Distinguish measuring a voltage from powering a load
Two equal resistors between 5 V and ground give 2.5 V if no current leaves their midpoint. Connecting another component adds a path: it is no longer the same circuit. That is why the connected device’s input resistance matters.
A practical approach is to choose the acceptable voltage drop first. If a particular load must change output by no more than 5%, calculate that requirement before trying arbitrary values.
The same ratio, different sensitivity
With R1 = R2 = 10 kΩ, unloaded output is 2.5 V. The resistance seen looking back from the output with the ideal voltage source set to zero is Rth = R1 ∥ R2 = 5 kΩ.
With RL = 10 kΩ: V = 1.6667 V
| R1 = R2 | Output | Drop |
|---|---|---|
| 10 kΩ | 1.6667 V | 33.33 % |
| 1 kΩ | 2.3810 V | 4.76 % |
Reducing both resistances tenfold reduces the load’s effect but raises unloaded current from 0.25 mA to 2.5 mA. The improvement has an electrical cost even though R2/(R1 + R2) is unchanged.
An error criterion you can test
For a drop ≤ 5%: RL ≥ 19 × Rth
For a drop ≤ 1%: RL ≥ 99 × Rth
With Rth = 5 kΩ, a 95 kΩ load causes exactly a 5% drop in this model. “High input impedance” becomes testable when compared with Rth and the error you can accept.
For a load with large current changes, consider a suitable regulated supply. Separating a signal from its load may call for a high-input-impedance, low-output-impedance stage, which also has current and voltage limits. This calculation does not select those components automatically.
The load here is a constant resistance. Sampling inputs, capacitance and changing current demand require further analysis. A correct unloaded voltage does not establish behavior during rapid changes.
Start with loading
The related experiment holds the supply and divider fixed to isolate one change: connecting RL. Then open the full tool and compare 10 kΩ + 10 kΩ with 1 kΩ + 1 kΩ.
What minimum load preserves 99% of the output with R1 = R2 = 1 kΩ?
Rth = 500 Ω. RL must be at least 99 × 500 = 49,500 Ω. This criterion covers resistive loading only, not resistor tolerance.
Does disconnecting the load eliminate divider current draw?
No. Current still flows from the supply through R1 and R2. With 5 V and two 10 kΩ resistors it is 0.25 mA.