Question 1

How do I calculate the output of a voltage divider?

Accepted Answer

For two resistors R1 (top) and R2 (bottom) the output taken across R2 is Vout = Vin × R2 / (R1 + R2). For example, 12 V with R1 = 10 kΩ and R2 = 10 kΩ gives 12 × 10000 / 20000 = 6 V.

Question 2

How do I pick R1 and R2 for a desired ratio?

Accepted Answer

Only the ratio matters for the no-load voltage: R2 / (R1 + R2) sets the fraction of Vin you keep. Fix one resistor and solve the other with R2 = R1 × Vout / (Vin − Vout) or R1 = R2 × (Vin − Vout) / Vout, then snap to a standard E24 value.

Question 3

Does connecting a load change the divider output?

Accepted Answer

Yes. A load resistor RL appears in parallel with R2, lowering the effective bottom resistance and pulling Vout down. The effect is small only when RL is much larger than R2 — a rule of thumb is RL at least 10 times R2.

Question 4

How much power do the divider resistors dissipate?

Accepted Answer

The current through an unloaded divider is I = Vin / (R1 + R2). Each resistor dissipates I squared times its resistance, so P1 = I² × R1 and P2 = I² × R2. Lower-value resistors waste more power but are less sensitive to loading.

Question 5

Can I use a voltage divider as a power supply?

Accepted Answer

No. A divider is only suitable for high-impedance references such as ADC inputs or bias points. Any real load draws current that changes the output, and the resistors waste power. For a stable supply rail use a regulator or a buck/boost converter instead.

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