (Vin / Vout − 1) — useful when you have a standard R2 in stock and need the series resistor.
Find R2: R2 = R1 × Vout / (Vin − Vout) — useful when you have a fixed R1 and need the bottom resistor.

Both are supported by the “Solve for” mode selector.

Worked example — 5 V to 3.3 V level shifter

A microcontroller with a 5 V GPIO drives an input rated only for 3.3 V. Using R1 = 2 kΩ and R2 = 3.9 kΩ:

Quantity	Formula	Result
Total R	2 kΩ + 3.9 kΩ	5.9 kΩ
Vout	5 V × 3.9 / 5.9	3.305 V
Current	5 V / 5.9 kΩ	0.847 mA
Power (R1)	(0.847 mA)² × 2 kΩ	1.43 mW
Power (R2)	(0.847 mA)² × 3.9 kΩ	2.80 mW
Total power	5 V × 0.847 mA	4.24 mW

Both resistors are well within their 1/8 W (125 mW) ratings. The divider current (0.847 mA) is roughly fifty times a typical GPIO input’s leakage current, so loading is negligible.

Loading effect and when to add a buffer

The formula above assumes the output is open-circuit (unloaded). Connecting a load resistance R_L in parallel with R2 reduces the effective bottom resistance to R2 ‖ R_L = R2 × R_L / (R2 + R_L), lowering Vout. To keep the error below 1%, make the divider current at least 100× the load current, or choose resistors whose parallel combination stays close to the intended R2. When driving an op-amp or comparator input, use a voltage follower (buffer) to eliminate loading entirely.

All calculations run client-side in your browser — no data is transmitted.