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Voltage Divider

Also known as potential divider

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

Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!

Learning zone

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
Vout=VinR2R1+R2V_{out} = V_{in} \frac{R_{2}}{R_{1} + R_{2}}
Where
  • VoutV_{out}= Output voltage
  • VinV_{in}= Input voltage
  • R1R_{1}= Upper resistance
  • R2R_{2}= Lower resistance
Missing one of these? Work it out first, then come back