Voltage Divider Calculator
Output voltage of a two-resistor divider, with current & power
Last updated September 6, 2026
Method: The standard two-resistor divider equation Vout = Vin × R2 ÷ (R1 + R2), rearranged algebraically to solve for Vin, R1 or R2, combined with Ohm's law (I = V ÷ R) and the power equation (P = I² × R) for each resistor.
Included: Output voltage, required supply voltage or resistor value, divider ratio, series current, the voltage drop and power in R1 and R2, the nearest E12 stock resistor, and the loaded output when you add a load resistance.
Not included: AC impedance and frequency effects, resistor tolerance and temperature drift, source impedance of the supply, wire and connector resistance, and semiconductor input bias currents. Results describe an ideal resistive divider.
⚡ Output voltage (Vout)
🔎 Current and power in each resistor
| Element | Resistance | Voltage | Current | Power |
|---|---|---|---|---|
| R1 (top) | 10 kΩ | 8.163 V | 816.33 µA | 6.664 mW |
| R2 (bottom) | 4.7 kΩ | 3.837 V | 816.33 µA | 3.132 mW |
| Total | 14.7 kΩ | 12 V | 816.33 µA | 9.796 mW |
Pick resistors rated for at least twice the power shown so they run cool.
📊 Same divider on other supply rails
| Vin | Vout | Current | Total power |
|---|---|---|---|
| 3.3 V | 1.055 V | 224.49 µA | 740.82 µW |
| 5 V | 1.599 V | 340.14 µA | 1.701 mW |
| 9 V | 2.878 V | 612.24 µA | 5.51 mW |
| 12 V | 3.837 V | 816.33 µA | 9.796 mW |
| 24 V | 7.673 V | 1.633 mA | 39.184 mW |
The ratio 0.3197 stays fixed, so Vout scales directly with Vin.
Ideal, unloaded result. Real resistors carry a tolerance (often 5% or 1%), and anything you connect to the output draws current and lowers Vout. A divider is a reference, not a power supply.
Voltage divider calculator: everything you need to know
A voltage divider calculator turns two resistors and a supply voltage into an output voltage. Feed 12 V into a 10 kΩ top resistor and a 4.7 kΩ bottom resistor and the output node sits at 3.837 V, with 0.816 mA flowing through both parts. This page also solves the problem backwards, so you can enter the voltage you need and get the resistor.
Two sister tools cover the neighboring questions: the Ohm's Law Calculator solves voltage, current, resistance and power for a single element when you know any two of them, and the Voltage Drop Calculator handles the unintended drop along a long wire run. Use this page when two resistors are deliberately splitting a known voltage; use Ohm's law when you are working with one component; use voltage drop when the wire itself is the problem.
How a voltage divider works
Two resistors in series carry exactly the same current, because there is nowhere else for that current to go. Ohm's law says the voltage across each resistor equals that shared current times its own resistance, so the supply voltage splits between the two parts in direct proportion to their resistances. Tap the node between them and you get a fraction of the input:
Vout = Vin × R2 ÷ (R1 + R2) R1 is the top resistor, between the supply and the output node. R2 is the bottom resistor, between the output node and ground. The fraction R2 ÷ (R1 + R2) is the divider ratio, always between 0 and 1. Rearranged for the other three unknowns:
Vin = Vout × (R1 + R2) ÷ R2
R1 = R2 × (Vin − Vout) ÷ Vout
R2 = R1 × Vout ÷ (Vin − Vout) Notice what is missing: absolute resistance never appears in the ratio, only the relationship between the two values. A 10 kΩ / 10 kΩ divider and a 1 MΩ / 1 MΩ divider both cut the voltage in half. What the absolute values do decide is how much current the divider wastes and how badly a load pulls the output down.
Worked example: 12 V split with 10 kΩ and 4.7 kΩ
Take a 12 V supply, a 10 kΩ resistor on top and a 4.7 kΩ resistor on the bottom. Work through it step by step:
- Total resistance: 10,000 + 4,700 = 14,700 Ω.
- Divider ratio: 4,700 ÷ 14,700 = 0.3197.
- Output voltage: 12 × 0.3197 = 3.837 V.
- Series current: 12 ÷ 14,700 = 0.000816 A, or 0.816 mA.
- Drop across R1: 0.816 mA × 10,000 Ω = 8.163 V. Add that to the 3.837 V across R2 and you get the full 12 V back, which is a good sanity check.
- Power in R1: P = I² × R = 0.000816² × 10,000 = 0.006664 W, or 6.664 mW.
- Power in R2: 0.000816² × 4,700 = 0.003132 W, or 3.132 mW.
- Total power: 12 V × 0.816 mA = 9.796 mW, which matches the two resistor figures added together.
Both parts are far under a tenth of a watt, so the common quarter-watt resistor is comfortable here. Everything above is what the calculator shows the moment the page loads.
Worked example: solving for a resistor instead
The more common real problem runs the other way: you know the voltage you need and have to find a resistor. Suppose you want 3.3 V from a 9 V battery and you have already picked R1 = 10 kΩ. Apply the rearranged formula:
R2 = R1 × Vout ÷ (Vin − Vout) = 10,000 × 3.3 ÷ (9 − 3.3) = 33,000 ÷ 5.7 = 5,789 Ω.
No one stocks a 5,789 Ω resistor. The nearest E12 value is 5.6 kΩ, which gives 9 × 5,600 ÷ 15,600 = 3.231 V, about 2% low. The next value up, 6.8 kΩ, gives 3.643 V, which is 10% high. So 5.6 kΩ is the better single-resistor choice, and if you need to land closer you either add a small resistor in series with it or move to the tighter E24 or E96 series. The calculator does this comparison for you whenever you solve for R1 or R2.
Common resistor pairs and their output voltage
The table below lists divider pairs you will actually find in a parts drawer, with the ratio each one produces and the resulting output on five common supply rails. The last column is the unloaded current at 12 V. All values come from the formula above.
| R1 / R2 | Ratio | 5 V in | 9 V in | 12 V in | 24 V in | mA at 12 V |
|---|---|---|---|---|---|---|
| 1 kΩ / 1 kΩ | 0.5000 | 2.500 V | 4.500 V | 6.000 V | 12.000 V | 6.000 |
| 10 kΩ / 10 kΩ | 0.5000 | 2.500 V | 4.500 V | 6.000 V | 12.000 V | 0.600 |
| 10 kΩ / 4.7 kΩ | 0.3197 | 1.599 V | 2.878 V | 3.837 V | 7.673 V | 0.816 |
| 10 kΩ / 2.2 kΩ | 0.1803 | 0.902 V | 1.623 V | 2.164 V | 4.328 V | 0.984 |
| 10 kΩ / 1 kΩ | 0.0909 | 0.455 V | 0.818 V | 1.091 V | 2.182 V | 1.091 |
| 4.7 kΩ / 10 kΩ | 0.6803 | 3.401 V | 6.122 V | 8.163 V | 16.327 V | 0.816 |
| 2.2 kΩ / 10 kΩ | 0.8197 | 4.098 V | 7.377 V | 9.836 V | 19.672 V | 0.984 |
| 22 kΩ / 10 kΩ | 0.3125 | 1.562 V | 2.812 V | 3.750 V | 7.500 V | 0.375 |
| 100 kΩ / 10 kΩ | 0.0909 | 0.455 V | 0.818 V | 1.091 V | 2.182 V | 0.109 |
Compare the last row against the 10 kΩ / 1 kΩ row. Both share the ratio 0.0909 and produce exactly the same output voltage, but scaling both resistors up by ten cuts the current from 1.091 mA to 0.109 mA. That is the whole trade-off in one table.
Current and wasted power by total resistance
A divider draws current continuously, whether or not anything is connected to its output. This table shows what a 12 V supply costs you at different total resistances, using I = Vin ÷ Rtotal and P = Vin × I.
| R1 + R2 | Current at 12 V | Power wasted | Practical note |
|---|---|---|---|
| 100 Ω | 120 mA | 1,440 mW | Needs 1 W parts, drains a battery fast |
| 1 kΩ | 12 mA | 144 mW | Stiff output, fine on mains-fed gear |
| 10 kΩ | 1.2 mA | 14.4 mW | The usual default for signal work |
| 100 kΩ | 0.12 mA | 1.44 mW | Good for battery monitoring |
| 1 MΩ | 0.012 mA | 0.144 mW | Very low drain, easily disturbed by noise |
Each step of ten in resistance cuts both the current and the wasted power by ten. Going too far the other way hurts as well: at 1 MΩ the divider is so weak that stray capacitance, board leakage and the input bias current of whatever it feeds start to matter.
Loading: the mistake that ruins most dividers
The formula assumes nothing is connected to the output. Connect something and that load sits in parallel with R2, lowering the effective bottom resistance and dragging the output down. The effective value is R2 × RL ÷ (R2 + RL), and the divider then runs on that number instead of R2. Here is what happens to a 12 V supply with two 10 kΩ resistors, which should give exactly 6.000 V:
| Load resistance | R2 in parallel | Actual Vout | Error |
|---|---|---|---|
| 1 kΩ | 909 Ω | 1.000 V | −83.33% |
| 4.7 kΩ | 3,197 Ω | 2.907 V | −51.55% |
| 10 kΩ | 5,000 Ω | 4.000 V | −33.33% |
| 47 kΩ | 8,246 Ω | 5.423 V | −9.62% |
| 100 kΩ | 9,091 Ω | 5.714 V | −4.76% |
| 1 MΩ | 9,901 Ω | 5.970 V | −0.50% |
The pattern is clear: the load needs to be roughly ten times R2 to keep the error under about 5%, and a hundred times R2 to get it under about 0.5%. Tick the load option in the calculator to run this check on your own numbers before you build anything.
How to use this calculator
Everything updates as you type, so you can sweep values and watch the output move. Work through it in this order:
- Choose what to solve for. Pick Vout when you already have both resistors, R1 or R2 when you know the voltage you want, or Vin when you are working out what supply a known divider was designed for.
- Set the resistance unit. Switch between Ω, kΩ and MΩ so you can type 4.7 rather than 4700. The unit applies to every resistance field at once.
- Enter the voltages. Use the quick buttons for the common rails of 3.3 V, 5 V, 9 V, 12 V and 24 V, or type any value.
- Enter the resistors. R1 is the top resistor next to the supply, R2 the bottom one next to ground. Getting these the wrong way round is the single most common error, and it gives you the complementary voltage.
- Add a load if there will be one. Tick the load box and enter the input resistance of whatever the output feeds to see how far the real output sags.
- Read the results. The headline shows the value you solved for, and the tiles below give the ratio, current, and the split of voltage between the two resistors. The tables show power per resistor and how the same divider behaves on other supply rails.
Who this calculator is for
- Hobbyists and makers scaling a sensor or battery voltage into the safe range of a microcontroller analog input.
- Electronics students checking homework on series circuits, divider ratios and power dissipation.
- Technicians working out why a measured node voltage does not match a schematic value.
- Audio and instrumentation builders setting attenuator ratios and bias points.
- Anyone repairing equipment who needs to confirm that a divider on a board is producing the reference voltage it should.
Key terms explained
- R1 (top resistor): sits between the supply and the output node. Increasing it lowers the output.
- R2 (bottom resistor): sits between the output node and ground. Increasing it raises the output.
- Divider ratio: R2 ÷ (R1 + R2), a number between 0 and 1 that multiplies the input voltage.
- Loading: the drop in output voltage caused by the current a connected circuit draws.
- Output impedance: R1 in parallel with R2, which tells you how stiff the output is. A 10 kΩ / 10 kΩ divider has a 5 kΩ output impedance.
- E12 series: the standard set of stock resistor values (10, 12, 15, 18, 22, 27, 33, 39, 47, 56, 68, 82 and their decades) used for 10% and many 5% parts.
- Potentiometer: a three-terminal variable resistor whose wiper forms an adjustable divider.
What changes the result the most
- The ratio, not the values: doubling both resistors leaves the output voltage untouched and halves the current.
- The load: by far the biggest source of real-world error, and the one the plain formula ignores completely.
- Resistor tolerance: two 5% parts can push a nominal 6.000 V anywhere between 5.700 V and 6.300 V.
- Supply stability: the output tracks the input exactly, so a battery sagging from 12 V to 11 V drops every divider output by the same 8.3%.
- Temperature: resistors drift with heat, and two parts of different types drift by different amounts, moving the ratio.
Tolerance: how accurate is the answer really?
The calculator gives the ideal figure. Real parts carry a tolerance band, and in a divider the two errors can work against each other. Take the 12 V, 10 kΩ / 10 kΩ divider that should output 6.000 V. If R1 lands 5% high at 10,500 Ω and R2 lands 5% low at 9,500 Ω, the output falls to 5.700 V. Reverse the errors and it rises to 6.300 V. That is a 10% window on a supposedly exact half-voltage. Swap in 1% parts and the same worst case narrows to 5.940 V to 6.060 V, a 2% window. If the divider feeds an analog-to-digital converter and you care about the reading, either buy 1% or 0.1% resistors, or measure the actual output once and calibrate it in software.
Where voltage dividers are used in practice
The circuit is simple enough to look academic, but it is everywhere. Common jobs include:
- Battery monitoring: a 100 kΩ / 22 kΩ divider scales 12 V down to 2.164 V for a 3.3 V analog input while drawing only 0.098 mA and wasting 1.180 mW.
- Logic level shifting: dropping a 5 V signal toward 3.3 V. With R1 = 10 kΩ the exact value is 19,412 Ω, so 18 kΩ gives 3.214 V and 20 kΩ gives 3.333 V.
- Sensor read-out: a thermistor or photoresistor forms one half of a divider, and the output voltage moves as the sensor resistance changes.
- Bias networks: setting the base voltage of a transistor or the mid-rail reference of an op-amp stage.
- Attenuators and volume controls: a potentiometer as a continuously variable divider on an audio signal.
- Test probes: a 10x oscilloscope probe is a divider designed to reduce the signal so the instrument sees a safe fraction of it.
Practical tips
- Start with R1 = 10 kΩ and solve for R2. It lands most dividers in the sensible 10 kΩ to 100 kΩ total range on the first try.
- Aim below your target. When the exact resistor does not exist, prefer the standard value that puts the output slightly low, so a tolerance stack does not overshoot the input limit of whatever it feeds.
- Combine resistors. Two values in series add, and two in parallel give R1 × R2 ÷ (R1 + R2), which lets you hit awkward numbers with parts you already have.
- Buffer the output. If the following stage draws real current, put a unity-gain op-amp between the divider and the load; it reads the divider without disturbing it.
- Double the power rating. Size resistors for at least twice the calculated dissipation so they stay cool and their value stays stable.
- Measure before you trust. Build the divider, measure the node with a meter, and remember that the meter itself is a load, usually around 10 MΩ, which is high enough to ignore on most dividers but not on a 1 MΩ pair.
Limitations and assumptions
- The formula assumes an unloaded output unless you tick the load option, and assumes the load is purely resistive.
- It assumes an ideal supply with zero source impedance, so a weak battery or a long feed wire will lower the real input voltage.
- It assumes exact resistances with no tolerance, temperature drift or aging.
- It covers DC or purely resistive AC only; with capacitance or inductance in the path the divider becomes frequency dependent.
- It does not size parts for voltage rating or surge, which matter on high-voltage dividers where a single resistor may need to be split into several in series.
- A divider is not a regulator: the output moves with the input and with the load, and it offers no protection or current limiting.
How it compares to related calculators
This page answers "what voltage appears between two series resistors?" If your question is different, another tool fits better:
- For a single component, or to convert between volts, amps, ohms and watts, use the Ohm's Law Calculator.
- For the unwanted drop along a cable run, use the Voltage Drop Calculator.
- To read the value of a resistor from its color bands before you use it here, use the Resistor Color Code Calculator.
- To turn a wattage into the current a circuit will draw, use the Watts to Amps Calculator.
- To price the energy a continuously powered circuit uses, use the Electricity Cost Calculator.
Sources
- International Electrotechnical Commission (IEC) - IEC 60063: Preferred number series for resistors and capacitors, the standard that defines the E12, E24 and E96 stock values used for the nearest-resistor suggestion.
- National Institute of Standards and Technology (NIST) - SI prefixes, for the kilo and mega prefixes used with Ω, and SI units for the volt, ampere, ohm and watt.
The divider equation itself is standard circuit theory derived from Ohm's law and Kirchhoff's voltage law, so it is deterministic and needs no external data.
💡 Good to know
Only the ratio sets the voltage
A 10 kΩ / 10 kΩ divider and a 1 MΩ / 1 MΩ divider both halve the input. The absolute values decide how much current you waste and how easily a load pulls the output down, not what the output voltage is.
A divider is a reference, not a power supply
Drawing current from the output changes the output. If the thing you are feeding needs more than about a tenth of the divider current, use a regulator or buffer it with an op-amp instead.
The two drops always add up to Vin
The voltage across R1 plus the voltage across R2 equals the supply voltage exactly. On a 12 V rail with 10 kΩ and 4.7 kΩ, that is 8.163 V plus 3.837 V. It is the fastest way to check your own arithmetic.
⚠️ Common mistakes & edge cases
Swapping R1 and R2
R1 is the top resistor next to the supply, R2 the bottom one next to ground. Swap them and you get the complementary voltage: a 10 kΩ / 4.7 kΩ divider on 12 V gives 3.837 V, but 4.7 kΩ / 10 kΩ gives 8.163 V.
Ignoring the load
The formula assumes nothing is connected. Hang a 1 kΩ load on a 12 V divider built from two 10 kΩ resistors and the output collapses from 6.000 V to 1.000 V. Check the load resistance before you trust the number.
Mixing up kΩ and Ω
Typing 4.7 when you meant 4,700 changes the current by a factor of a thousand. The ratio may look right while the power figures are wildly wrong. Set the unit selector first, then type the numbers.
Using a divider to power something
A divider cannot supply current. Feeding an LED, a relay or a module from one gives an output that sags to almost nothing under load and resistors that run hot. Use a regulator instead.
Forgetting the power rating
Small resistors are usually fine, but the numbers scale fast. A pair of 100 Ω resistors on 12 V draws 60 mA and burns 360 mW in each resistor, which will cook a standard quarter-watt part.
Expecting an exact resistor value to exist
Solving for R2 often returns something like 5,789 Ω, which nobody stocks. Take the nearest standard value, recheck the output it actually produces, and confirm it still fits your tolerance budget.
❓ Frequently asked questions
What is the voltage divider formula?
The two-resistor voltage divider formula is Vout = Vin × R2 ÷ (R1 + R2), where R1 is the top resistor between the supply and the output node and R2 is the bottom resistor between the output node and ground. The same current flows through both resistors, so the supply voltage splits in proportion to the resistances.
How do I calculate the output voltage of a voltage divider?
Add the two resistances, divide R2 by that total to get the divider ratio, then multiply by the input voltage. With 12 V across a 10 kΩ top resistor and a 4.7 kΩ bottom resistor, the ratio is 4,700 ÷ 14,700 = 0.3197, so Vout = 12 × 0.3197 = 3.837 V.
How do I choose R1 and R2 for a target voltage?
Pick one resistor first, usually R1, then solve for the other. To drop 12 V to 5 V with R1 = 10 kΩ, R2 = R1 × Vout ÷ (Vin − Vout) = 10,000 × 5 ÷ 7 = 7,143 Ω. Only the ratio sets the output voltage, so a 1 kΩ / 714 Ω pair gives the same 5 V but wastes ten times more current.
How much current flows through a voltage divider?
The unloaded divider current is Vin ÷ (R1 + R2). A 12 V supply across 10 kΩ plus 4.7 kΩ pushes 12 ÷ 14,700 = 0.816 mA through both resistors, and that current is the same in R1 and R2 because they are in series.
How much power does each resistor dissipate?
Use P = I² × R for each resistor. At 0.816 mA, a 10 kΩ top resistor dissipates 6.664 mW and a 4.7 kΩ bottom resistor dissipates 3.132 mW, for 9.796 mW total. Choose parts rated for at least twice that figure so they run cool.
Why is my measured output lower than the calculated value?
Almost always because the circuit you connected to the output is drawing current. Any load sits in parallel with R2 and lowers the effective bottom resistance. A 6 V divider built from two 10 kΩ resistors on 12 V collapses to 4.000 V with a 10 kΩ load and to 1.000 V with a 1 kΩ load.
Can a voltage divider power a circuit?
No. A divider is a reference, not a supply. Its output voltage only holds while the load draws far less current than the divider itself, and every milliamp the load takes drags the output down. Use a regulator, a buck converter, or a proper power supply when the load needs real current.
How do I pick the total resistance for a divider?
Balance wasted current against loading. A small total resistance holds its voltage well under load but burns current continuously: 12 V across 200 Ω total draws 60 mA and wastes 720 mW. A large total resistance saves power but is easily pulled off by the load and by input bias currents. For general use, a total between 10 kΩ and 100 kΩ is a common compromise.
What is the difference between a voltage divider and Ohm's law?
Ohm's law (V = I × R) links voltage, current and resistance in a single element. The divider formula is just Ohm's law applied twice to two resistors in series: the same current runs through both, so each resistor takes a share of the supply voltage proportional to its resistance.
How accurate is a voltage divider with 5% resistors?
Worst case, the errors stack. Two nominally equal 10 kΩ resistors with 5% tolerance can produce anywhere from 5.700 V to 6.300 V from a 12 V supply instead of the nominal 6.000 V. With 1% resistors the same divider stays between 5.940 V and 6.060 V, which is why precision work uses 1% or 0.1% parts.
Can I use a voltage divider to read a battery voltage with a microcontroller?
Yes, that is one of its best uses. A 100 kΩ / 22 kΩ divider scales a 12 V battery to 2.164 V, safely inside a 3.3 V analog input, and draws only 0.098 mA. Because the analog-to-digital converter draws almost no steady current, the loading error stays small, though very high resistances can slow the sample-and-hold settling.
Does a voltage divider work with AC?
A purely resistive divider splits an AC waveform by the same ratio it splits DC, so the output is a scaled copy of the input. Once capacitance or inductance is involved, the split becomes frequency dependent and you have to work with impedance instead of plain resistance. This calculator assumes DC or a purely resistive AC divider.
What happens if I swap R1 and R2?
You get the complementary ratio. A 10 kΩ / 4.7 kΩ divider on 12 V outputs 3.837 V, while swapping the resistors to 4.7 kΩ / 10 kΩ outputs 8.163 V. The two outputs always add up to the supply voltage, because the two resistors share it.
Can I use a potentiometer as a voltage divider?
Yes. A potentiometer is a continuously adjustable divider: the wiper splits the track into R1 above and R2 below, and turning it sweeps the ratio from 0 to 1. The same formula and the same loading rules apply, so the wiper voltage sags whenever the following stage draws current.
Is this voltage divider calculator free?
Yes. It runs entirely in your browser with no sign-up and no limits, works on phone, tablet or desktop, and nothing you type is sent anywhere. Solve for Vout, Vin, R1 or R2 as many times as you like.