Voltage divider calculator

A voltage divider calculator works out the output voltage of two resistors in series across a supply. The output across the bottom resistor is V_out = V_s × R₂ ÷ (R₁ + R₂), so 10 kΩ over 20 kΩ on a 5 V supply gives 3.33 V. Our Circuit Builder has a voltage divider mode that gives the output, the current and the power in each resistor, with the working, and shows what a load or a sensor does to the output.

The Circuit Builder opens on its build grid. To switch it to dividers, choose Guided circuits, then Voltage divider. In the UK the same circuit is usually called a potential divider.

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How to use the voltage divider calculator

1. Under Calculate, choose Guided circuits, then pick Voltage divider from the list. It starts on 5 V across 10 kΩ and 20 kΩ, and Load an example has a light sensor too.
2. Under Inputs, set the Supply voltage and R1 (top resistor).
3. For the bottom part, choose Resistor and type R2, or choose Thermistor or LDR and set the temperature or light level with the box or the slider.
4. Tick Connect an output load to put a load across the output, and type its resistance.
5. Read the output voltage, the supply current and the resistance below the tap under Readings, with the voltage, current and power of each part. Working shows every step.

Each resistance takes a unit such as Ω, kΩ or MΩ. If a resistor takes more power than the Resistor power rating, 0.25 W to start with, it is flagged. The diagram updates as you type, and Show current flow animates the current. Practice and help sets divider questions, loaded and unloaded, including choosing R2 for a target output.

The voltage divider formula

The same current flows through both resistors, I = V_s ÷ (R₁ + R₂), so each one takes a share of the supply in proportion to its resistance. The bigger resistor always gets the bigger share. The two voltages add up to the supply: V₁ + V_out = V_s.

Worked examples you can check in the calculator

Example 1: 5 V down to 3.3 V

A 5 V signal has to be scaled down for a 3.3 V input. With R1 = 10 kΩ and R2 = 20 kΩ:

These are the values the mode starts on. The calculator gives 3.3333 V, a current of 166.67 µA, and 1.6667 V across R1. The ratio is what matters: 1 kΩ and 2 kΩ give the same 3.33 V, with ten times the current.

Example 2: 12 V to about 5 V

With R1 = 4.7 kΩ and R2 = 3.3 kΩ on a 12 V supply:

Type 12, 4.7 kΩ and 3.3 kΩ. The calculator gives 4.95 V and 1.5 mA, with 7.05 V across R1.

Example 3: choosing R2 for a target output

A 9 V supply, R1 = 10 kΩ, and you want 3.0 V out. R1 has to drop the other 6.0 V, so:

The calculator works forwards only: it has no box for a target output. Find R2 by hand, then type it in. 5 kΩ gives exactly 3.0 V. A standard 5.1 kΩ resistor gives 3.0397 V.

Try a divider question in the course, then press Mark:

How R2 changes the output

Here x is R2 in kΩ and y is the output voltage in volts. When R2 equals R1, the output is half the supply. The output rises quickly at first, then levels off towards 5 V but never reaches it. So the output is not proportional to R2: doubling R2 from 10 kΩ to 20 kΩ only takes it from 2.5 V to 3.33 V.

What a load does to a voltage divider

Anything connected to the output is in parallel with R2, which lowers the resistance below the tap and pulls the output down. Tick Connect an output load on the 5 V, 10 kΩ, 20 kΩ divider and add a 10 kΩ load:

The output falls from 3.33 V to 2.0 V. With a 1 MΩ load instead, the calculator gives 3.3113 V, almost unchanged. The rule of thumb: keep the load resistance much bigger than R2, or use smaller resistors in the divider, at the cost of more current. A divider is good for scaling a signal, not for powering something.

Sensor circuits: thermistors and LDRs

Swap R2 for a sensor and the divider turns temperature or light into a voltage. Choose Thermistor or LDR under the bottom part. In the calculator's models, the thermistor is 10 kΩ at 25 °C and the LDR is 15 kΩ at 10 lux, and both fall as it gets warmer or brighter.

With the sensor at the bottom, the output falls as it gets warmer or brighter. For an output that rises instead, swap the sensor and the fixed resistor, so the sensor is on top. The calculator always puts the sensor at the bottom, so for the swapped version, use Vout = Vs × R1 ÷ (R1 + R_sensor) by hand. Real thermistors and LDRs vary, so the resistances here are the model's values, not a datasheet's.

Potential divider (the UK name)

In UK schools this circuit is called a potential divider, because it divides the potential difference of the supply between the two components in proportion to their resistances. The potential divider equation is the same formula, usually written V_out = V_in × R₂ ÷ (R₁ + R₂), or as a ratio: V₁ ÷ V₂ = R₁ ÷ R₂.

Potential dividers at GCSE

AQA's GCSE Physics specification says that a thermistor's resistance falls as it gets warmer, and an LDR's resistance falls as the light gets brighter. It requires their applications: a thermostat for the thermistor, and switching lights on when it gets dark for the LDR. Both work as potential dividers: the sensor's changing resistance changes its share of the supply voltage, and that voltage switches the rest of the circuit.

Potential dividers at A-level

AQA's A-level specification covers the potential divider used to supply a constant or variable potential difference from a power supply, with examples using variable resistors, thermistors and LDRs, and expects students to design potential divider circuits for given outcomes. It is part of the first-year (AS) content. A potentiometer, a variable resistor with a sliding contact, is a potential divider in a single component: moving the slider changes the output smoothly from zero up to the full supply voltage.

Exam-style example: a temperature sensor

A 6.0 V supply is connected across a fixed 4.0 kΩ resistor in series with a thermistor, and the output is taken across the thermistor. The thermistor has a resistance of 2.0 kΩ at 20 °C and 0.50 kΩ at 60 °C. Find the output voltage at each temperature.

As it warms up, the thermistor's resistance falls, so it takes a smaller share of the 6.0 V and the output falls from 2.0 V to 0.67 V. For a circuit that switches a fan on when it gets hot, you want an output that rises with temperature, so take the output across the fixed resistor instead: at 60 °C it gets 6.0 × 4.0 ÷ 4.5 = 5.33 V.

Common mistakes

• Putting the wrong resistor on top of the fraction. The output is across R2, so R2 goes on top: Vout = Vs × R2 ÷ (R1 + R2).
• Mixing Ω and kΩ. 4.7 kΩ and 3300 Ω in the same sum must both be in the same unit.
• Ignoring the load. A load across R2 lowers the output. Combine it with R2 in parallel first.
• Thinking the output is proportional to R2. Doubling R2 does not double the output; it depends on R2 ÷ (R1 + R2).
• Getting the sensor direction backwards. Warm a thermistor and its resistance drops, so with the thermistor at the bottom the output drops too.

Practise circuits in a short timed paper

Potential dividers come up in A-level Physics: Year 1 practice, and sensor circuits with thermistors and LDRs in GCSE Physics Higher. Each daily paper is a short timed paper of 8 questions worth 25 to 35 marks, and the papers mix topics. For resistors in parallel on their own, see the parallel resistor calculator guide.

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