Audited 29 Jul 2026·Last updated 28 Oct 2026·6 citations·Tier 1·0 uses

Voltage Divider Calculator (with Load)

Work out a resistor divider's output voltage, or the R2 for a target voltage, plus its output impedance, current draw and how much a real load pulls it down.

Voltage Divider Calculator

What are you solving for?
The supply rail feeding the top of R1. Assumed stiff — a real source's own output impedance adds to R1 and this calculator does not model it.
V
The resistor between the supply and the output tap. Swapping R1 and R2 inverts the answer, so check which is which before you read the result.
Ω
The resistor between the output tap and ground. The output voltage is the voltage across this one. Ignored when you are solving for R2.
Ω
Used only in the solve-for-R2 mode. It must be below the input voltage — a resistive divider can only step down. Going up needs a boost converter or a transformer.
V
Whatever the divider output feeds — an ADC input, a transistor base, a meter. Enter 0 for no load at all, which gives the open-circuit output. This field is where most divider designs come unstuck.
Ω
Output voltage (no load)
3.8367
The open-circuit voltage at the tap, Vin × R2 ÷ (R1 + R2), in volts, with nothing connected.
R2 in use
4,700 Ω
Divider ratio
31.97
Quiescent current
8.163 × 10⁻⁴ A
Power burned in the divider
0.0098 W
Output impedance
3,197.2789 Ω
Output voltage with the load connected
3.8367 V
Output droop under load
0.00
Is the loading acceptable?
No load connected, so this is the open-circuit output. The divider's own output impedance is 3.1973 kΩ (R1 in parallel with R2) — anything you connect that is not comfortably above about ten times that will pull the output down. Enter the load resistance to see by how much.

Background.

A microcontroller's analog input tops out at 3.3 V and you want it to read a 12 V battery. The usual answer is two resistors in series across the battery, with the input wired to the joint between them. That is a voltage divider. Its output is a fixed fraction of the input, R2 ÷ (R1 + R2), where R1 runs from the supply down to the tap and R2 from the tap to ground. With 10 kΩ on top and 4.7 kΩ below, 12 V becomes 3.8367 V, a little too much for that input, which is exactly the kind of thing worth knowing before you solder.

The same pair of resistors sets the output of adjustable regulators through their feedback pins, biases transistor stages, gives comparators a threshold, shifts 5 V logic down to 3.3 V, and, with one resistor swapped for a thermistor, light-dependent resistor or potentiometer, turns a physical quantity into a voltage. The formula is identical in every case, and so is its blind spot: it assumes nothing draws current from the tap. Connect anything and it sits in parallel with R2, the effective lower resistance falls, and the output sags. A divider that reads perfectly on a multimeter can read a quarter low once the real circuit is attached.

That is why the load is an input here, not an afterthought. Next to the ideal output you get the divider's output impedance, the output with your load connected, the percentage it sagged and a plain verdict on whether that is acceptable. You also see the current the pair draws at all times and the power it turns into heat, which are the two prices you pay for a stiffer output.

You can check a divider you already have, or design one by giving R1 and the voltage you want and letting R2 be solved. The model assumes a stiff supply, steady or slowly changing voltages, and resistors at exactly their nominal values; the FAQs below cover what changes when those assumptions fail.

What is voltage divider calculator?

A voltage divider is two resistances in series across a voltage source, with the output taken from the node between them. Because there is only one path, the same current flows through both resistors, and by Ohm's law that current is I = Vin ÷ (R1 + R2). The output is the voltage dropped across the lower resistor, I × R2, which gives Vout = Vin × R2 ÷ (R1 + R2). Only the ratio of the two values matters to this unloaded voltage, so 10 kΩ over 4.7 kΩ and 10 MΩ over 4.7 MΩ give the same answer.

The ratio does not tell the whole story, because the two pairs behave very differently once something is connected. Seen from the tap, a divider behaves like an ideal source of Vout in series with a resistance equal to R1 and R2 in parallel, its Thévenin equivalent. Whatever you connect forms a second divider with that internal resistance. A potentiometer is the same circuit with a sliding tap, and a sensor divider is the same circuit with one resistor that changes with temperature, light or position.

How to use this calculator.

  1. Choose the mode: the output voltage if you already have both resistors, or the lower resistor R2 if you know the voltage you want and have picked R1.
  2. Enter the input voltage across the whole divider, the supply rail or battery feeding the top of R1, not the output you are aiming for.
  3. Enter R1, the upper resistor between the supply and the tap, and either R2, the lower resistor between the tap and ground, or the target output voltage. Swapping R1 and R2 gives a plausible but wrong answer: 12 V across 4.7 kΩ over 10 kΩ gives 8.16 V instead of 3.84 V.
  4. Enter the resistance of whatever the tap drives: the input resistance of an ADC, meter, comparator or transistor stage from its datasheet. Leave it at 0 if nothing is connected; 0 here means an open circuit, not a short to ground.
  5. Read the loaded output and the droop beside the ideal figure, and check the verdict. If the droop is too large, scale both resistors down by the same factor, which keeps the voltage and cuts the output impedance, or buffer the tap with an op-amp follower.
  6. In design mode, round the solved R2 to a resistor you can buy, switch to the first mode and run it again with that value to see the voltage you will actually get.

The formula.

V_out = V_in · R₂/(R₁+R₂) · R₂ = R₁·V_out/(V_in−V_out) · R_th = R₁∥R₂

Unloaded output: Vout = Vin × R2 ÷ (R1 + R2). Rearranged for design, R2 = R1 × Vout ÷ (Vin − Vout). The denominator shows why a divider can only step down: as the target approaches the supply, Vin − Vout shrinks toward zero and R2 heads for infinity; past it, R2 would be negative. A target at or above the input voltage is therefore refused, and a boost converter, charge pump or transformer is needed instead.

The current through the pair is I = Vin ÷ (R1 + R2), and it flows whether or not anything is connected. The power it dissipates is P = Vin × I = Vin² ÷ (R1 + R2), all of it as heat in the two resistors.

The output impedance is R_th = R1 × R2 ÷ (R1 + R2), the two resistors in parallel, which is what the tap looks like with the supply replaced by a short. A useful identity follows from the divider equation: R_th = R1 × (Vout ÷ Vin). With a load R_L connected, R2 is replaced by R2 in parallel with R_L and the divider equation is applied again. The same result can be written as Vout × R_L ÷ (R_th + R_L), and the droop as a fraction of the unloaded output is R_th ÷ (R_th + R_L). That last expression shows that the sag depends on one number only, the ratio k = R_L ÷ R_th. The verdict bands translate directly: under 1% droop needs k above 99, under 5% needs k above 19, and k below 4 means a sag of 20% or more, at which point the load is setting the voltage as much as the resistors are. The familiar "load at least ten times the output impedance" rule gives 1/11, about 9.1%, which is only good enough for rough work.

Resistor tolerance is not included, because each part is taken at its nominal value. Its effect is easy to estimate: the fractional error in Vout is roughly (1 − Vout/Vin) multiplied by the combined tolerance of the two resistors, so a divider that drops most of the supply is the most sensitive.

A worked example.

Example

The default divider takes a 12 V rail through R1 = 10 kΩ and R2 = 4.7 kΩ, and its tap feeds an input with 10 kΩ of resistance. Open-circuit, the output is 12 × 4,700 ÷ 14,700 = 3.8367 V, a ratio of 31.97%. The pair draws 12 ÷ 14,700 = 0.8163 mA, shown as 8.163 × 10⁻⁴ A, and dissipates 12 × 0.0008163 ≈ 9.8 mW. The output impedance is 10,000 × 4,700 ÷ 14,700 = 3,197.2789 Ω, or about 3.2 kΩ. Now attach the 10 kΩ load. In parallel with R2 it makes an effective lower arm of 4,700 × 10,000 ÷ 14,700 = 3,197.2789 Ω, the same figure as the output impedance only because the load happens to equal R1, and the output becomes 12 × 3,197.2789 ÷ (10,000 + 3,197.2789) = 2.9072 V. That is a droop of (3.8367 − 2.9072) ÷ 3.8367 = 24.23%. The load is only 10,000 ÷ 3,197.2789 = 3.1277 times the output impedance, and 1 ÷ (1 + 3.1277) gives the same 24.23%, so the verdict is severe loading. Nothing here is faulty. A multimeter on the tap with the load disconnected would show 3.84 V and suggest all is well; the 0.93 V shortfall appears only when the real input is connected, and it would shift again if that input's resistance changed.

load Resistance10,000
modeoutputVoltage
r24,700
input Voltage12
r110,000

Frequently asked questions.

Why does my divider read correctly on a multimeter but low in the circuit?
A typical digital multimeter presents about 10 MΩ, over 3,000 times the 3.2 kΩ output impedance of a 10 kΩ / 4.7 kΩ divider, so it lowers the reading by only about 0.03%. The real input may be thousands of times lower than the meter. Enter its resistance as the load to see the voltage the circuit really receives.
How do I pick actual resistor values once I know the ratio?
Choose by output impedance against wasted current. Scale 10 kΩ / 4.7 kΩ down to 1 kΩ / 470 Ω and the same 10 kΩ load costs 3.10% instead of 24.23%, but the pair draws 8.163 mA and burns about 98 mW. Scale up to 100 kΩ / 47 kΩ and the drain falls to 81.6 µA while the droop with that load rises to 76.18%.
What resistors give 3.3 V from a 12 V supply?
With R1 = 10 kΩ, design mode returns R2 = 3,793.1034 Ω. The nearest E24 value, 3.9 kΩ, gives 3.3669 V, about 2% high; the E96 value 3.83 kΩ gives 3.3232 V. Output impedance is 10 kΩ × 3.3 ÷ 12 = 2,750 Ω, fine for a high-impedance ADC or comparator input but useless for powering a 3.3 V device.
Can I power a microcontroller or sensor module from a divider?
No. A divider's output depends on the current drawn from it, so its voltage is right at only one load current. A microcontroller's supply current jumps by large factors between sleep and active modes, and the rail would swing with it. Dividers are for signals feeding high-impedance inputs. To supply power, use a linear regulator or a switching converter, which actively holds its output steady.
How much do resistor tolerances shift the output?
With ±5% parts on 10 kΩ / 4.7 kΩ, the worst cases, R2 high with R1 low and the reverse, move the output by +6.93% and −6.68%. With ±1% parts the spread shrinks to about ±1.36%, and with ±0.1% to about ±0.14%. Matched resistor networks on one substrate also drift together with temperature, so their ratio holds even better.
Why should I worry about output impedance when feeding an ADC?
A sampling ADC briefly connects a small internal capacitor to its input, and that capacitor must charge through the divider's output impedance within the sampling window. Too high an impedance and readings come out low or noisy. Datasheets specify a maximum source impedance; AVR microcontrollers, for example, are designed for sources of about 10 kΩ or less. A capacitor from the tap to ground supplies that charge for slow signals.
How long will a divider take to drain a battery?
The quiescent current never stops. The 10 kΩ / 4.7 kΩ pair on 12 V draws 0.8163 mA, about 19.6 mAh a day, enough to empty a 2,000 mAh pack in around 102 days with nothing else running. A 100 kΩ / 100 kΩ monitor on a 4.2 V lithium cell draws 21 µA, about 0.5 mAh a day. For long standby, switch the divider off with a transistor between readings.
Can a resistor divider shift 5 V logic down to 3.3 V?
For slow signals, yes: 1 kΩ over 2 kΩ turns 5 V into 3.333 V, with an output impedance of about 667 Ω. Fast edges are the problem, because that impedance and the input capacitance of the receiving pin form a low-pass filter that rounds the edges. For high-speed buses or anything driven in both directions, a dedicated level-shifter chip is safer.
Does the formula hold for AC signals and oscilloscope probes?
At low frequencies, yes. Stray capacitance at the tap forms a low-pass filter with the output impedance: 3.2 kΩ and 20 pF cut off near 2.49 MHz. That is why a 10:1 scope probe, 9 MΩ in series with the scope's 1 MΩ input, includes a compensation capacitor, adjusted so that the capacitive ratio matches the resistive one and the division holds at every frequency.
What limits a divider on a high-voltage rail?
Two separate ratings. Power: the 10 kΩ / 4.7 kΩ ratio across 400 V would dissipate 10.88 W, which is why high-voltage dividers use megohm values. Voltage: every resistor has a maximum working voltage, independent of wattage, such as 200 V for a Vishay MRS16, 350 V for an MRS25, and 75 V to 500 V for chip sizes from 0402 to 2512. Long upper arms are split into several series resistors.

References& sources.

  1. [1]OpenStax (Rice University), "University Physics Volume 2", §10.2 "Resistors in Series and Parallel". Equation 10.2 gives the series rule RS = R1 + R2 + … = Σ Ri and the common-current constraint the divider relation is derived from; Equation 10.3 gives the parallel rule used for the loaded case and states that the parallel equivalent "is less than the smallest of the individual resistances". Retrieved and read 2026-07-29.
  2. [2]"Series and parallel circuits", reference article, retrieved and read 2026-07-29. Independent statement of both rules used here: R = Σ Ri in series, R = (Σ 1/Ri)⁻¹ in parallel with the two-resistor form R = R1R2/(R1 + R2) — the expression this page uses for the divider's Thevenin output impedance and for the loaded lower arm. Agrees with OpenStax term for term.
  3. [3]Vishay BCcomponents, "MRS16, MRS25 Professional Thin Film Leaded Resistors", document 28724, revision 07-Mar-16. Read directly from the PDF on 2026-07-29. Source of the maximum-operating-voltage figures quoted in the high-voltage-divider answers: Umax AC/DC of 200 V (MRS16) and 350 V (MRS25), alongside rated dissipation P70 of 0.4 W and 0.6 W and a peak permissible film temperature of 155 °C — a voltage limit that is specified separately from, and often binds before, the power limit.
  4. [4]Vishay, "D/CRCW e3 Standard Thick Film Chip Resistors", document 20035, revision 14-Apr-2026. Read directly from the PDF on 2026-07-29. Gives the surface-mount maximum operating voltages used on this page — 75 V (0402 and 0603), 150 V (0805), 200 V (1206, 1210, 1218), 400 V (2010), 500 V (2512) — together with rated dissipation quoted as P70 and a 155 °C film limit.
  5. [5]IEC 60063:2015, "Preferred number series for resistors and capacitors", third edition, 27 March 2015. Defines the E24 (±5 %) and E96 (±1 %) value sets referred to when turning a solved R2 into a resistor you can actually buy. Retrieved 2026-07-29 via the reference summary of the series values and their standard tolerances.
  6. [6]Vishay Draloric / Beyschlag / BCcomponents, "Resistor Color Card", document VMN-MS6212-1501 (2015). Read directly from the PDF on 2026-07-29. Prints the E24 and E96 preferred-value tables under the heading "According to IEC 60063" — the source used to check that a solved R2 such as 3,793 Ω has a nearest stock value of 3.9 kΩ in E24 and 3.83 kΩ in E96.

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V_out = V_in · R₂/(R₁+R₂) · R₂ = R₁·V_out/(V_in−V_out) · R_th = R₁∥R₂
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