Voltage divider: 5.00 V with 4.7kΩ / 10kΩ → 3.40 V
Worked answer for a 5.00 V input across R1 = 4.7kΩ (top) and R2 = 10kΩ (tapped, bottom), measured across R2.
Output voltage 3.40 V Vout = Vin · R2 / (R1 + R2) · ideal, unloaded
| Input voltage (Vin) | 5.00 V |
| R1 — top resistor | 4.7kΩ |
| R2 — bottom (tapped) resistor | 10kΩ |
| Output Vout = Vin · R2 / (R1 + R2) | 3.40 V |
| Divider current I = Vin / (R1 + R2) | 0.34 mA |
| Total power P = Vin · I | 1.7 mW |
R2 is the tapped (bottom) resistor — Vout is the voltage measured across it. This figure assumes an ideal, unloaded divider: connecting a load draws extra current through R1 and lowers Vout, so the heavier the load relative to R2, the more the output sags.
Different values? Change Vin, R1 or R2 in the interactive tool:
Open the Voltage Divider Calculator →Disclaimer: This is an ideal, unloaded divider. Any real load on the output draws current and lowers Vout; for a stable reference use a buffer (op-amp) or pick R2 much smaller than the load resistance. Verify against your circuit.