Voltage Divider Calculator

The voltage divider formula

A voltage divider is two resistors in series across a supply, with the output taken across the lower one, and its output is Vout = Vin × R₂ / (R₁ + R₂): the output is the same fraction of the input as R₂ is of the total resistance. R₁ sits between the input and the output, R₂ between the output and 0 V. For the values loaded above, 5 V across 10 kΩ and 15 kΩ, Vout = 5 × 15 ÷ (10 + 15) = 3 V, three fifths of the supply.

Enter any three of the input voltage, the output voltage, R₁ and R₂, and choose Solve for this on the fourth. Solving for R₂ answers the question most divider designs start with, which resistor gives this output, and solving for the input reads a supply back from a measured output. Beside the answer the calculator shows the current through the divider, the division ratio Vout ÷ Vin, the voltage across R₁, the power in each resistor, the output resistance, and the output with a load connected. The load starts at 10 MΩ, about what a basic digital voltmeter presents, so the page opens on what a meter across R₂ would read. Every figure except that last one describes the divider with nothing connected to its output.

Why the supply divides in the ratio of the resistors

With nothing drawing on the output, one current flows through both resistors, and in series their resistances add. By Ohm’s law that current is I = Vin / (R₁ + R₂), and the voltage across R₂ is I × R₂, which is the formula above. Kirchhoff’s voltage law closes the loop: the two drops, I × R₁ and I × R₂, add up to the supply. For the values above the current is 5 V ÷ 25 kΩ = 200 µA, R₁ drops 2 V and R₂ the remaining 3 V. The Ohm’s law calculator works each of those drops on its own.

The same current in both resistors also gives the second form that OCR’s A level formula booklet prints beside the first, V₁ / V₂ = R₁ / R₂: each resistor takes a share of the supply in proportion to its resistance. Only the ratio matters, so 1 kΩ over 1.5 kΩ and 100 kΩ over 150 kΩ give the same 3 V from 5 V. What the size of the resistors decides is how much current the divider wastes and how well it copes with a load.

Worked example: a 9 V battery divider, with and without a load

Two 10 kΩ resistors in series across a 9 V battery make a divider. Find its output, the current and power, and what happens when a 10 kΩ load is connected to it.

• Vout = 9 V × 10 kΩ ÷ (10 kΩ + 10 kΩ) = 4.5 V, half the supply.
• The divider draws 9 V ÷ 20 kΩ = 450 µA, and each resistor dissipates (450 µA)² × 10 kΩ = 2.025 mW.
• Its output resistance is 10 kΩ × 10 kΩ ÷ (10 kΩ + 10 kΩ) = 5 kΩ.
• With the load connected: 4.5 V × 10 kΩ ÷ (10 kΩ + 5 kΩ) = 3 V, a third below the unloaded answer.

The same 3 V comes from treating the load as part of the circuit: in parallel with R₂, the lower half of the divider becomes 5 kΩ, and 9 V shared between 10 kΩ and 5 kΩ gives 3 V. A basic 10 MΩ voltmeter across the same output reads 4.498 V, only 0.05 percent low, which is why a meter reads a divider almost exactly while a real load may pull it well down. Enter 9 V, 10 kΩ and 10 kΩ above and the load readout shows 4.498 V at the starting 10 MΩ and 3 V with the load set to 10 kΩ, and the calculator warns that the load has moved the output.

Loading a divider: the output resistance decides

Seen from its output, a divider behaves like a source of the unloaded Vout behind a resistance of R₁ and R₂ in parallel, Rout = R₁ × R₂ / (R₁ + R₂), which is its Thévenin equivalent. A load RL across the output therefore receives VL = Vout × RL / (RL + Rout). The output stays within 1 percent of Vout only while the load is at least 99 times the output resistance, and within 10 percent while it is at least 9 times. For the two 10 kΩ resistors that means a load of 495 kΩ or more for 1 percent and 45 kΩ or more for 10 percent. The calculator warns once the load pulls the output more than 1 percent below Vout.

Smaller resistors make a stiffer divider, at a price. Swap the two 10 kΩ resistors for 100 Ω ones and the output is still 4.5 V, and the same 10 kΩ load now leaves it at 4.48 V, only 0.5 percent low. But the divider draws 45 mA instead of 450 µA, and each resistor dissipates 0.2 W, most of a quarter-watt rating. Choosing a divider is choosing between those two: resistors small enough that the load barely matters, and large enough that the current and heat stay small.

Potential dividers with sensors

Potential divider is the name UK A level physics uses for the same circuit, and AQA’s course asks for it with variable resistors, thermistors and light-dependent resistors (LDRs). The resistance of a thermistor falls as its temperature rises, and the resistance of an LDR falls as the light on it gets brighter. Put either sensor in a divider with a fixed resistor and its output voltage follows the temperature or the light, so with the sensor as R₁ the output rises as it warms or brightens, and with the sensor as R₂ the output falls. Swapping the two resistors is how a circuit is made to switch on in the dark rather than the light. A potentiometer is a divider in one part: the wiper splits a single track into R₁ and R₂, and turning it sets the ratio.

What this calculator does not cover

Every resistor here is exact. Real parts carry a tolerance, and with ±5 percent resistors the 5 V divider the page opens on could give anywhere from about 2.88 V to 3.12 V, so a divider that has to set a precise threshold needs 1 percent parts or a trimming potentiometer. The supply is ideal too: a real source has internal resistance, which sits in series with R₁ and lowers the output.

The load is assumed to be a resistor. An LED, a motor or a transistor has no single resistance, and an LED needs one series resistor rather than a divider, which the LED resistor calculator sizes. A capacitor across R₂ turns the divider into a low-pass filter whose time constant uses R₁ × R₂ / (R₁ + R₂) as its resistance, which the RC time constant calculator takes from there. For more than two resistors, a load built from several parts, or a current divider, use the resistor network calculator, which gives the current and voltage in every resistor of a series chain of parallel groups.

Common mistakes

• Putting the wrong resistor on top of the fraction. The output is across R₂, so R₂ goes on top. For the values above, R₁ on top gives 5 × 10 ÷ 25 = 2 V, which is the voltage across R₁, not the output.
• Mixing kilohms and ohms. The ratio only cancels when both resistors are in the same unit: 10 kΩ with 15 Ω is a ratio of 10,000 to 15, not 10 to 15. The calculator converts each field, and its working prints both resistors in one unit.
• Forgetting the load. The formula is for a divider with nothing connected. Anything across the output that is not far larger than the output resistance pulls it down, which the load readout shows.
• Using a divider as a power supply. Its output changes whenever the load does, so it suits signals and reference voltages feeding high-resistance inputs, not circuits that draw real current.
• Ignoring the power. Small resistors on a high voltage run hot. Each resistor dissipates I²R, and the calculator warns when either goes past the 0.25 W of a quarter-watt part.

Converting units first? Use the resistance and voltage conversion tables.