formula.expert

Voltage Divider

Also known as potential divider

Vout=VinR2R1+R2V_{out} = V_{in} \frac{R_{2}}{R_{1} + R_{2}}

Worked example: 12 V over 10 kΩ + 5 kΩ → 4 V — press Try an example to run it live, then adjust anything.

Enter your known values, leave one input blank, and solves for the missing one. Tap a variable’s symbol to see what it means, with a typical value. Try different units for next level excitement!

Here the solver did the work — could you?

Dividers →

UniversityCircuits & Electrical Power

Test your skills in the Exam Room: new numbers every attempt — free lessons for students, no sign-up, just pure learning. Find 1 more lesson on this formula.

See your Report Card
Compete with your friends
share your results
Learning zone

Voltage Divider explained

R₁R₂VinVout

Series resistors carry the same current, so each takes a share of the voltage in proportion to its resistance — the divider is Ohm's law applied twice. Twelve volts across 10 kΩ over 5 kΩ puts 12 × 5/15 = 4 V at the junction. It is the most-used circuit in electronics: bias networks, level shifters, the feedback string that sets a regulator's output, and every potentiometer, which is just a divider with a movable tap.

The trap is loading. This formula assumes nothing draws current from the tap; connect a load comparable to R₂ and the output sags, because the load parallels R₂. The rule of thumb is to make the divider current at least ten times the load current — but not so low-resistance that it wastes power. That trade-off is why high-impedance dividers pair with op-amp buffers, and why measuring a 1 MΩ divider with a cheap 1 MΩ-input meter reads badly wrong.

Voltage Divider formula

Vout=VinR2R1+R2V_{out} = V_{in} \frac{R_{2}}{R_{1} + R_{2}}
Where
  • VoutV_{out}= Output voltage (V)
  • VinV_{in}= Input voltage (V)
  • R1R_{1}= Upper resistance (Ω)
  • R2R_{2}= Lower resistance (Ω)

Missing one of these? Work it out first, then come back