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Circuit SimulatorBeta

Voltage Divider

Two resistors in series split the source voltage. The point between them sits at a lower voltage, in proportion to R2's share of the total resistance.

Schematic of the Voltage Divider circuit

Numbers from the simulation

Computed by the simulator from the part values in the schematic.

Voltage across R2 (Vout)
2.88 V
Current through R1 and R2
612 µA

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How it works

The same current flows through R1 and R2 because they are in series: I = Vin / (R1 + R2) = 9 V / 14.7 kΩ.

The voltage across R2 is that current times R2, so Vout = Vin × R2 / (R1 + R2). Work it out, then compare it with the simulated number above.

The formula holds while the load on Vout draws much less current than the divider itself. A heavy load pulls Vout down.

Try changing

  1. Change R2 to 10 kΩ. With equal resistors, Vout is half of 9 V.
  2. Put a 4.7 kΩ resistor in parallel with R2 as a load and watch Vout drop.
  3. Set the source to 12 V. The ratio of Vout to Vin stays the same.

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